#groups-rings-fields | Mathematics | Page 597

oblique river Jul 6, 2021, 9:36 PM

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plug in (0,0) and all that's left is psi(0,0)

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i.e. the constant term of psi

maiden ocean Jul 6, 2021, 9:37 PM

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Oh yeah duh product rule

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Ok

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So right yeah

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and then when we pass to the local ring f is gonna look like 0 and we know that phi is non-zero at 0

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so you get y = gx = - phi psi^{-1} x

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(abusing notation a bit bit yea)

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then M_P is generated by (x, y) = (x, gx) = (x) and thus is principal and so O_X, P is local with non-zero principal maximal ideal and thus dedekind

oblique river Jul 6, 2021, 9:41 PM

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yeah that sounds right ot me

maiden ocean Jul 6, 2021, 9:41 PM

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epic

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thanks buncho nozoomi

oblique river Jul 6, 2021, 9:41 PM

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np!

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how is szamuely going overall?

maiden ocean Jul 6, 2021, 10:16 PM

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Sorry wifi lmfao

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its going well!

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I think

chilly ocean Jul 6, 2021, 11:05 PM

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is any (finite?) symmetric group a non trivial product?

woven delta Jul 6, 2021, 11:20 PM

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No, the only nontrivial normal subgroup of the symmetric groups for n>=5 is A_n

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@chilly ocean

chilly ocean Jul 6, 2021, 11:22 PM

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what about infinite ones?

woven delta Jul 6, 2021, 11:29 PM

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Apparently for the symmetric group on countably many elements there are only 2 nontrivial normal subgroups (according to Wikipedia). These are the set of permutations that fix all but finitely many elements, as well as the even permutation subgroup of that

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I think this is pretty interesting, I wonder if the reason for this is because the thing that determines if 2 elements are congugate in this group are the cardinality of the set of elements that are fixed as well as the cardinality of the set of elements that are not fixed

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Actually probably it's a bit more complicated than that

oblique river Jul 6, 2021, 11:36 PM

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is it?

woven delta Jul 6, 2021, 11:36 PM

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I think you just need a bijection between orbits

oblique river Jul 6, 2021, 11:36 PM

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that seems reasonable to me

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yeah, I agree

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it's basically just cycle type decomposition

vestal snow Jul 7, 2021, 2:50 AM

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Can someone explain to me what I have to prove here?

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The book only defines the Zariski tangent space for local rings and schemes, and in this exercise, A is an arbitrary and "the Zariski tangent space of A" makes no sense

oblique river Jul 7, 2021, 2:54 AM

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Its common to just use A to refer to Spec A

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I think thats what theyre doing

vestal snow Jul 7, 2021, 2:55 AM

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Oh okay

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But they need to specify a point right?

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Can you write out exactly what the exercise wants me to prove?

oblique river Jul 7, 2021, 2:59 AM

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The point is the point m

vestal snow Jul 7, 2021, 3:00 AM

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What does the "cut out" part mean?

oblique river Jul 7, 2021, 3:00 AM

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Imagine like R^2 and some curve through a point P

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R^2 has a tangent space at P

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And so does the curve

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And the tangent spacr of the curve is a subspace of the tangent space of R^2

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That’s the picture

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R^2 is spec A, P is m, and the curve is spec A/f

vestal snow Jul 7, 2021, 3:16 AM

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oblique river R^2 is spec A, P is m, and the curve is spec A/f

So the Zariski tangent space of A at m is Hom(m/m^2, A/m)

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And it wants me to consider the tangent space cut out by f in the tangent space of A

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What exactly is that?

oblique river Jul 7, 2021, 3:18 AM

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Write down the tangent space of A/f at m

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And there should be a natural map

vestal snow Jul 7, 2021, 3:21 AM

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The tangent space of A/f at m is just Hom(m_f/m_f^2, (A/f)/(m_f) where m_f is the image of m in A/f

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Of course, (A/f)/m_f = A/m

oblique river Jul 7, 2021, 3:22 AM

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Write down a specific example where there’s easy geometry you can see

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Like the picture I gave earlier

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Consider R^2 and the origin and the curve y=0

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A = R[x,y]

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Or C instead of R if you prefer

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m = (x,y)

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f = f(x,y) = y

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Now you can see that A/f = R[x] and its spec is just the x axis

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Sitting inside R^2

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And the tangent spacr has dimension one less than the tangent space of R^2

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Now write down all the homs and keep track kf all the maps

vestal snow Jul 7, 2021, 3:25 AM

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I think you're misunderstanding my question

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You know how we say that f cuts out a portion of Spec A which is Spec A/f

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Here it's saying that f cuts out a portion of the tangent space of Spec A which is ...?

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I don't know what ... is referring to

oblique river Jul 7, 2021, 3:27 AM

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The tangent space of A/f

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The tangent space of A/f is a subspace of the tangent space of A (at a point)

vestal snow Jul 7, 2021, 3:27 AM

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oblique river The tangent space of A/f

That's what we have to prove lol

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Like what is ... exactly

oblique river Jul 7, 2021, 3:28 AM

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No, what you have to prove is that the tangent space of A/f is cut out by f

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In some sense

vestal snow Jul 7, 2021, 3:28 AM

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Yes

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But what does it mean by "the tangent space of A cut out by f"?

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Like what is this thing as a set/topological space/scheme

oblique river Jul 7, 2021, 3:29 AM

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It’s just describing an algebraic operation

vestal snow Jul 7, 2021, 3:30 AM

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Do you know of a proof?

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I think it would be better if I just see it

oblique river Jul 7, 2021, 3:30 AM

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Like, f mod m^2 is an element of m/m^2

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And you can quotient by it

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And get a map on tangent spaces

vestal snow Jul 7, 2021, 3:31 AM

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Essentially I need to prove A = B but I don't have a description for B

oblique river Jul 7, 2021, 3:31 AM

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And that inclusion is exactly the inclusion of the tangent space of A/f

vestal snow Jul 7, 2021, 3:31 AM

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*a formal description

oblique river Jul 7, 2021, 3:31 AM

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I am giving you one lol

vestal snow Jul 7, 2021, 3:31 AM

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The one you are giving me is A

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which is exactly what we have to prove

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Unless i'm misunderstanding

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which is possible

oblique river Jul 7, 2021, 3:32 AM

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No…

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There is the tangent space of A/f at m

vestal snow Jul 7, 2021, 3:32 AM

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Yes

oblique river Jul 7, 2021, 3:33 AM

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And there is the subspace of Hom(m/m^2, A/m) induced from the quotient of m/m^2 by the element f mod m^2

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m/m^2 —> m/(f + m^2)

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Induces a map on homs going the other way

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Those are the two things

vestal snow Jul 7, 2021, 3:34 AM

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Ah okay

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Now I understand

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I think

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Let me write this down

oblique river Jul 7, 2021, 3:34 AM

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oblique river Like, f mod m^2 is an element of m/m^2

This is what i was saying here

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So like stepping back for a sec, this is an occasion where i think you should try to fill in the holes a little bit yourself

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There is not much that “subspace of tangent space of A cut out by f” can mean interms of like

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Algebraic operations

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Also, assuming the problem is correct (which is a good assumption)

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Even if you dont know what B is exactly

vestal snow Jul 7, 2021, 3:36 AM

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oblique river m/m^2 —> m/(f + m^2)

Does this mean m/m^2 --> (m/m^2)/(f+m^2)?

oblique river Jul 7, 2021, 3:36 AM

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You know what A is and you know A = B

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So you can use that to sort of back-solve for B

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And a good way to do this is to write down explicit examples

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When youre doinf research, youre not always going to have someone who can tell you these things

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And learning how to fill in holes on your own is more important than just trying to learn a bunch of statements

oblique river Jul 7, 2021, 3:38 AM

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vestal snow Does this mean m/m^2 --> (m/m^2)/(f+m^2)?

Those are the same

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M^2 + f contains m^2

vestal snow Jul 7, 2021, 3:38 AM

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Yeah I usually try and do that but this particular exercise was a bit confusing

oblique river Jul 7, 2021, 3:38 AM

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So once you wuotient m by m^2

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You can quotient further by m^2 + f

vestal snow Jul 7, 2021, 3:39 AM

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Mostly because the book just introduced (co)tangent spaces and then used a couple of new vocab at once

azure plinth Jul 7, 2021, 4:13 AM

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Might seem dumb but I am stuck with this one: Write $x_1^2x_2+x_2^2x_3+x_3^2x_1$ in terms of elementary symmetric functions in $x_1,x_2,x_3$.

cloud walrusBOT Jul 7, 2021, 4:13 AM

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AidenM27

azure plinth Jul 7, 2021, 4:14 AM

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I got the representation in that form for $\sum_{i\neq j}x_i^2 x_j$ for $i,j\in{1,2,3}$, but have had no luck with this special case of it, so to speak

cloud walrusBOT Jul 7, 2021, 4:14 AM

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AidenM27

hidden haven Jul 7, 2021, 5:04 AM

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azure plinth Might seem dumb but I am stuck with this one: Write $x_1^2x_2+x_2^2x_3+x_3^2x_1$...

Hint: terms of the form a²b appear in (a+b)³

azure plinth Jul 7, 2021, 5:05 AM

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Oh. Boy. Thanks a bunch, I will go down that route mate

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Ah I got stuck again in that train of thought. What I did in my first attempt was this

hidden haven Jul 7, 2021, 5:12 AM

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Do you know how to write Σᵢ xᵢ³ in terms of elementary symmetric polynomials? You'd need that for what I'm doing

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You can try to figure that out separately, then plug things into the above

azure plinth Jul 7, 2021, 5:13 AM

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$s_1=x_1+x_2+x_3, s_2=x_1x_2+x_2x_3+x_1x_3, s_3=x_1x_2x_3\implies s_1s_2=x_1^2x_2+x_1x_2x_3+x_1^2x_3+x_1x_2^2+x_2^2x_3+x_1+x_2+x_3+x_1x_2x_3+x_2x_3^2+x_1x_3^2)\implies s_1s_2=3s_3+x_1x_2^2+x_1^2x_3+x_2x_3^2+(x_1^2x_2+x_2^2x_3+x_3^2x_1)$

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Ah I can attempt that but what I am concerned about is ending up with the terms $x_1x_2^2, x_2x_3^2, x_3x_1^2$, which I cannot seem to resolve to arrive at an answer

cloud walrusBOT Jul 7, 2021, 5:14 AM

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AidenM27

hidden haven Jul 7, 2021, 5:15 AM

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Wait so did you just get s_1s_2 = 3s_3 + twice the polynomial that you need?

azure plinth Jul 7, 2021, 5:17 AM

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Not twice actually

hidden haven Jul 7, 2021, 5:17 AM

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And the intermediate calculations seem wrong, everything in between should be degree 3 homogeneous

azure plinth Jul 7, 2021, 5:17 AM

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Ah that's a typo

cloud walrusBOT Jul 7, 2021, 5:18 AM

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AidenM27

hidden haven Jul 7, 2021, 5:18 AM

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oh damn I didn't realize that the original thing isn't completely symmetric

azure plinth Jul 7, 2021, 5:18 AM

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Exactly, that's what happened to me as well

hidden haven Jul 7, 2021, 5:19 AM

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But you can't write non symmetric polynomials as products and sums of symmetric ones

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For anything written in terms of symmetric polynomials, you could apply a permutation on the variables and the expression should remain unchanged

azure plinth Jul 7, 2021, 5:20 AM

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Right, that's what struck me now

hidden haven Jul 7, 2021, 5:20 AM

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catThink

azure plinth Jul 7, 2021, 5:21 AM

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Yeesh, gonna have to take it up with the prof then. Some problem set (not for grades though). I will let you know if I make any progress

unique juniper Jul 7, 2021, 7:39 AM

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there is an algorithm to write it in terms of elementary symmetric functions

maiden ocean Jul 7, 2021, 7:50 AM

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so im kind of confused here

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K(X) is the field of fractions of O_X, P

latent anvil Jul 7, 2021, 7:50 AM

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Yup

maiden ocean Jul 7, 2021, 7:51 AM

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so i think here we need K(X) a finite extension of k(t)

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but im not sure why thats true?

latent anvil Jul 7, 2021, 7:51 AM

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hmm

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So we have a dvr A = O_X,p

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And we're taking the field of fractions K = K(X)

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Yeah?

maiden ocean Jul 7, 2021, 7:51 AM

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Right

latent anvil Jul 7, 2021, 7:52 AM

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So choose t as in there

maiden ocean Jul 7, 2021, 7:52 AM

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in K(X)?

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or O_X, p

latent anvil Jul 7, 2021, 7:52 AM

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No this is in A

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Yeah O_X,p

maiden ocean Jul 7, 2021, 7:52 AM

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Ok yea

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generator for the maximal ideal m

latent anvil Jul 7, 2021, 7:52 AM

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What do elements of A look like?

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They're like a t^n for some unit a

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Yeah?

maiden ocean Jul 7, 2021, 7:52 AM

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I think so?

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OK yea

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i remember this

latent anvil Jul 7, 2021, 7:52 AM

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Yup

maiden ocean Jul 7, 2021, 7:53 AM

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so then the field of fractions is like at^n/bt^m with b(p) non-zero

latent anvil Jul 7, 2021, 7:54 AM

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Yeah

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So why is this finite over k(t)...

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The only variance is like, a unit

maiden ocean Jul 7, 2021, 7:54 AM

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er just backing up for a sec:

latent anvil Jul 7, 2021, 7:54 AM

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yup

maiden ocean Jul 7, 2021, 7:54 AM

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are the a, b in k?

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shouldnt they be in O(X)?

latent anvil Jul 7, 2021, 7:54 AM

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They are not in k

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If they were in k we'd have K = k(t)

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They're units of A

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So functions on X which don't vanish at p

maiden ocean Jul 7, 2021, 7:55 AM

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yeah so how exactly do we think of at^n/bt^m as an element of k(t) since the coefficients themselves arent in k

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Unless im being really dumb

latent anvil Jul 7, 2021, 7:56 AM

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It isn't an element of k(t), it's an element of K(X)

maiden ocean Jul 7, 2021, 7:56 AM

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Oh wait

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Im dumb

latent anvil Jul 7, 2021, 7:56 AM

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?

maiden ocean Jul 7, 2021, 7:56 AM

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Right it just contains k(t) by taking the case where a and b are constant functions right

latent anvil Jul 7, 2021, 7:56 AM

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Yup

maiden ocean Jul 7, 2021, 7:56 AM

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okay that makes more sense lmfao

latent anvil Jul 7, 2021, 7:56 AM

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That's the way we extend it

maiden ocean Jul 7, 2021, 7:56 AM

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exactly

latent anvil Jul 7, 2021, 7:57 AM

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And I agree that we want to show K/k(t) is finite

maiden ocean Jul 7, 2021, 7:57 AM

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so we need finiteness and then char 0 will imply by the primitive element thm that K(X) = k(t)(s) = k(t, s)

latent anvil Jul 7, 2021, 7:57 AM

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We're in char 0 so separable isn't an issue

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yup

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Oh maybe this is easy

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hmm

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I'm thinking we can choose finitely many functions generating the coordinate of X overall

maiden ocean Jul 7, 2021, 7:58 AM

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Ummm i think

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that makes sense

latent anvil Jul 7, 2021, 7:58 AM

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Hm but when we localize to get O_X,p we like, lose some amount of finiteness

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Okay so backing up a sec

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Do you know Zariski's lemma?

maiden ocean Jul 7, 2021, 7:58 AM

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if X is integral then O(X) is finitely generated

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as a k[x] module

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Uh no i do not

latent anvil Jul 7, 2021, 7:59 AM

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oh, I meant finitely generated as a k algebra

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Zariski's lemma says that a field which is a finitely generated algebra over another field is actually finite over that field

maiden ocean Jul 7, 2021, 7:59 AM

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wait yes i have heard this

latent anvil Jul 7, 2021, 7:59 AM

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Nice!

maiden ocean Jul 7, 2021, 7:59 AM

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This is a problem in the AM chapter that i paused AM doing becuase it was boring opencry

latent anvil Jul 7, 2021, 7:59 AM

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So we can make the goal a little easier

maiden ocean Jul 7, 2021, 7:59 AM

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chapter 5

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Ok yeah

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we need to show that its finitely generated as a k(t) algebra

latent anvil Jul 7, 2021, 8:00 AM

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so choose finitely many generators f1,...,fn for the coordinate ring of X

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These are generating it as an algebra over k

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Hm

maiden ocean Jul 7, 2021, 8:01 AM

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hmmCat

latent anvil Jul 7, 2021, 8:01 AM

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I thought I saw an easier proof but it didn't work

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So lets continue

maiden ocean Jul 7, 2021, 8:01 AM

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so ur claim that when we write 1 = a1f1 + ... + anfn the ai are constant?

latent anvil Jul 7, 2021, 8:01 AM

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No

maiden ocean Jul 7, 2021, 8:01 AM

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and can be regarded as elements of k?

latent anvil Jul 7, 2021, 8:01 AM

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That would be generating it as a module

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Generating it as an algebra means that any element is a polynomial in f1,...,fn

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With k coefficients

maiden ocean Jul 7, 2021, 8:02 AM

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wait O(X) is literally a quotient of k[x] im silly

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Ok yes

latent anvil Jul 7, 2021, 8:02 AM

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No!

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the only (reduced) quotients of k[x] are an affine line or finitely many points

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A curve like y^2 = x^3 is not a quotient of k[x]

maiden ocean Jul 7, 2021, 8:03 AM

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uhhh

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oh wait right fuck

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i mixed it up its a quotient of k[x, y]

latent anvil Jul 7, 2021, 8:04 AM

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Why?

maiden ocean Jul 7, 2021, 8:04 AM

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umm we're working with integral plane curves arent we

latent anvil Jul 7, 2021, 8:04 AM

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Take the curve parameterized by (t^3, t^4, t^5)

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Oh it said plane curve?

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I don't see where it says that

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The result you're trying to show is about plane curves

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But it doesn't restrict X to be such

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Right?

maiden ocean Jul 7, 2021, 8:05 AM

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er affine curve

latent anvil Jul 7, 2021, 8:05 AM

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So the thing is, we know X can be embedded as a closed subset of some affine space

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yeah?

maiden ocean Jul 7, 2021, 8:05 AM

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omg wait theres like 6 notions of dimension at play here that im mixing up

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integral affine curve means K(X) is of transcendence degree 1 over k

latent anvil Jul 7, 2021, 8:06 AM

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It's the vanishing locus of an ideal I of k[f1,...,fn] and then the coordinate ring is k[f1,...,fn]/I

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Oh!

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I hadn't thought about that

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I think that's important here

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It says K(X)/k(t) is algebraic

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But why finite...

maiden ocean Jul 7, 2021, 8:08 AM

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Ok so wait to clarify

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affine plane curve means that its a quotient of k[x, y]

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integral affine curve means that K(X) has transcendence degree 1 over k

latent anvil Jul 7, 2021, 8:08 AM

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Yes

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Equivalently, any prime ideal of the coordinate ring is either 0 or maximal

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Yeah?

maiden ocean Jul 7, 2021, 8:09 AM

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and specifically for plane curve it has to be defined as the locus of zeroes of a single polynomial f in k[x, y]

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but an integral affine curve is not necessarily either a quotient of k[x, y] or the zero of a single function

latent anvil Jul 7, 2021, 8:09 AM

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Yes, that is correct

maiden ocean Jul 7, 2021, 8:10 AM

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latent anvil Jul 7, 2021, 8:10 AM

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maiden ocean and specifically for plane curve it has to be defined as the locus of zeroes of ...

This follows from just being a curve in k[x, y], fyi. Things of dimension n-1 are given by a single equation

latent anvil Jul 7, 2021, 8:10 AM

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maiden ocean

This stuff is super hard without examples. Make sure you have things in your back pocket you can check definitions against

maiden ocean Jul 7, 2021, 8:11 AM

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I think I will need to sleep on this and come back tomorrow

latent anvil Jul 7, 2021, 8:11 AM

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hahaha

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goodnight

maiden ocean Jul 7, 2021, 8:11 AM

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This is weird because like

latent anvil Jul 7, 2021, 8:11 AM

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,ti moth

cloud walrusBOT Jul 7, 2021, 8:11 AM

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latent anvil Jul 7, 2021, 8:11 AM

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,ti moth in shambles

cloud walrusBOT Jul 7, 2021, 8:11 AM

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The current time for Moth is 04:11 AM (EDT) on Wed, 07/07/2021.
shamrock is 3 hours behind, at 01:11 AM (PDT) on Wed, 07/07/2021.

latent anvil Jul 7, 2021, 8:11 AM

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go to sleep

maiden ocean Jul 7, 2021, 8:11 AM

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there are certain terms here in each definition that i associate mentally with things

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and then when theyre involved in other terms they dont mean the things i associate them with

latent anvil Jul 7, 2021, 8:12 AM

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So keep in mind that like

maiden ocean Jul 7, 2021, 8:12 AM

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and i get mixed up opencry

latent anvil Jul 7, 2021, 8:12 AM

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"affine integral curve"

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Is two adjectives and a noun

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Don't think of it as a unit

maiden ocean Jul 7, 2021, 8:12 AM

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Yeah so like

cloud walrusBOT Jul 7, 2021, 8:12 AM

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Member selection timed out.

latent anvil Jul 7, 2021, 8:12 AM

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"plane curve" means "curve whcih is embedded in the plane"

maiden ocean Jul 7, 2021, 8:12 AM

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curve means it is defined as the zeroes of m polynomials in n variables

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just in general

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?

latent anvil Jul 7, 2021, 8:13 AM

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that's what a variety is

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A curve is one of those which is 1 dimensional

maiden ocean Jul 7, 2021, 8:13 AM

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In what sense

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like defined by 1 function or transcendence degree 1

latent anvil Jul 7, 2021, 8:14 AM

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The dimension of a variety is defined by looking at the longest chain of irreducible subvarities (or dually primes of the coordinate ring), because intuitively to drop down an irreducible variety you need to decrease the dimension by at least 1

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It is a result that this can be calculated via transcendence degree for integral things

latent anvil Jul 7, 2021, 8:14 AM

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maiden ocean like defined by 1 function or transcendence degree 1

Defined by n functions tells you nothing about dimension!

maiden ocean Jul 7, 2021, 8:14 AM

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Ok i see

latent anvil Jul 7, 2021, 8:15 AM

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It is unfortunate for this explanation that being defined by 1 equation is actually a big deal

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But it means the opposite of what you're thinking. It means that your codimension is 1

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The codimension of a variety X in A^n is just n - dim X

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Think about concrete examples here

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A single equation in R^3 gives a plane, or a paraboloid, or a cone

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not a line!

maiden ocean Jul 7, 2021, 8:17 AM

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so:
variety - solutions to m functions in n variables
affine variety - variety where the m functions and their combinations are the only functions for which the points are solutions
integral - coordinate ring is integral
curve - transcendence degree 1 (using this defn cause its the only one the book supplied)
plane curve - curve given by 1 function in 2 variables

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so affine integral curve implies its an affine variety of dimension 1 with integral coordinate ring

latent anvil Jul 7, 2021, 8:17 AM

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Okay sorry back up

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I was being imprecise about variety vs affine variety

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And using them interchangeably

maiden ocean Jul 7, 2021, 8:18 AM

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yeah i know theyre different like

latent anvil Jul 7, 2021, 8:18 AM

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I'm not sure what your definition of affine variety means here

maiden ocean Jul 7, 2021, 8:18 AM

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if I = (f1, ..., fm) then V(I) is the set of solutions the variety

latent anvil Jul 7, 2021, 8:18 AM

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Your definition of a variety is the correct definition of affine variety

maiden ocean Jul 7, 2021, 8:18 AM

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and affine is when the V(I) being a solution to some arbitrary f implies f in I

latent anvil Jul 7, 2021, 8:18 AM

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A general variety is something which locally looks like an affine variety

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Like projective space

maiden ocean Jul 7, 2021, 8:18 AM

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That is uh

latent anvil Jul 7, 2021, 8:18 AM

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It's locally like affine space, but globally weird

maiden ocean Jul 7, 2021, 8:19 AM

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not the way szamuely defined it

latent anvil Jul 7, 2021, 8:19 AM

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wtf

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I need to see that definition, then

maiden ocean Jul 7, 2021, 8:19 AM

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Oh wait im dumb

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ok so he actually calls V(I) affine closed sets and then its an affine variety when the thingy holds

latent anvil Jul 7, 2021, 8:19 AM

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Which thingy?

maiden ocean Jul 7, 2021, 8:19 AM

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uh the solutions to the functions in the ideal are the affine closed set

latent anvil Jul 7, 2021, 8:20 AM

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So like, I = (f1,...,fn)? Or what?

maiden ocean Jul 7, 2021, 8:20 AM

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and the affine closed set is an affine variety when those solutions also define the ideal in the sense that the ideal contains the only functions for which the points are all solutions

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Yeah

latent anvil Jul 7, 2021, 8:20 AM

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This does not make sense to me

maiden ocean Jul 7, 2021, 8:20 AM

#

#

latent anvil Jul 7, 2021, 8:21 AM

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I'm very uncertain about the case he has in mind

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Does he give a non-example

maiden ocean Jul 7, 2021, 8:22 AM

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uhhh like

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I = ((x - a)^2)

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wouldnt work

latent anvil Jul 7, 2021, 8:22 AM

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Yeah, but what does that have to do with X?

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X doesn't have the data of I

maiden ocean Jul 7, 2021, 8:22 AM

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because every point in X is a solution to f = x - a but f is not in I

latent anvil Jul 7, 2021, 8:23 AM

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like he writes a condition on I and then says it's about X

maiden ocean Jul 7, 2021, 8:23 AM

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I think hes saying its an affine variety when it does have the data

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like when you dont just get that I defines V(I) but that V(I) defines I

latent anvil Jul 7, 2021, 8:23 AM

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If X = V(x) and Y = V(x^2), then X = Y and X is an affine variety but Y is not

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pandaHmm

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I find this at best unclear

maiden ocean Jul 7, 2021, 8:24 AM

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opencry

latent anvil Jul 7, 2021, 8:24 AM

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You can always just take the radical of the ideal

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And get a thing making the set an affine variety

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Bizarre

maiden ocean Jul 7, 2021, 8:24 AM

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yeah i think thats what hes saying

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Idk

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Gamerm oment

latent anvil Jul 7, 2021, 8:24 AM

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Whatever

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Go to bed

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I'll think about your problem

maiden ocean Jul 7, 2021, 8:24 AM

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ok fine sadcat

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cyberbullied by baby algebraic geometry

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sad

latent anvil Jul 7, 2021, 8:24 AM

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lol

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Ah

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I think I have a proof

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Something slightly more general

thorny flame Jul 7, 2021, 9:33 AM

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How do I approach the question?
x^(25)=2mod(133)

hidden haven Jul 7, 2021, 9:37 AM

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More appropriate for #elementary-number-theory

unique juniper Jul 7, 2021, 12:34 PM

#

#

for part (d)

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idrk how to do it

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ive got this: $\alpha\beta = -(\alpha^2 + \alpha) = -(\beta^2 + \beta)$ and $\zeta^{p-1}$ is a part of alpha if p=1 (4) and in beta if p=3 (4)

cloud walrusBOT Jul 7, 2021, 12:36 PM

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Yes ツ

lapis aurora Jul 7, 2021, 2:30 PM

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unique juniper Jul 7, 2021, 2:39 PM

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all the seperable elements + all elements in K will make up L

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since L is algebraic and normal over K

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(i think)

dusty river Jul 7, 2021, 2:42 PM

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That's not true, L could have non separable elements

unique juniper Jul 7, 2021, 2:43 PM

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oh

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i thought normal extensions were seperable extensions

dusty river Jul 7, 2021, 2:44 PM

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normal + separable = Galois

lapis aurora Jul 7, 2021, 2:44 PM

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if it helps, i managed to show that K_s / K is separable and K_i / K is purely inseparable

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and L/K_s should also be purely inseparable

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mhm, guess i will take a look at galois extensions in my book but this exercise is from the chapter before that when they werent introduced yet

dusty river Jul 7, 2021, 2:49 PM

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nice pfp btw

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KEK

lapis aurora Jul 7, 2021, 2:49 PM

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thanks :)

unique juniper Jul 7, 2021, 2:51 PM

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an element is separable if the minimum polynomial that it generates is separable right?

dusty river Jul 7, 2021, 2:51 PM

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yep

unique juniper Jul 7, 2021, 2:55 PM

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maybe char != 0 is important

rustic crown Jul 7, 2021, 2:55 PM

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yep

dusty river Jul 7, 2021, 2:56 PM

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char = 0 makes the problem trivial

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because everything is separable

rustic crown Jul 7, 2021, 2:56 PM

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(if you're working over fields :P)

unique juniper Jul 7, 2021, 2:56 PM

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oh yea

dusty river Jul 7, 2021, 3:03 PM

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Suppose c is in L - K_s. Let f(x) be its minimal polynomial over K. By pure inseparability of L/K_s, there is some d such that c is a p^nth root of d, where p is the characteristic. The powers of x that occur in f(x) should then all be multiples of p^n. Then replacing the coefficients of f(x) with their p^nth roots, and x^kp^n with x^k, we should get a polynomial polynomial for c, separable over K_i. Idk how to proceed though lol

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actually idk if my claims about p^n are legit but I hope they are stare

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And I think this only shows that L/K_i is separable? Is that even right? lol

lapis aurora Jul 7, 2021, 3:06 PM

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yea

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i also got L/K_i is separable

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my argument was a little different tho

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my problem is that L/K_i separable doesnt really help me so far xd

dusty river Jul 7, 2021, 3:08 PM

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yeah opencry

frank fiber Jul 7, 2021, 4:15 PM

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left derived functors of a exact functor $F$ are defined by $L_iF(A)=H_i(F(P))$ where $A$ is an object and $P$ is a proyective resolution of $A$, but the proyective resolutions are exact secuences and so $F(P)$ is exact, this implies that $L_iF(A)=H_i(F(P))=0$ ?

cloud walrusBOT Jul 7, 2021, 4:15 PM

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Or x1

wraith obsidian Jul 7, 2021, 4:16 PM

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Sounds right to me

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I mean left derived functors are constructed to measure failure of exactness of F

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so if F is exact, they should vanish

frank fiber Jul 7, 2021, 4:19 PM

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thanks

maiden ocean Jul 7, 2021, 6:39 PM

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So contextually here n is the number of indeterminates in k[x1, ..., xn] for k algebraically closed and m is the number of generators of the ideal

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and the affine closed set is just the points that are solutions to all generators

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im kind of confused as to the claim here because if n = 2, m =1 implies V(I) is a plane curve then the affine closed set in A^2 thats a union would have to be the case m = 2 right?

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But then you're adding more generators which means you'll have less points in your affine closed set

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i just can't come up with an example where the intersection of 2 plane curves is a plane curve + finite points

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or is it that the plane curve example requires both generators to be non constant and then you can consider like f = (x - 2)(x - 3) with y always 0 or something

latent anvil Jul 7, 2021, 6:43 PM

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The union of two plane curves is a plane curve

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yeah?

maiden ocean Jul 7, 2021, 6:43 PM

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So then f is a finite set of points, g is a plane curve where x and y are both non constant and you take the intersection of (f) and (g)

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yeah V(I1) cup V(I2) = V(I1 cap I2)

latent anvil Jul 7, 2021, 6:44 PM

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If you have curves f = 0 and g = 0, the union is fg = 0. Now take the intersection of the curves fg = 0 and fh = 0 for appropriately chosen f, g, h

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So I'm thinking like, two perpendicular lines each union the same parabola

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intersect the two sets and you get the parabola and the single intersection point of the lines

maiden ocean Jul 7, 2021, 6:45 PM

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Uhhh

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would it be ok if u drew this

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cause im not sure i get what u mean exactly

latent anvil Jul 7, 2021, 6:45 PM

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I'll just write it or

maiden ocean Jul 7, 2021, 6:45 PM

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like on sketch.io or smth

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that works too

latent anvil Jul 7, 2021, 6:45 PM

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Take A = V((y-x^2-1)y) and B = V((y-x^2-1)x)

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The intersection is V(y-x^2-1) union V(x, y)

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So this is two plane curves which intersect at the union of a plane curve and a point

maiden ocean Jul 7, 2021, 6:48 PM

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Oh I see

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(y - x^2 - 1) gets you a parabola

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and then x and y respectively get you vertical and horizontal lines

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and their intersection is the parabola plus (0, 0)

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?

latent anvil Jul 7, 2021, 6:49 PM

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💯

maiden ocean Jul 7, 2021, 6:49 PM

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Ok this makes sense

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Its weird to think about but you can get very "non curve like" plane curves by taking products of various functions

latent anvil Jul 7, 2021, 6:49 PM

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maiden ocean im kind of confused as to the claim here because if n = 2, m =1 implies V(I) is ...

Taking a union of a thing given by m and k generators gives you something with at most mk generators, by multiplying each pair of generators and looking at the ideal generated by all of that. In this case, unioning keeps you at 1*1 = 1 generator

maiden ocean Jul 7, 2021, 6:49 PM

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which is the same thing as just taking the union of their curves

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I see hmmCat

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catFone

lapis aurora Jul 7, 2021, 6:50 PM

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@dusty river @unique juniper update on my galois theory question from earlier: i managed to write a proof by showing that L/K_s(K_i) is both separable and purely inseparable, and therefore the fields are the same. thanks again for the help :)

maiden ocean Jul 7, 2021, 6:50 PM

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Moth is now slightly less owned then they were yesterday

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this conversation really helped sham i think i kinda get how to construct examples now

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thanks nozoomi

latent anvil Jul 7, 2021, 6:52 PM

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Yay!

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It's funny that you started asking about the same stuff at the moment I opened discord on my phone

maiden ocean Jul 7, 2021, 6:52 PM

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I summoned you smugsmug

ionic drum Jul 7, 2021, 8:52 PM

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anybody here familiar with representations of algebraic groups?

maiden ocean Jul 7, 2021, 9:10 PM

#

#

Shouldnt f circ (h1, h2) be 0

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since Im (h1, h2) has to be in X and by definition points of X are mapped to 0 by f

fossil shuttle Jul 7, 2021, 9:26 PM

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frank fiber thanks

(for i >0)

amber stag Jul 8, 2021, 1:19 AM

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maiden ocean since Im (h1, h2) has to be in X and by definition points of X are mapped to 0 b...

What’s the domain of (h1, h2) when you talk about its image

maiden ocean Jul 8, 2021, 1:19 AM

#

doesnt it have to be X

amber stag Jul 8, 2021, 1:20 AM

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If we restrict the domain to Y, then yes, the composition is 0 on Y

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Since it’s a multiple of h

maiden ocean Jul 8, 2021, 1:21 AM

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in the defn i got the domain was Y specifically

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hmmCat

amber stag Jul 8, 2021, 1:21 AM

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Yeah, then the composition is indeed 0 on Y

maiden ocean Jul 8, 2021, 1:22 AM

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weird

amber stag Jul 8, 2021, 1:22 AM

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Y is the set of zeros of h, and the composition is a multiple of h, so

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Anyway, it’s not hard to give an example. f = x1-x2, h = x1-x2^2, h1 = x1, h2 = x2^2

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Haven’t touched algebraic geometry for ages I feel I explained badly lmao

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f circ (h1, h2) isn’t 0 as a polynomial (whose domain is K^2)

maiden ocean Jul 8, 2021, 2:56 AM

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@oblique river i was thinking about the thing we talked about yesterday and i realized that im not actually sure why when you localize f is 0?

#

#

repasting the image for context but like

#

Why would the image of f be 0 in O_X, p?

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and if its not how does the y = gx claim work?

latent anvil Jul 8, 2021, 3:08 AM

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isn't f identically 0 on X? Like it's the zero element of O_X(X)?

oblique river Jul 8, 2021, 3:23 AM

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I agree with shamrock here I think

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X is sort of defined as like "the thing where f is 0"

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so in O_X, f should be 0

maiden ocean Jul 8, 2021, 2:58 PM

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Oh wait yeah im dumb

#

Lmfao

chilly ocean Jul 8, 2021, 2:58 PM

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stare

maiden ocean Jul 8, 2021, 4:09 PM

#

#

Im confused about the part where the morphism from U to V(f) is defined

#

like t and s are functions so i dont really see whats happening...

oblique river Jul 8, 2021, 4:58 PM

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It does seem to be going backward to me

#

Or something

light tusk Jul 8, 2021, 6:15 PM

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I think instead of "$(t,s) \mapsto (x,y)$ defines an isomorphism", the author means "the homomorphism given by sending $t \mapsto x$ and $s \mapsto y$ and extending linearly to all of $K(X)$". But they wrote a pair as if it were part of a cartesian product, ie the underlying space.

cloud walrusBOT Jul 8, 2021, 6:15 PM

#

Apopheniac

light tusk Jul 8, 2021, 7:07 PM

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If $u\in U$, then $\rho:U \to V(f)$ is given by $\rho(u) = (t(u),s(u)) \in \mathbf{A}_k^2$.

Then $f(t(u),s(u)) = 0$ by definition of $f$, so $(t(u),s(u))\in V(f)$.

This is locally (U is open) given by polynomials, hence a morphism $\rho: U \to V(f)$.

cloud walrusBOT Jul 8, 2021, 7:07 PM

#

Apopheniac

latent anvil Jul 8, 2021, 8:39 PM

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~~book is bad~~

maiden ocean Jul 9, 2021, 12:48 AM

#

light tusk If $u\in U$, then $\rho:U \to V(f)$ is given by $\rho(u) = (t(u),s(u)) \in \math...

late but this makes sense!! thank you

timber sun Jul 9, 2021, 1:32 PM

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I have trouble with the only if direction of this statement. I Know that $\alpha$ is injective and $\sigma$ is surjective, as their composition is the identity. But how do i get the isomorphism? It feels like im missing something simple.

cloud walrusBOT Jul 9, 2021, 1:32 PM

#

Cloud63

wraith obsidian Jul 9, 2021, 1:42 PM

#

timber sun I have trouble with the only if direction of this statement. I Know that $\alph...

Hint: σα is an idempotent endomorphism of P.

timber sun Jul 9, 2021, 4:28 PM

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wraith obsidian Hint: σα is an idempotent endomorphism of P.

thanks i got it

valid shoal Jul 9, 2021, 7:07 PM

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hi! im trying to understand U(1) in the context of astrophysics, usually its followed by the word gauge. someone already explained abelian groups in a more philosophical context to me, but i'm still struggling with understanding its meaning in my field and in the papers ive seen it in -- i can send screenshots of the sentences in the papers if that's helpful

valid shoal Jul 9, 2021, 8:01 PM

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got it, so what would U(1)x gauge boson Z’ or U(1)Lµ−Lτ bosons look like?

prisma ibex Jul 9, 2021, 10:56 PM

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so what you've written looks like electroweak theory already

#

for U(1) there is only one gauge boson, the photon

#

The U(1) gauge field, the electromagnetic field, is a gauge field because it's invariant under these phase rotations by U(1)

#

certainly you can do this globally without much issue: for example you can negate the sign on all electric charges and physics doesn't change so much

#

but you can also do this locally: you can apply any phase rotation at any point of space and let these phase rotations vary smoothly, then the physics is still fine

maiden ocean Jul 10, 2021, 1:31 AM

#

kay posting this here in case someone else sees this

#

I'm very confused by the part of the example where they talk about taking the tensor product of C with itself

#

I dont understand why we're taking A/M otimes C rather than A otimes C

#

Like points of X_C over a maximal ideal m of X are going to be maximal ideals m1 and m2 with i^1(m_i) = m

#

the only thing i can think of is that for some reason here we can say that quotient by tensor product = tensor product with quotient which isnt usually true and im not sure why it would be here...

#

But even then like.

#

I guess the idea is that we have i: A -> A otimes C and for each point p in X corresponding to a maximal ideal m it induces a map A/M -> A/M otimes C so C -> C otimes C? and choices of such a map look like choices of a maximal ideal m_i in A otimes C to quotient by so having two natural maps C -> C otimes C means that there are two ideals m_i with spec(i)(m_i) = m?

latent anvil Jul 10, 2021, 3:14 AM

#

@maiden ocean I'm having trouble understanding what your q is

maiden ocean Jul 10, 2021, 4:40 AM

#

latent anvil <@358385343629885440> I'm having trouble understanding what your q is

I dont understand what C tensored with itself has to do with anything

#

Im just confused as to how it relates to the points over m

amber stag Jul 10, 2021, 5:45 AM

#

So M is a closed point in X. If we want to write M as spec( ), what should we put in the parentheses?

#

Or what’s the coordinate ring of M?

bleak perch Jul 10, 2021, 7:53 AM

#

Can anyone help with this problem? I'm in an introductory abstract alg course and we just got to groups. I assume that I would need to use some group properties, maybe the exponent rules, however I am stuck.

upper pivot Jul 10, 2021, 7:55 AM

#

what did you try?

carmine fossil Jul 10, 2021, 7:58 AM

#

If you want a refresher on exponents,
(x^m)^r=x^(mr)

#

And x^a x^b=x^(a+b)

bleak perch Jul 10, 2021, 8:03 AM

#

Do I then just directly assume x = y to be true and try to manipulate the exponents? I'm really not sure

upper pivot Jul 10, 2021, 8:03 AM

#

so try manipulating the equations given, so for a hint

#

like try combining two equations in some way

bleak perch Jul 10, 2021, 8:09 AM

#

Since x^n is equal to y^n, could i create an equation where its just x^n - x^n = 0, and work from there?

dusty river Jul 10, 2021, 8:09 AM

#

write x = x^1

#

see what you can do from here

bleak perch Jul 10, 2021, 8:11 AM

#

Ok thanks for the help people, I'll try to dig into it some more

snow cedar Jul 10, 2021, 8:18 AM

#

Hello, I am working on a simple proof from Dummit and Foote abstract algebra (3e). The problem with solution goes like this:

#

My question is: Is it fair to assume that matrix multiplication is associative? I obviously know that this is true from LA, but the textbook has not covered associativity of matrices yet, and feel like the "correct" proof should use the definition of multiplication of 2x2 matrices since that was clearly defined in the problem.

upper pivot Jul 10, 2021, 8:22 AM

#

I mean if it bothers you, you could go ahead and prove associativeness once

#

but in general, you really shouldnt worry about this

quaint tree Jul 10, 2021, 8:22 AM

#

Matrix multiplication is associative.

#

I'm assuming the book asked you to do this problem because matrices are a common example of a group.

#

(That said I haven't read D&F yet so maybe I'm mistaken)

dusty river Jul 10, 2021, 8:25 AM

#

Not sure what you were saying about 2x2 matrices in your question. You can prove associativity in general

snow cedar Jul 10, 2021, 8:33 AM

#

Thanks for the replies. There is no official solutions (as far as I am aware), so I have a hard time understanding if something is fair game or not. The solution is straightforward if we just assume associativity. I just find it strange that they included the definition of matrix multiplication if we aren't even going to use it in the proofs.

dusty river Jul 10, 2021, 8:35 AM

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It seems that matrix multiplication was defined before, but the problem is saying that "by PQ we mean matrix multiplication of P and Q which you already know"

#

In which case either they are assuming you already know it in which case it's fair game to use basic properties of it, or they defined it somewhere earlier

bleak perch Jul 10, 2021, 9:31 AM

#

bleak perch Can anyone help with this problem? I'm in an introductory abstract alg course an...

referring to this problem again, I took x^n = y^n and multiplied both sides by the inverse x^-n to obtain (x^-n)x^n = (x^-n)y^n, since multiplying an element by its inverse returns the neutral element e_g, i get e_g = x^-n y^n. Is this logical so far? and am I on the right track?

hidden haven Jul 10, 2021, 9:32 AM

#

This is correct but I don't think it'll lead anywhere

bleak perch Jul 10, 2021, 9:33 AM

#

hmm..

hidden haven Jul 10, 2021, 9:33 AM

#

What I was hinting at was starting at x, and have a chain of equalities ending at y. The starting point is x = x^1 = ....
Can you use something given in the problem statement now, to get further equalities?

bleak perch Jul 10, 2021, 9:36 AM

#

x = x^1 = y^1 ?

#

I'm not sure

hidden haven Jul 10, 2021, 9:39 AM

#

You are given that there exist s and r such that rm+sn = 1

bleak perch Jul 10, 2021, 9:54 AM

#

Do i try to rearrange exponents to follow the linear combination? I just don't see how having a chain of equalities helps here

hidden haven Jul 10, 2021, 9:54 AM

#

x = x^1 = x^(rm+sn). Do you see it now?

#

You will have to use the given facts that x^m = y^m, x^n = y^n

#

And exponent laws

bleak perch Jul 10, 2021, 9:58 AM

#

oh wow.. that makes sense.. I was staring at my paper for 30 minutes and just wasn't able to catch that.. It's my first proof heavy course, any advice on how to get better at proofs other than practicing problems? or is that it lol

hidden haven Jul 10, 2021, 9:59 AM

#

Yeah I guess practising problems is the way to go, it's about learning to put together all the given facts

bleak perch Jul 10, 2021, 10:33 AM

#

Hey Moldilocks, I think i figured out the proof, how do I use the LaTeX bot so I can run it by you to see if its logically sound?

hidden haven Jul 10, 2021, 10:35 AM

#

If you just type latex code it will catch it (it detects the dollar signs) or you can start your message with ,tex to explicitly tell it tex your message

bleak perch Jul 10, 2021, 10:35 AM

#

Ok cool

#

$x^1 = x^{rm + sn} = x^{rm} \cdot x^{sn} = (x^m)^r \cdot (x^n)^s$, since $x^m = y^m \land x^n = y^m$, we replace x with y, such that; $(y^m)^r \cdot (y^n)^r = y^{mr} \cdot y^{ns} = y^{mr + ns} = y^1 = y$

cloud walrusBOT Jul 10, 2021, 10:39 AM

#

bigheavyballs

bleak perch Jul 10, 2021, 10:40 AM

#

forgot to start with x instead of x^1

hidden haven Jul 10, 2021, 10:40 AM

#

Yep, nice

hidden haven Jul 10, 2021, 10:40 AM

#

bleak perch forgot to start with x instead of x^1

I don't think anyone would mind that lol

bleak perch Jul 10, 2021, 10:40 AM

#

awesome man. Thank you so much, not a better feeling than getting a proof right

hidden haven Jul 10, 2021, 10:42 AM

#

True catKing

chilly ocean Jul 10, 2021, 10:50 AM

#

moldi for honorable

wooden ember Jul 10, 2021, 10:56 AM

#

im a little confused here: i did exercise 4,a) using the permutation representation of the action,but apparently that was not the way to go considering question 5. I don't understand because the group action can only be considered a homomorphism (and hence have a kernel) if you look at its permutation representation. The kernel of the action as a map from G x A to A doesn't mean anything right?

#

Okay nvm i really can't read, they defined the kernel again a little differently for an action...

#

i read through the theory a few days ago and forgot about that detail

hidden haven Jul 10, 2021, 11:00 AM

#

I'm guessing the kernel is defined as the kernel of the group homomorphism that the action corresponds to, from G to Aut(A)? catThink ie set of all group elements that act the same way as identity

wooden ember Jul 10, 2021, 11:01 AM

#

no see i thought that's what it meant but they had simply defined it as ${g\in G : \forall a\in A, ga=a }$ i just didnt see that

#

but the definitions are equivalent

cloud walrusBOT Jul 10, 2021, 11:02 AM

#

Little Narwhal

hidden haven Jul 10, 2021, 11:02 AM

#

That is what I said stareFlushed

wooden ember Jul 10, 2021, 11:02 AM

#

no what i mean is they didnt explicitly state it like you did

#

and exercise 5 is meant to make you see it's the same

hidden haven Jul 10, 2021, 11:03 AM

#

Ahh I see, didn't read that

wooden ember Jul 10, 2021, 11:03 AM

#

"permutation representation" is the group homomorphism you're mentioning

hidden haven Jul 10, 2021, 11:04 AM

#

Right

wooden ember Jul 10, 2021, 11:04 AM

#

also did you mean to say Aut(A)?

#

because isnt that only defined if A is a group?

#

in here they have it from $G \to S_A$

cloud walrusBOT Jul 10, 2021, 11:05 AM

#

Little Narwhal

hidden haven Jul 10, 2021, 11:06 AM

#

Automorphism is a general term, not just used for groups

#

A set automorphism is just a bijection from the set to itself

wooden ember Jul 10, 2021, 11:06 AM

#

i see

hidden haven Jul 10, 2021, 11:07 AM

#

The collection of automorphisms of anything always form a group though

wooden ember Jul 10, 2021, 11:07 AM

#

but if G were to act on another group H wouldnt wouldnt Aut(H) be more restrictive than the set of permutations on H?

#

oh i guess not

hidden haven Jul 10, 2021, 11:08 AM

#

How would a group act on another group?

wooden ember Jul 10, 2021, 11:08 AM

#

S_H is always a subset of Aut(H) right?

hidden haven Jul 10, 2021, 11:08 AM

#

Do you mean act on the underlying set of H?

wooden ember Jul 10, 2021, 11:08 AM

#

yeah

hidden haven Jul 10, 2021, 11:08 AM

#

wooden ember S_H is always a subset of Aut(H) right?

The other way around, for structures. A group a structure, on a set

wooden ember Jul 10, 2021, 11:08 AM

#

ah but then Aut(H) would refer to automorphisms on the underlying set

#

i see

hidden haven Jul 10, 2021, 11:08 AM

#

wooden ember ah but then Aut(H) would refer to automorphisms on the underlying set

Yep

wooden ember Jul 10, 2021, 11:09 AM

#

yeah so isnt using Aut a little confusing

#

idk the S notation just feels less confusing to me

hidden haven Jul 10, 2021, 11:09 AM

#

Usually you wouldn't make a distinction between a group and it's underlying set, but it's always useful to keep it in mind

#

Well mostly it is clear from the context, if you see Aut(A) and nothing is known about A other than it being a set, then you can be sure that it's just S_A

wooden ember Jul 10, 2021, 11:10 AM

#

fair enough

hidden haven Jul 10, 2021, 11:10 AM

#

Also you would often see Aut_Set(A) or Aut_Grp(G)

wooden ember Jul 10, 2021, 11:10 AM

#

right

hidden haven Jul 10, 2021, 11:10 AM

#

If disambiguation is needed

wooden ember Jul 10, 2021, 11:11 AM

#

the notation is just longer then

hidden haven Jul 10, 2021, 11:11 AM

#

Yeah lol

#

But for sets yeah sym(A) or S_A is more standard it's just that Aut is also kind of universal notation

#

You can take Aut for any kind of structure, graphs, categories, rings, fields, top spaces etc

wooden ember Jul 10, 2021, 11:12 AM

#

true

#

is there a meaningful way in which a group can act on any of these structures, not just the underlying set?

hidden haven Jul 10, 2021, 11:14 AM

#

I guess you could define a group G acting on a structure A to be any group homomorphism G → Aut(A)

#

But that's then also the same as a homomorphism G→Aut(underlying set of A) by composing with the inclusion

wooden ember Jul 10, 2021, 11:15 AM

#

the inclusion?

hidden haven Jul 10, 2021, 11:15 AM

#

wooden ember S_H is always a subset of Aut(H) right?

The opposite of this

wooden ember Jul 10, 2021, 11:15 AM

#

right

hidden haven Jul 10, 2021, 11:16 AM

#

We do study groups acting on other things but ig the theory from groups acting on sets is sufficient for most purposes

#

You just have an additional condition that the action of each element of the group is an automorphism of the structure which can be studied separately

wooden ember Jul 10, 2021, 11:17 AM

#

alright

chilly ocean Jul 10, 2021, 11:36 AM

#

group actions catThink

#

geometry is group actions

chilly ocean Jul 10, 2021, 12:30 PM

#

group action on a non concrete category

#

Are all (locally?) small categories equivalent to a subcategory of Set?

dusty river Jul 10, 2021, 12:34 PM

#

small and locally small would both be necessary

#

sorry locally would be

chilly ocean Jul 10, 2021, 12:37 PM

#

do you know counterexample for small?

dusty river Jul 10, 2021, 12:38 PM

#

Set

chilly ocean Jul 10, 2021, 12:38 PM

#

Set is a subcategory of Set tho

dusty river Jul 10, 2021, 12:39 PM

#

I think you are confusing necessary and sufficient

chilly ocean Jul 10, 2021, 12:40 PM

#

huh?

#

im asking if there's a small category not equivalent to a subcat of Set

dusty river Jul 10, 2021, 12:41 PM

#

ah yeah

chilly ocean Jul 10, 2021, 12:41 PM

#

cmon barbie lets go party, ah ah ah yeah

dusty river Jul 10, 2021, 12:42 PM

#

wait nvm

#

I confused myself KEK

chilly ocean Jul 10, 2021, 12:42 PM

#

stare

dusty river Jul 10, 2021, 12:49 PM

#

I think for any given C you can do something with hom sets, since they determine C upto isomorphism

#

Map every object c to product of hom(a,c) (a varies). Every morphism a→b determines a map from every hom(x,a) to hom(x,b), so you get a morphism from every hom(x,a) to product_x of hom(x,b)'s, and this you can compose with the projections from product_x hom(x,a) to each hom(x,a) to get a morphism from product to product

#

Let me check if this works KEK

#

nvm that actually gives you a lot of morphisms from product for a to product of b stare not a single one

#

ah this works I think. What I have described on the morphisms is essentially defining a map on each coordinate. So the functor is $F: C \to Set$, on objects, $c\mapsto\prod_{x\in \text{obj} C} \hom(x,a)$, and the morphism $f:a\to b$ maps to the morphism which acts on the coordinate corresponding to each $x$ by post composition. That it is a functor is obvious. Faithfulness comes from $Ff = Fg\Rightarrow\hom(-,f)=\hom(-,g)\Rightarrow f=g$

cloud walrusBOT Jul 10, 2021, 1:07 PM

#

Moldilocks

thorny flame Jul 10, 2021, 1:09 PM

#

What does this mean?

scarlet estuary Jul 10, 2021, 1:09 PM

#

wtf

dusty river Jul 10, 2021, 1:09 PM

#

I am guessing it is a typo and it is supposed to be (12)

scarlet estuary Jul 10, 2021, 1:09 PM

#

i mean <(1, 2)> is probably the subgorup generated by (1, 2)

#

except thats silly

#

yeah (1 2) makes more sense

#

but i have no clue why theyd denote it .H

#

$\langle (1 2) \rangle$ indicates the subgroup generated by $(1 2)$, and $\leq S_3$ denotes it being a subgroup of $S_3$

cloud walrusBOT Jul 10, 2021, 1:11 PM

#

Namington

scarlet estuary Jul 10, 2021, 1:11 PM

#

and theyre using <> instead of langle rangle because laziness i guess

lavish spoke Jul 10, 2021, 1:11 PM

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my abstract algebra course this year used notation like (1, 2), i prefer it tbh

scarlet estuary Jul 10, 2021, 1:11 PM

#

huh i really dislike that

#

maybe thats just me but like

#

that makes me think its storing elements

#

like a vector

#

permutations dont store elements

#

they move them

dusty river Jul 10, 2021, 1:12 PM

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product group lol

scarlet estuary Jul 10, 2021, 1:12 PM

#

commas "separate" them but permutations arent about "separating things"

#

theyre about "moving" things

#

feels counterintuitive

#

that might just be my bias though

lavish spoke Jul 10, 2021, 1:12 PM

#

i just think its more readable personally

dusty river Jul 10, 2021, 1:14 PM

#

cloud walrus **Moldilocks**

@chilly ocean

chilly ocean Jul 10, 2021, 1:16 PM

#

i have to reread it carefully, am noob at category theory

dusty river Jul 10, 2021, 1:19 PM

#

I was being a bit silly. The functor I have defined is exactly the composition of all the covariant hom functors with the "product over objects of c" multifunctor

#

https://math.stackexchange.com/questions/2997256/is-every-small-category-isomorphic-to-a-subcategory-of-the-category-of-relations also the answers on this do an embedding into Set, similar idea but they use coproduct instead of product cocatThink

thorny flame Jul 10, 2021, 1:35 PM

#

scarlet estuary i mean <(1, 2)> is probably the subgorup generated by (1, 2)

I see. Can you show me what it looks like?

scarlet estuary Jul 10, 2021, 1:37 PM

#

{e, (1, 2)}

thorny flame Jul 10, 2021, 1:39 PM

#

{e, 1, 2}?

scarlet estuary Jul 10, 2021, 1:40 PM

#

no

#

the subgroup has two elements

#

e, the identity permutation
(1, 2), the permutation representing swapping 1 and 2

#

this is isomorphic to every other group of 2 elements of course

#

we could use the symbols 0 and 1 instead, in which case we'd call it Z/2Z

thorny flame Jul 10, 2021, 1:43 PM

#

That I understood

#

However how does it show for groups is bewildering

delicate bloom Jul 10, 2021, 2:08 PM

#

you're being too vague

wooden ember Jul 10, 2021, 2:20 PM

#

scarlet estuary theyre about "moving" things

ngl we should just write cycles as circles with arrows on them

thorny flame Jul 10, 2021, 2:27 PM

#

What does this mean?

hidden haven Jul 10, 2021, 2:34 PM

#

Is 𝕌 defined somewhere?

chilly ocean Jul 10, 2021, 2:35 PM

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wild guess - group of units?

thorny flame Jul 10, 2021, 3:08 PM

#

Math group says it's the set of reversible remainders.

hidden haven Jul 10, 2021, 3:19 PM

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Modulo some fixed integer?

#

I guess then U = Z/nZ. U* = group of invertible things in this, and f is then a homomorphism from U* to itself

thorny flame Jul 10, 2021, 3:21 PM

#

What?

hidden haven Jul 10, 2021, 3:25 PM

#

What does reversible remainders mean?

thorny flame Jul 10, 2021, 3:27 PM

#

Trying to understand. Dm'd the guy who answered

hidden haven Jul 10, 2021, 3:28 PM

#

That's just every number

#

gcd(a,a+1) = 1 for all a

thorny flame Jul 10, 2021, 3:29 PM

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True, It's a function on modulo groups

#

Apperantly

hidden haven Jul 10, 2021, 3:29 PM

#

Right, Z/nZ right?

#

Integers modulo some number n

thorny flame Jul 10, 2021, 3:29 PM

#

IDK what Z/nZ means

hidden haven Jul 10, 2021, 3:29 PM

#

0,1,...,n-1

thorny flame Jul 10, 2021, 3:29 PM

#

Ohh. THat's what you meant

#

Sorry

hidden haven Jul 10, 2021, 3:30 PM

#

So in that set there are some things which don't have inverses

#

Like if n=4, then 2 doesn't have an multiplicative inverse

thorny flame Jul 10, 2021, 3:31 PM

#

right

hidden haven Jul 10, 2021, 3:31 PM

#

A cross in superscript means you take the subset of things which are invertible

#

Which is the same as taking the set of things that have gcd 1 with n

#

(by Chinese remainder theorem)

#

And then this becomes a group under multiplication

#

So you can talk about group homomorphisms

chilly ocean Jul 10, 2021, 5:32 PM

#

Freyd-Mitchell's embedding theorem says every small abelian category is equivalent to a (full) subcategory of the category of left-modules over a ring.

#

I wonder: is there something fundamentally very different about non-small abelian categories? Or are they also in some sense "like" categories of left-modules over a ring

hidden haven Jul 10, 2021, 5:55 PM

#

If you use Tarski's axiom (any set can be contained in some universe) then any abelian category C is small with respect to some universe U. You can then take R-mod_U to be the category of all small (wrt U) R modules, and then C has a full and exact embedding into R-mod_U

#

The problem with non small categories in general is that they could be too large, so we are just enlarging our universe here

#

I am not sure how this would work without Tarski's axiom though, I think then you'd just have to leave it at "this only works for small categories"

#

Also why are you asking meme math in #groups-rings-fields instead of #category-theory? cocatThink

#

@chilly ocean

chilly ocean Jul 10, 2021, 6:03 PM

#

well categories are algebraic structures, and so are modules lol

hidden haven Jul 10, 2021, 6:04 PM

#

What does algebraic mean to you cocatThink

chilly ocean Jul 10, 2021, 6:08 PM

#

actually no, categories aren't, they're partial algebras tho

hidden haven Jul 10, 2021, 6:08 PM

#

KEK

bleak perch Jul 11, 2021, 2:50 AM

#

Does anyone have any advice as to how to prove 4d? My first assumption was to prove it via double inclusion using the assumption that 4b and 4c are true. Is this on the right track?

amber stag Jul 11, 2021, 3:11 AM

#

I think 4b and 4c are more like hints.

#

To prove lhs \subset rhs, we can express a symmetry in terms of tau and sigma by studying what the images of -1,0,1 are.

bleak perch Jul 11, 2021, 3:22 AM

#

As in try to show that there is a bijection? like an identity function?

hidden haven Jul 11, 2021, 3:25 AM

#

You have to show that every symmetry has one of those 2 forms. Take a symmetry y. Look at b and c and try to use a similar strategy like terrific said (you only need to know where 0 and 1 go)

bleak perch Jul 11, 2021, 3:26 AM

#

so every symmetry has the form of either i + 1 or -(i + 1)?

hidden haven Jul 11, 2021, 3:27 AM

#

Nope, the forms are given to you in d, there are infinitely many

#

0 and 1 can go to a lot of things

#

And you'll be able to prove that if 2 symmetries do the same thing to 0 and the same thing to 1, they will be equal everywhere

bleak perch Jul 11, 2021, 3:30 AM

#

hmm i see

#

Why is it that I only need to prove what happens to 0 and 1?

#

If thats the case can't I plug in 0 to each symmetry and see its output? because if thats the case sigma^1 == sigma^1 o tau for 0

hidden haven Jul 11, 2021, 3:37 AM

#

0 alone isn't enough

#

But 0 and 1 together determine the symmetry

#

Proving that images of 0, 1 are enough: ||Symmetries preserve distances. Given a number x, you can write its distance from 1 and from 0 on the number line. Then apply the symmetry, and these distances should stay the same. Solving these equations you will get a unique choice for x which could satisfy these. This proves that there is at most 1 symmetry that has the given effect on 0 and 1||

bleak perch Jul 11, 2021, 3:47 AM

#

So given x, its distance to 1 should be x - 1 = 0, x = 1 , distance to 0 should be x - 0 = 0, x = 0. If I apply then the symmetries sigma^i and sigma^i o tau, to the values of x, they should be the same?

hidden haven Jul 11, 2021, 3:49 AM

#

Why =0?

#

The distance to 1 should just be x-1

bleak perch Jul 11, 2021, 3:49 AM

#

ah my mistake youre right

hidden haven Jul 11, 2021, 3:50 AM

#

Also don't apply symmetries of the given form, assume that you are given some arbitrary symmetry y

#

So you can write (distance between x and 1) = d(x,1) = x-1, and similarly d(x,0) = x

#

Preserving distances means that if
d(a,b) = n,
Then d(y(a), y(b)) = n as well. ie distance between a pair is the same as the distance between its image

bleak perch Jul 11, 2021, 3:57 AM

#

I suppose a follow up question would be how can some arbitrary symmetry y be related to sigma/tau, is that something i need to explicitly find? Why even define arbitrary y? Because if we found preserved distances with y how can we know for sure that it translates to sigma/tau. I hope I'm not completely missing some symmetry axiom that would answer my question lol

hidden haven Jul 11, 2021, 4:00 AM

#

Well suppose 0 goes to a, 1 goes to b. Then can you come up with a symmetry made from the sigma and tau that does exactly that to 0 and 1?

carmine fossil Jul 11, 2021, 4:07 AM

#

bleak perch I suppose a follow up question would be how can some arbitrary symmetry y be rel...

Suppose you have a symmetry s.
Then either one of the two statements is true:
s(1)=s(0)-1,
s(1)=s(0)+1

bleak perch Jul 11, 2021, 4:10 AM

#

hmmm.. Thanks for the help you two. I'm gonna try to work on it some more by myself and ask questions if I need help again

carmine fossil Jul 11, 2021, 4:12 AM

#

||The idea is to consider the symmetry t= (\sigma)^(-s(0)) s. This new symmetry t has this property that t(0)=0 and t(1) is either 1 or -1. Now use b and c parts to find t and hence s||

#

||Note that everything here is forced by the fact that s is a symmetry||

bleak perch Jul 11, 2021, 6:49 AM

#

How exactly did you come up with that specific symmetry?

#

I understand everything up until that point tbh

carmine fossil Jul 11, 2021, 6:52 AM

#

I didn't like that s(0) was nonzero

#

So I made a symmetry such that it will be zero

#

Which turned out to work perfectly

dusty river Jul 11, 2021, 6:53 AM

#

Try to picture what this is saying. It says that all symmetries of Z are either translations, or translations + reflection

#

From this you can try to guess what the symmetry would be in terms of σ and τ

carmine fossil Jul 11, 2021, 6:54 AM

#

So,If you negate the translation,you get identity or reflection

bleak perch Jul 11, 2021, 6:57 AM

#

Right, but since (\sigma) is raised to zero wouldn't this imply that t = 1?

carmine fossil Jul 11, 2021, 6:58 AM

#

\sigma is raised to -s(0)

#

not 0

#

Applying $\sigma^{-s(0)}$ moves everything to the left by s(0) units

cloud walrusBOT Jul 11, 2021, 6:58 AM

#

Buncho Dragons

bleak perch Jul 11, 2021, 6:59 AM

#

but isn't s(0) equal to 0 based on 4b/4c?

dusty river Jul 11, 2021, 7:00 AM

#

That's not what 4b and 4c say

#

They say that if s(0) = 0 then certain things happen not that s(0) = 0

bleak perch Jul 11, 2021, 7:13 AM

#

So the symmetry t, because the way it was set up, t(0) = 0, but now we need to figure out whether t(1) = 1 or -1?

dusty river Jul 11, 2021, 7:15 AM

#

Yeah

bleak perch Jul 11, 2021, 7:18 AM

#

I have a feeling its 1, but cant prove it

dusty river Jul 11, 2021, 7:18 AM

#

It can be either. You have to relate it to s

bleak perch Jul 11, 2021, 7:43 AM

#

I'm not sure... I just don't know how to to relate it to s, when the mapped values of s are unknown

hidden haven Jul 11, 2021, 7:55 AM

#

s(1) and t(1) will be related

#

And s(1) is related to s(0) as ||s(1) = s(0) ± 1||

trim grove Jul 11, 2021, 10:01 AM

#

the group of order 4 non cyclic can be of type $D_2$ , but further how to procede?

cloud walrusBOT Jul 11, 2021, 10:01 AM

#

honey99

unique juniper Jul 11, 2021, 10:51 AM

#

do you know the sylow theorems?

#

steady axle Jul 11, 2021, 11:21 AM

#

can someone take a look at my solution to this problem and tell me if its ok?

#

Consider ideal $J=(x^2+x+1,5)$ of $Z[x]$. Note that $(x^4+x^2+1)\subset I$ in $Z[x]$ as $(x^2+x+1)(x^2-x+1)=x^4+x^2+1$. So $I=J/(x^4+x^2+1)$ is an ideal of $A$. $A/I\cong Z[x]/J$ by fourth isomorphism theorem. We note that $J$ is kernel of $Z[x]\rightarrow \frac {Z/5Z[x]}{(x^2+x+1)}$. The RHS is degree 2 extension of $F_5$ - a field of degree 25.

cloud walrusBOT Jul 11, 2021, 11:21 AM

#

bert

steady axle Jul 11, 2021, 11:28 AM

#

also, $R[x]/((x^2+1)(x^2+x+1))\cong C \times C$ right?

cloud walrusBOT Jul 11, 2021, 11:28 AM

#

bert

hidden haven Jul 11, 2021, 11:31 AM

#

Yes, by the Chinese remainder theorem

thorny flame Jul 11, 2021, 12:59 PM

#

Is there a way to calculate this?

hidden haven Jul 11, 2021, 1:00 PM

#

Calculate as in simplify?

thorny flame Jul 11, 2021, 1:00 PM

#

yeah

hidden haven Jul 11, 2021, 1:00 PM

#

Or numerically calculate

thorny flame Jul 11, 2021, 1:00 PM

#

both

hidden haven Jul 11, 2021, 1:00 PM

#

I don't think you can simplify the expression

#

Numerically it is x^(a^b)

#

So you compute the 2 exponents

thorny flame Jul 11, 2021, 1:01 PM

#

I see

scarlet estuary Jul 11, 2021, 2:15 PM

#

whys this in #groups-rings-fields

chilly ocean Jul 11, 2021, 2:16 PM

#

thonk

thorny flame Jul 11, 2021, 2:25 PM

#

Sorry, it was the part I had stuck in an abstract algebra in permutations

#

I was looking for pemutation to timer of 5^25

#

I've got a diffrent question anyway

#

How do I calculate? I'm keep getting 8 as a solution

delicate bloom Jul 11, 2021, 2:27 PM

#

24=25-1 = -1 mod 5 so just need to know if 101 is even or odd

thorny flame Jul 11, 2021, 2:27 PM

#

Sorry. wrong one

#

delicate bloom Jul 11, 2021, 2:28 PM

#

do you know euler's theorem?

thorny flame Jul 11, 2021, 2:28 PM

#

yeah

#

Do I need to implement it?

delicate bloom Jul 11, 2021, 2:28 PM

#

with the chinese remainder theorem too, probably

thorny flame Jul 11, 2021, 2:29 PM

#

Okay

delicate bloom Jul 11, 2021, 2:29 PM

#

unless you see some other trick, it might have an easy way out

thorny flame Jul 11, 2021, 2:30 PM

#

Right. I'll try it

#

How do I break it down?

#

Chinese remainder theorem has a few broken modulos and their remainder and tries to find a common modulo

delicate bloom Jul 11, 2021, 2:34 PM

#

solve it mod 4 and mod 25

thorny flame Jul 11, 2021, 2:34 PM

#

ohh. I thought of using their primes 2 and 5

#

that was the mistake

delicate bloom Jul 11, 2021, 2:34 PM

#

that's a good strategy too sometimes, but here you don't have to

#

@thorny flame did you get it? Also for this sort of stuff you should probably be using #elementary-number-theory instead

thorny flame Jul 11, 2021, 2:47 PM

#

no I got 8 instead of 24

#

posted there now

frank fiber Jul 11, 2021, 4:15 PM

#

Let $A$ and $B$ abelian categories, what condition should a functor $F:A \rightarrow B$ satisfy for its left derived functor $L_i F$ exist?

#

F must be aditive or right exact?

cloud walrusBOT Jul 11, 2021, 4:16 PM

#

Or x1

sturdy marsh Jul 11, 2021, 5:08 PM

#

F must be additive, and you need enough projectives @frank fiber

#

you can get away with "enough acyclics" if defined properly

frank fiber Jul 11, 2021, 5:40 PM

#

thanks

chilly radish Jul 11, 2021, 9:02 PM

#

By the fundamental theorem of finitely generated abelian groups, determining the number of groups of a given order just comes down to a combinatorial factoring problem right?

hollow imp Jul 11, 2021, 9:07 PM

#

the number of abelian groups only

#

you can't say anything about non-abelian groups with that theorem

oblique river Jul 11, 2021, 9:09 PM

#

just to add on to this, we dont have a good way of answering the full question "how many groups are there of a given order"

#

we do know like, groups of order p^k where p is a prime number and k < 8

#

but i dont think we know like, how many groups of order 5^8 there are

chilly radish Jul 11, 2021, 9:19 PM

#

Oh yea

#

I meant abelian groups

#

the word just slipped from my fingers

quaint tree Jul 11, 2021, 9:20 PM

#

For small orders, it often comes down to trying things with Sylow's theorems, showing that certain groups are impossible, factoring the order and looking for simple groups whose orders are factors and doing direct and semidirect products, etc.

#

At least in my experience doing problems like "find all groups of order 2223" as homework problems

oblique river Jul 11, 2021, 9:21 PM

#

yeah you can do things in large swaths too

chilly radish Jul 11, 2021, 9:21 PM

#

quaint tree At least in my experience doing problems like "find all groups of order 2223" as...

I've had enough of those problems for the rest of my life

oblique river Jul 11, 2021, 9:21 PM

#

like "groups of order pq"

#

there are either 1 or 2 depending on p mod q (wlog p > q)

#

there are some interesting patterns if you look at like

#

groups of order p^k for some fixed k

chilly radish Jul 11, 2021, 9:22 PM

#

we've done all of the standard results in my group theory class

quaint tree Jul 11, 2021, 9:22 PM

#

chilly radish I've had enough of those problems for the rest of my life

I dunno, these can be kind of fun, at least the first dozen or so.

chilly radish Jul 11, 2021, 9:22 PM

#

The first dozen or so is the keyword

oblique river Jul 11, 2021, 9:22 PM

#

for example, for p > 3, the number of groups of order p^6 is 3p^2 + 39p + 344 + 24 gcd(p - 1,3) + 11 gcd(p-1,4) + 2 gcd(p-1,5)

quaint tree Jul 11, 2021, 9:23 PM

#

oblique river for example, for p > 3, the number of groups of order p^6 is 3p^2 + 39p + 344 + ...

How do you prove that. That's more advanced than what my "Abstract Algebra I" course covered.

oblique river Jul 11, 2021, 9:23 PM

#

no idea I just read it on groupprops

#

haha

quaint tree Jul 11, 2021, 9:24 PM

#

Oh yeah that's the other method of finding all groups of a given order

#

The most powerful method mathematicians have

#

Look it up on the internet

#

And hope someone's already done it

oblique river Jul 11, 2021, 9:24 PM

#

https://www.jstor.org/stable/2006106

A complete list of the groups of order $p^6$, where $p$ denotes an odd prime number, is given using P. Hall's concept of isoclinism.

#

there's the paper

chilly radish Jul 11, 2021, 9:28 PM

#

On an unrelated note, i'm currently trying to prove that $F_2$ contains $F_n$ for all $n\in \mathbb N$. I have the subgroups
$$H_n = \langle {b^iab^{-i}: i \in [n]}}\rangle$$
I want to prove that this is subgroup is free. I already have a surjective homomorphism $f:F_n\rightarrow H_n$ from the universal property of the free group. I'm wondering if it would be easier to prove injectivity of this homomorphism, or try to show that $H_n$ satisfies the universal property itself. Or maybe show that every word in the generators of $H_n$ has a unique reduced form and from there show injectivity

cloud walrusBOT Jul 11, 2021, 9:31 PM

#

ShiN
Compile Error! Click the errors reaction for more information.
(You may edit your message to recompile.)

chilly ocean Jul 11, 2021, 9:33 PM

#

\bN

oblique river Jul 11, 2021, 9:34 PM

#

beez nuts

chilly radish Jul 11, 2021, 9:34 PM

#

chilly ocean \bN

?

oblique river Jul 11, 2021, 9:43 PM

#

That’s why youre getting a compile error

#

\N is not a default command

chilly radish Jul 11, 2021, 9:45 PM

#

I changed it but it didn't recompile

oblique river Jul 11, 2021, 9:45 PM

#

Do you not need braces around the N?

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It should recompile if you edit

chilly radish Jul 11, 2021, 9:47 PM

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oblique river Do you not need braces around the N?

No

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If it's a single char latex will just take it as input

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I think it timed out on the editing or smth

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I omit braces all the time, saves a lot of time typing

oblique river Jul 11, 2021, 9:48 PM

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Oh I thought that there might be an issue with spacing

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I do things like 2^4 without braces

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And single-character subscripts

chilly radish Jul 11, 2021, 9:49 PM

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Nah latex doesn't care about spaces

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Actually not doing a space might make it think it's a different command

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And cause an error

oblique river Jul 11, 2021, 9:50 PM

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That’s right

chilly radish Jul 11, 2021, 9:52 PM

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cloud walrus **ShiN** Compile Error! Click the <:errors:504266556054962196> reaction for more...

Anyways can anyone help

barren sierra Jul 11, 2021, 10:32 PM

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The answer is I think yes but I can't think of an example. Can you have a homomorphism that doesn't preserve commutativity? So let's say we have groups G and H, G abelian. Now say we have a homomorphism f: G -> H. Can we have f(x + y) = f(x) + f(y) and f(y + x) = f(y) + f(x) and x + y = y + x but f(x) + f(y) =/= f(y) + f(x)???

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Also this would never be the case for isomorphisms right?

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Due to isomorphisms being bijections, specifically due to then being injective.

chilly ocean Jul 11, 2021, 10:38 PM

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f(x) + f(y) = f(x + y) = f(y + x) = f(y) + f(x)

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@barren sierra

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this has nothing to do with f being injective

chilly radish Jul 11, 2021, 10:42 PM

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the image of an abelian group under homomorphism must be abelian

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because of what tterra said

barren sierra Jul 11, 2021, 10:49 PM

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Oh

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Yes duh 💀

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Idk what I was thinking

chilly ocean Jul 11, 2021, 10:50 PM

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you probably mixed up well defined (a = b implies f(a) = f(b)) with injective (f(a) = f(b) implies a = b)

barren sierra Jul 12, 2021, 12:02 AM

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Yea

thorny flame Jul 12, 2021, 4:21 PM

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Can 2 infinite sub groups be isomorphic?
Can for example a subgroup of Q/Z be isomorphic to Z?

kind temple Jul 12, 2021, 4:23 PM

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thorny flame Can 2 infinite sub groups be isomorphic? Can for example a subgroup of Q/Z be is...

any non-trivial subgroup of (Z,+) is isomorphic to (Z,+)

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so yes. just gave you countably many instances

thorny flame Jul 12, 2021, 4:24 PM

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right but that's due to them having the same infinite

kind temple Jul 12, 2021, 4:24 PM

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due to them being infinite and cyclic

thorny flame Jul 12, 2021, 4:25 PM

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what if you have 2 groups A and B
one is |A| = aleph_0; |B| = Aleph_3

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is A isomorphic to B?

kind temple Jul 12, 2021, 4:25 PM

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ok then they can’t be isomorphic

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there literally isn’t a bijection on the level of sets

thorny flame Jul 12, 2021, 4:25 PM

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I see

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|Q| = aleph_1
|Z| = aleph_0
ASFAIK

kind temple Jul 12, 2021, 4:26 PM

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?

thorny flame Jul 12, 2021, 4:26 PM

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Z is the intergers

kind temple Jul 12, 2021, 4:26 PM

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Q and Z both have the same cardinality

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|R| is aleph_1

thorny flame Jul 12, 2021, 4:27 PM

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Oh...

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so what is aleph_0?

kind temple Jul 12, 2021, 4:27 PM

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|N|

hollow imp Jul 12, 2021, 4:27 PM

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Countable

thorny flame Jul 12, 2021, 4:29 PM

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ohh

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So how do I prove that there's no subgroup of Q/Z that is isomorphic to Z?

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multiplication and sum

chilly ocean Jul 12, 2021, 4:37 PM

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R and R^2 are isomorphic as groups and these two are isomorphic to those, so this doesn't work. a better example is Z and Z^2.

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don't worry it's not intuitive at all why they should be isomorphic KEK

kind temple Jul 12, 2021, 4:37 PM

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huh what the

chilly ocean Jul 12, 2021, 4:37 PM

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axiom of choice fuckery

kind temple Jul 12, 2021, 4:38 PM

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dude u gotta be joking rn

south storm Jul 12, 2021, 4:38 PM

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If you want to prove that (Z,+) cannot be embedded into (Q,+)/Z you can use that Z is a torsion free group and that Q/Z is a torsion group. That means all elements of Q/Z have finite order and all but the identity of Z have infinite order.

chilly ocean Jul 12, 2021, 4:38 PM

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kind temple dude u gotta be joking rn

https://math.stackexchange.com/questions/1744066/are-bbbr-and-bbbr2-isomorphic-what-about-bbbz-and-bbbz2

Mathematics Stack Exchange

Are $\Bbb{R}$ and $\Bbb{R^2}$ isomorphic? What about $\Bbb{Z}$ and ...

I'm asked to prove or disprove whether there are any isomorphisms between $\Bbb{R}$ and $\Bbb{R^2}$, as well as between $\Bbb{Z}$ and $\Bbb{Z^2}$.

It's all very confusing since I understand inject...

south storm Jul 12, 2021, 4:38 PM

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And order has to be preserved under embedding

kind temple Jul 12, 2021, 4:39 PM

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i quit

chilly ocean Jul 12, 2021, 4:39 PM

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lol

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math is eternal suffering

south storm Jul 12, 2021, 4:40 PM

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Are Q_p and Q_p^n under addition also isomorphic?

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thinkies

chilly ocean Jul 12, 2021, 4:40 PM

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ask merosity

kind temple Jul 12, 2021, 4:40 PM

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wait that’s R and R^2 as vector spaces over Q

chilly ocean Jul 12, 2021, 4:41 PM

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"in particular as additive groups"

urban ice Jul 12, 2021, 4:41 PM

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Hang on, isn't ℚ/ℤ isomorphic to [0,1) ∩ ℚ ?

south storm Jul 12, 2021, 4:42 PM

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I think so

urban ice Jul 12, 2021, 4:42 PM

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ok, good

south storm Jul 12, 2021, 4:42 PM

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(Mod 1)

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When adding I mean

south storm Jul 12, 2021, 4:43 PM

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kind temple wait that’s R and R^2 as vector spaces over Q

If for some field F one has that G over F and G‘ over F are linear isomorphic as vector spaces, this implies that their additive groups are isomorphic.

urban ice Jul 12, 2021, 4:43 PM

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ℤ/ℤ ≅ {0}
multiply by ℤ and thus ℤ ≅ {0}
easy 😎

south storm Jul 12, 2021, 4:44 PM

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Funny enough here this is actually valid if you take the coset appropriately

kind temple Jul 12, 2021, 4:45 PM

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south storm If for some field F one has that G over F and G‘ over F are linear isomorphic as...

yes i see that but i’m still not convinced that R over R and R^2 over R are isomorphic

south storm Jul 12, 2021, 4:45 PM

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Oh wait right lol

kind temple Jul 12, 2021, 4:45 PM

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as additive groups

chilly ocean Jul 12, 2021, 4:45 PM

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additive groups doesn't care about underlying field...

south storm Jul 12, 2021, 4:45 PM

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But it‘s over Q

kind temple Jul 12, 2021, 4:45 PM

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unless i’m o

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mehhh

south storm Jul 12, 2021, 4:46 PM

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With R there’s a dimension problem but the dimension of R over Q and R^2 over Q is the same, though infinite

kind temple Jul 12, 2021, 4:46 PM

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right

urban ice Jul 12, 2021, 4:47 PM

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sometimes i hate axiom of choice because
"that can't possibly be true, right?"
"nope... well, except with AC"
"huh... and so what does the counterexample look like?"
"no idea"

chilly ocean Jul 12, 2021, 4:47 PM

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stareFlushed

south storm Jul 12, 2021, 4:48 PM

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I mean there‘s a lot of counterintuitive things that are true, intuition isn’t perfect

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That‘s not only a thing about the axiom of choice

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Like that (R,+) and (R^2,+) are isomorphic is also counterintuitive

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catThimc

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Oh wait lmao

south storm Jul 12, 2021, 4:52 PM

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south storm Like that (R,+) and (R^2,+) are isomorphic is also counterintuitive

This follows from AC

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And isn’t provable in ZF

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https://mathoverflow.net/questions/25375/ac-in-group-isomorphism-between-r-and-r2

MathOverflow

AC in group isomorphism between R and R^2

Using the axiom of choice, one can show that $\mathbb{R}$ and $\mathbb{R}^2$ are isomorphic as additive groups. In particular, they are both vector spaces over $\mathbb{Q}$ and AC gives bases of t...

chilly ocean Jul 12, 2021, 4:53 PM

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cotteppa

south storm Jul 12, 2021, 4:53 PM

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cocatThink

urban ice Jul 12, 2021, 4:53 PM

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I was confused about how R was a vector space over Q, but an MO post rightfully noted that a basis can be uncountable

kind temple Jul 12, 2021, 4:54 PM

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south storm Like that (R,+) and (R^2,+) are isomorphic is also counterintuitive

this is now my least favorite thing in math

south storm Jul 12, 2021, 4:54 PM

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Why

chilly ocean Jul 12, 2021, 4:54 PM

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how could you not dislike this sully

south storm Jul 12, 2021, 4:54 PM

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I don’t dislike this

chilly ocean Jul 12, 2021, 4:54 PM

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this plays with your emotions

south storm Jul 12, 2021, 4:54 PM

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It‘s interesting that it is true

kind temple Jul 12, 2021, 4:54 PM

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it’s so gross

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like why are they the same

south storm Jul 12, 2021, 4:55 PM

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From my experience, if you look at the gross maths there‘s a lot of interesting there

chilly ocean Jul 12, 2021, 4:55 PM

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i am very familiar with gross math

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opencry

kind temple Jul 12, 2021, 4:55 PM

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sure it’s interesting and all, it just makes me wanna vomit

chilly ocean Jul 12, 2021, 4:55 PM

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(im joking)

south storm Jul 12, 2021, 4:55 PM

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I don’t find it gross tbh

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This is just that R and R^2 are bijective on steroids

kind temple Jul 12, 2021, 4:56 PM

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jk but idk. i just don’t like that R and R^2 are isomorphic

urban ice Jul 12, 2021, 4:56 PM

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Doesn't this also mean that (ℝ,+) and (ℝ^ℝ,+)

south storm Jul 12, 2021, 4:56 PM

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I think R^R has bigger cardinality

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Because the power set of R can be injected into it

urban ice Jul 12, 2021, 4:56 PM

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right

south storm Jul 12, 2021, 4:57 PM

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So they are not isomorphic

urban ice Jul 12, 2021, 4:57 PM

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thank god

south storm Jul 12, 2021, 4:57 PM

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I think it follows though that (R^R,+) and ((R^n)^(R^n),+) are isomorphic

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hyperhonk

kind temple Jul 12, 2021, 4:57 PM

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if X is countable is R^X still in bijection with R?

south storm Jul 12, 2021, 4:57 PM

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Yes I think so

kind temple Jul 12, 2021, 4:58 PM

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aw bruh

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R^N and R are isomorphic then (just assuming, have no idea)

south storm Jul 12, 2021, 4:58 PM

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I‘m not sure whether that is, I guess some nonstandard analyst would know

urban ice Jul 12, 2021, 5:00 PM

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let's pull out cardinal arithmetic

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yeah that works

hidden haven Jul 12, 2021, 5:01 PM

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For countable N yeah

south storm Jul 12, 2021, 5:01 PM

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I think it is true that AC is equivalent to that for nonempty X, X^n and X are bijective iff X is infinite

hidden haven Jul 12, 2021, 5:02 PM

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south storm I think it is true that AC is equivalent to that for nonempty X, X^n and X are b...

Nope

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|X^X| > |X| if |X| > 1

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You need bounds on the exponent

south storm Jul 12, 2021, 5:02 PM

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For n a finite ordinal bigger 0

hidden haven Jul 12, 2021, 5:02 PM

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The bound is exactly the cofinality of |X| assuming generalised continuum hypothesis

urban ice Jul 12, 2021, 5:04 PM

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|R^N|=|R|^|N|=(2^ℵ₀)^ℵ₀=2^(ℵ₀^2)=2^ℵ₀=|R|

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i have no idea what they did in the third case but whatever

hidden haven Jul 12, 2021, 5:07 PM

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It's very weirdly written

urban ice Jul 12, 2021, 5:07 PM

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This is kind of horrendous, but fun at the same time

hidden haven Jul 12, 2021, 5:08 PM

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urban ice i have no idea what they did in the third case but whatever

That's not supposed to be read as a sequence

south storm Jul 12, 2021, 5:08 PM

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hidden haven Nope

https://en.m.wikipedia.org/wiki/Tarski's_theorem_about_choice

Tarski%27s_theorem_about_choice

hidden haven Jul 12, 2021, 5:08 PM

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You separately so all of those

urban ice Jul 12, 2021, 5:08 PM

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Indeed, but I have no ideas how they got their equalities

south storm Jul 12, 2021, 5:08 PM

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And X bijective with X^2 trivially implies for all finite ordinals n>0 X bijective with X^n

urban ice Jul 12, 2021, 5:08 PM

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specifically c^ℵ₀ = ℵ₀^ℵ₀

hidden haven Jul 12, 2021, 5:09 PM

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south storm And X bijective with X^2 trivially implies for all finite ordinals n>0 X bijecti...

Yeah I was assuming n to be want ordinal

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Rather than finite

urban ice Jul 12, 2021, 5:09 PM

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my bad

south storm Jul 12, 2021, 5:09 PM

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hidden haven Yeah I was assuming n to be want ordinal

Yeah I said that it‘s finite here #groups-rings-fields message

hidden haven Jul 12, 2021, 5:10 PM

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urban ice specifically c^ℵ₀ = ℵ₀^ℵ₀

I see, mb

south storm Jul 12, 2021, 5:10 PM

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south storm https://en.m.wikipedia.org/wiki/Tarski%27s_theorem_about_choice

Also funny quote from here

urban ice Jul 12, 2021, 5:11 PM

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lmaooo

south storm Jul 12, 2021, 5:11 PM

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Lol

hidden haven Jul 12, 2021, 5:12 PM

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urban ice specifically c^ℵ₀ = ℵ₀^ℵ₀

You do pretty much what you did lol 2^N ≥ (2^N)^N ≥ c^N ≥ N^N ≥ 2^N

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Well the first 2 are just equalities so I didn't need to do the last 2

quaint tree Jul 12, 2021, 5:15 PM

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south storm Also funny quote from here

Clearly Frechet didn't live in an era with publish-or-perish