groups-rings-fields

[eager willow]

however, given that I am translating from a language I do not speak, and that the application of this lemma is only towards rings which are indeed freely generated by such x_i, this may be an assumption I or the original author failed to include

oops!

[cobalt heath]

Ah, freely generated. Hmm

Well I mean, x_n x_m = 2 x_{min(n, m)} already means it is not freely generated tho

[eager willow]

sorry, as an abelian group it is freely generated

[cobalt heath]

Ah, as an abelian group. Duh me not catching that

[crystal vale]

Fix m, n ∈ N. Consider the map f : Z × Z → Z defined by f(x, y) = mx + ny.
Identify the image and kernel of f.

The image will be multiple of gcd(m,n), right?

But I am not sure about the kernel

[cobalt heath]

I guess proof by support argument might work then.

[eager willow]

yes, it's the multiplies of gcd by bezout's identity.

[crystal vale]

Thank you, what about the kernel?

[eager willow]

assuming m and n, x,y are all nonzero then mx + ny = 0 implies y/x = -m/n, so they're lattice points of a fixed rational slope y/x =-m/n

for example if m is divisible by n then y is divisible by x, but there's not too much else to say in general. Maybe you could avoid the cases of things being zero by saying the vector dot product mx + ny = (x, y)*(m, n) = 0 so the vector (x, y) is orthogonal to the vector (m, n)

[crystal vale]

So how can I write a kernel in this case

[cobalt heath]

Uhh about that.. 💀
I guess it depends on how clean you want it

[crystal vale]

How I express this relation x/y = -n/m ?

[cobalt heath]

Idk, will you get penalty if you wrote like, {(x, y) \in Z * Z | mx + ny = 0}

[crystal vale]

But then it will be obvious they expect more specific I think, maybe this one is correct

[cobalt heath]

Yeah, true. But then, I am at loss of ideas

Maybe cancel out gcd(m, n)? idk

[eager willow]

it's Z generated by the vector (-n/gcd(m,n), m/gcd(m,n))

[cobalt heath]

Oh

[crystal vale]

Okay, thank you

[crystal vale]

Z generated means cyclic group generated by (-n/gcd(m,n), m/gcd(m,n) ) ?

[eager willow]

yes

[crystal vale]

How do you get it from x/y = -n/m ?

[eager willow]

which integer points fall on the line (-nt, mt) in R^2?

[crystal vale]

If t = p/q then q must be divide n and m and t cannot be irrational

[wraith swan]

Got an exercise to find any isomorphism G -> G x H with non trivial groups. But since G x H has a piecewise operation wouldn't the element in H that gets picked be independent of the one in G? And since order should be preserved... Wait is this even possible for finite groups? Should i be looking at infinite groups?

Well of course as I wouldn't have a bijection between the underlying sets

[winged void]

I guess the total must be 66

Is it possible if I send it to you to check it if you have time

[rocky cloak]

That seems right

[winged void]

Sure but how did you check it so fast

And you did not even look to what I did

[winged void]

I was cheking my answer

i guess its more than 66 actually i change my word

but not so sure

i will send what i have

give me a second

[rocky cloak]

Well, I just calculated it for myself and got 66, so it seemed right

[winged void]

this is what i have

[winged void]

this is what i did

there is only one thing that im not sure about i multiplied by (ab)(cd) with half

[rocky cloak]

Yeah, that looks good

[winged void]

but i did not do that with (abc)(de)

so im not really sure if i should also there

cuz if i did then i get 56

[rocky cloak]

Well the reason you divide by 2 is because (ab)(cd) = (cd)(ab)
But there is no such overcounting for (abc)(de)

[winged void]

sure but (abc)(de) = (de)(abc)

right

so why here i do not divide with 2

[rocky cloak]

Yes, but you never count the second case

You only look at things of the form (abc)(de)

[winged void]

ah i see

but i mean when i say 2c2

i do count it in a way

or i do not really get what you exactly mean

because like in the case (ab)(cd) i did 5c2 * 3c2

[tribal moss]

That overcounts, because you get (12)(34) and (34)(12) counted one each.

[winged void]

sure but that is a similar case

in (abc)(de)

i mean (123)(45)= (45)(123)

or am i wrong

[tribal moss]

The notation "(45)(123)" does not look like "(abc)(de)".

So it wouldn't get counted in the first place, and therefore there's no overcounting to correct for.

[winged void]

i know but i mean in permutation since the support is disjunct then there will be no problem if you did (abc)(de) or (de)(abc)

[tribal moss]

Sure, if you really want to, you can do somthing like "count all of the ways to write (ab)(cde), then also count all of the ways to write (abc)(de), then add them together, and then divide by 2 to correct for overcounting". That would just be really roundabout for no good reason.

[winged void]

i know i get that but i mean did not i count them both when i said 5c3 * 3c2

[tribal moss]

No?

(Um, what exactly is it you're counting with "5c3 * 3c2" and how would that work?)

[winged void]

there are 5 elements in S5 im to make my two cycles in the form of (abc)(de). I will make the first cycle 5c3 then the second 3c2

so im not overcounting actually btw

[tribal moss]

I don't get what "3c2" is doing there.

[winged void]

i mean 2c2

sorry

[still locust]

what s the problem?

[winged void]

my mistake

[tribal moss]

Actually (5 choose 3)(2 choose 2) undercounts, because you need to make a choice of which way to cycle your three elements.

[winged void]

yes i mean * 2

for sure because (abc) and (acb)

[tribal moss]

Counting homomorphisms from C6 into S5.

[rocky cloak]

What you're counting is ways to assign numbers to a, b, c, d, e. There's no way to pick a, b, c, d and e such that
(abc)(de) Looks like (12)(345)

[still locust]

oh ok

[winged void]

oh that is true

oh i see what you mean

you can compare with making a team

like make a team of 10 from 5

10c5 /2

because you did not specify which team first

[still locust]

you just had to find element with order = 6 no ?

and you take the application 1 -> this element

idk

there are only (a b c) (d e) as you said if you look the decomposition of a permutation and to count it idk

i think you should just count the first (a b c) because the (d e) are fixed if you have the (a b c)

and there are 5* 4* 3/3=20

oh you had to watch the element whose order divide 6 also

[winged void]

that is true

agree

[crystal vale]

Any hint?

[still locust]

only 1 for identity , 5* 4/ 2 = 10 for (a b), 5* 4* 3 / 3 = 20 for (a b c), also 20 for (a b c) (d e) (because it is in bijection with (a b c)) , and for (a b) ( c d) (you have 5* 4 / 2 * ( 3 * 2 ) / 2 =30 choice for ( (a b) , (c d)) so 15 choice for {(a b), (c d)} in bijection with (a b)(c d))

the last one is hard to put in bijection with smth

[tribal moss]

Is that even necessarily true if G is infinite? I think this would be a counterexample: the free group on two generators, and the automorphism that interchanges the generators.

On the other hand, if G is known to be finite, it is enough to show that x \mapsto xf(x^-1) is injective, which is straightforward.

[still locust]

i think the sum of the generator is fixed (for your counterexample)

[tribal moss]

The free group is not abelian, so f(ab)=ba is not a fixed point.

Simpler counterexample, though: (Z,+) and negation. :-D

[still locust]

ah yeah you're right

[rocky cloak]

The best exercises are those where you have to prove something that's false /s

[wraith swan]

When someone only writes "Q", the group of rational numbers, do they mean the additive group?

[tardy hedge]

A homomorphism between groups G1 , G2 only illuminates structure of G1 if the map has nonzero kernel right?

[rocky cloak]

Typically yes, but could vary by context

[wraith swan]

Ok thank you

[tough raven]

Let R = Z[X1, X2, …] and f(p) := Σ_{m =/= n} min(m,n) p_m,n + Σ_m 2 m p_m,m where p_m,n = coefficient of X_m X_n in p.
Then the hypothesis is true but not the conclusion.

[south patrol]

Q can be viewed in many ways

But the usual multiplication doesn't turn it into a group

(As 0 has no inverse)

[eager willow]

Hey thanks! I've been assuming a different hypothesis now, because the Russian author is really using the lemma in far less generality than what my translation suggests. It is useful in a calculation of a ring of modular representations, actually.

here is a much much weaker version of the lemma, which applies to the scenario that the author was using, and which is potentially correct now!

the classic Russian style is of course to leave the actual hypotheses of the lemma as an exercise to the reader

[still locust]

If you take R= Z[X] without the constant, X^i are a Z basis ?

[eager willow]

yes, I would consider that a ring with a Z linear basis

[still locust]

okay

[eager willow]

wait you're right the x_i together with 1 should be a Z linear basis

[still locust]

it is hard to find a function f i think with this basis it is not possible (X^4 = X * X^3 = X^2 * X^2) and there is a contradiction

[eager willow]

yep, it's not gonna be well defined. Really you should have in mind quotients R = Z [X1, X2, ...] by relations in the form XiXj = sum ci Xi

e.g. Z[X] = Z[X1, X2, ....] / (XiXj - X(i+j)) is an incompatible set of relations with the hypothesis

[tardy hedge]

This true? Basic question yeah but im tryna review fundamentals

[agile burrow]

Not necessarily. If the map has zero kernel then the image is isomorphic to G1. So if subgroups of G2 satisfy some property, then G1 satisfies it as well. For example, if G2 is abelian then so is G1.

Idk if that's the kind of answer you're looking for though

[still locust]

you take Xi*Xj = 2X_(min(i,j)) ?

[tardy hedge]

It is! Thank you for that!

[eager willow]

this is an example, yes. Note that if you believe the lemma, the conclusions then imply f(Xi) = i.

Another example is to quotient by only by XiXj = 2X(min(i, j)) when i =/=j or when i = j > 1, as well as the relation X1^2 = X2

[tardy hedge]

Oh. Sort of

Subgroups of G aren’t necessarily related to the overall structure of G in some way right?

I have more thinking to do but

But normal subgroups are^?

Normal subgroups are somehow related to the overall structure of G but just any old subgroup is not

[eager willow]

is there a native speaker who might confirm that my original translation is correct and it's simply not a truthful theorem?

oh wait I think that f(xs) = s was something I forgot to include! This might mean in turn that the lemma is correct, as it excludes the counterexample of @tough raven

[eager willow]

there's this word образующими that feels ambiguous to whether or not these are generators for a ring ( without unit? ) or a basis for an abelian group

[tardy hedge]

Hi

[hidden haven]

Hello how's your day been

[dull ginkgo]

is it sufficient to state that (x_k) and (e_n,m) together generate both rings

[eager willow]

Careful. What are (x_k) in the left hand side ring? All elements on the left hand side are matrices, all elements on the right hand side are polynomials where matrices are coefficients

[dull ginkgo]

Both are in some manner of speaking semigroup rings (the former is a quotient though) so the generator sets commute. There are natural injections from K into M_n(K), K into K[x_1...x_n]

[eager willow]

Maybe just write an explicit isomorphism using how the rings are defined

[dull ginkgo]

Explicitly stating it is basically using that

it fries my bit to say isomorphism here because they seem literally equal when trying to describe them so you have to be a bit careful

[eager willow]

They are not literally equal. One construction has elements as matrices and one has elements as polynomials

[dull ginkgo]

I know lol, but the matrix ring is a finitely generated module/algebra over R, generated by matricies who are 1 in only a single entry

e_n,m is the matrix that is all 0 except for a 1 at (n,m)

and R embeds into it via the sum over diagonal entries

I guess you'd have to like, pass the generators through the injections lol

By the constructions of the two "functorial" ring extensions they don't really mingle too much and give an iso

Actually I can probably do a shortcut

[tribal moss]

Hmm, of course writing it down explicitly will work, but could we g.e. show something like each of them is iso to the tensor product of M(R) and R[x1,...,xr]?

[dull ginkgo]

M_n(R) maps into M_n(R[x_1...x_n])
Extend that monomorphism to a map from M_n(R)[x_1...x_n] to M_n(R[x_1...x_n]) sending x_k to it's respective generator in M_n(R[x_1...x_n])

This map is both injective and surjective and gives an explicit isomorphism

[eager willow]

yep, it should be easy enough to show for a two sided R algebra S that M(S) = M(R)(x) S and that S[x1, ... , xn] = R[x1, ... , xn] (x)S

[dull ginkgo]

though I know the tensor product I haven't gotten that far in jacobson

so I want to use the provided assumptions

[eager willow]

wait is R commutative? otherwise the tensor thing might be annoying to work with

[dull ginkgo]

R is not commutative no

just realized that

[eager willow]

yeah probably avoid the tensor product then. Might still work but probably harder to deal with.

[dull ginkgo]

So that universal property doesn't work too nicely

[tribal moss]

Ah shoot then.

[dull ginkgo]

The generator route is rather easily due to generator set commutivity and functor shit

the problem is idk how to state the noncommutative polynomial ring universal property

Probably for maps R into K, then for any x in R's ring centralizer in K, then there's an extension from R[X]

[tribal moss]

Hmm, you'd want the images of x to be in the center of K, I expect.

(Or that centralizer will work too)

[eager willow]

Just send a matrix (pij) of polynomials to the sum of eijpij, and a polynomial sum M_i x^i to the matrix M_ix^i treating x^i as a scalar. These are clearly inverse morphisms to one another

[dull ginkgo]

that's basically what I'm doing above

just skipping over the "morphisms of generator" crap

[dull ginkgo]

trying to process if the centralizer would be a subring

[Abel subgroup by default lmao]
abx = axb = xab
0x = x0 = 0
1x = x1 = 0

we good

UGGGHHHH

[tribal moss]

I'm with datorangeguy: using the concrete representations of matrix and polynomial rings we already know and love will be much easier than juggling universal properties here.

[dull ginkgo]

yeah I just thought I could shortcut it

[kind temple]

what about induction on r?
it seems like it wouldn't be too difficult to show the case when r = 1.
for the inductive step, we have
M_n(R)[x1,...,xr,xr+1] = (M_n(R)[x1,...,xr])[xr+1] = M_n(R[x1,...,xr])[xr+1] =? M_n(R[x1,...,xr,xr+1])

[dull ginkgo]

turns out proving it's an iso from the universal property just does the same thing anyway

[dull ginkgo]

God

is it acceptable for me to skip like a third of the verification bullshit by saying it's an ass algebra over the direct sum

[dull ginkgo]

should I do all of these

[edgy jackal]

is there a meaningful difference between saying a polynomial splits in a field vs a polynomial is reducible over a field?

[tough raven]

IG you want to add 1_R to the basis, apart from the x_i?

[tough raven]

Yes — X^3 - 1 is reducible over R (the field of real numbers) but does not split over R.

[dull ginkgo]

Splitting is being “fully” reducible basically

[edgy jackal]

yes i think i see it now

thanks!

i have to understand the galois correspondence and its consequences by wednesday

as you can see im quite a ways away 😭

[tough raven]

How is (x_i) a basis when x_0 = 0 (I don't know any Russian to be clear) 💀?

[tough raven]

My example is excluded by the (x_i) being a Z-linear basis rather than generating R as a ring, which makes a world of difference.

[eager willow]

No even if the Russian is saying generated as a ring, it also includes specifically f(xs)= s in addition to f(xnxm)=2min(n,m).

here is the current version I am trying to prove, still having trouble

the extent of the inductive argument I've puzzled out so far

I'm including some other stuff like the coefficients being positive and a base case x1x2 = 2x1, because this is known of the rings I am concerned with.

[eager willow]

Wait I think I got it!

[tough raven]

You in fact get 4 = Σ_{s >= n} 2 c_s, right? (Strengthening your inductively proved conclusion to x_m x_n = 2 x_min(m,n) even for m = n.)

[eager willow]

my base case is m = 0, so no, I can't divide by m

[tough raven]

By the linear independence of the x_i.

[eager willow]

I can't resolve the x_m^2 term

tempting though!

[tough raven]

.

Is x_n^2 = 2x_n known to be false in your ring?

[eager willow]

sometimes!

You can check that we can just set x_1^2 = x_2 but x_ix_j = 2x_min(i, j) in all other cases and you get a ring satisfying all of the hypotheses

and in fact this ring occurs in nature! It is a certain subring of representations of the Klein 4 group in characteristic 2

[tough raven]

It satisfies f(x_i x_j) = 2 x_min(i,j) even when i = j?
Because I think you have used that in your first displayed equation.

Second*

[eager willow]

yes this is true of all i, j

[tough raven]

OK.

[eager willow]

problem is I can't divide by m in that second equation

and that should be the scalar 2min(i, j) not 2x_min

[tough raven]

Well, the "at most two values" part is also wrong then (for m = 0).

[tough raven]

Also please continue, don't let me stop you.

[eager willow]

hmm, that's a good catch

[tough raven]

Can you add m = 1 to your hypotheses i.e. x_1 x_n = 2x_1 for n >= 1? Or just n = 2?

[eager willow]

I did add this for n = 2. This is a big problem to try to add for all n though

[edgy jackal]

could i bounce an idea i have for a proof off of one of yall so i know if im at least headed in the right direction
the statement is that for F a field, there is no sequence of nontrivial ideals $I_n \subset I_{n+1} \subset \cdots \subset F[x]$. My idea is this: $F$ is a field, so $F[x]$ is a PID. Then take some ideal $I = <f(x)> \subset F[x]$. If $I$ is to be proper, we know $f(x)$ can't be a unit, but I further claim that this polynomial must in fact be irreducible, and that this fact should further imply that $I$ has no proper subideals.
is this the right approach? or should i look elsewhere

[eager willow]

Again, it's a ring of representations. I can't easily compute infinitely many tensor (Kronecker) products

But I can do an arbitrary amount of computations for as strong of a base case as needed

[cloud walrus]

Leninade

[eager willow]

are you trying to prove the Hilbert basis theorem?

[edgy jackal]

ngl chief ive never heard of that

[tough raven]

The statement is not true: consider (x^n) contained in (x^{n-1}) contained in … contained in (x) contained in F[X].

Oh, n is inside n+1.

Yeah, this is basically the conclusion of the Hilbert Basis Theorem.

But you can prove it much more easily in this case.

[eager willow]

that's not an ascending chain, it's true, although the statement should be more like 'there is no strictly increasing chain of ideals' or 'every ascending chain eventually becomes constant'

[tough raven]

f need not be irreducible, but it does have a prime factorisation.
Can you use that to bound the number of ideals which contain I = ‹f› as a subset of them?

[eager willow]

I think you have the right idea: let I be an ideal. It is generated by f. If J is generated by g, then I < J is the same thing as f being an element of J i.e. f is divisible by g

||but f is of finite degree||

[edgy jackal]

i like these ideas. thanks yall!

[tough raven]

I think the situation of the last paragraph forces x_1^2 = 2 x_1: if x_1^2 = x_2 then f(x_n x_1^2) = 4 (n >= 2) whereas x_n x_1^2 = 2x_1.

[dull ginkgo]

let me check with the chat, this is a corolllary of Hilbert Basis Theorem right?

[eager willow]

that's what I was looking at actually. I know again from modular representation theory a specific ring in which x_1^2 = x_2, but my casework toward x_nx_1 = x_2 contradicts this

[tough raven]

It more or less is the Hilbert Basis Theorem for F[X], F a field.

[dull ginkgo]

Didn't want to explicitly say it was Noetherian

[eager willow]

this is why the inductive step is proven for m = 0

[dull ginkgo]

what can we say if (f(x)) is strictly contained in (g(x))?

||If we have a strictly increasing sequence of ideals I_n in F[X], then each is generated by a minimal polynomial f_n(X). if (f_n(X)) lies in (f_{n+1}(X)) then we have f_n(X) | f_{n + 1}(X). Thus the degrees are strictly decreasing, but they must terminate at some point as the degrees must be positive by infinite descent.||

[dull ginkgo]

The only way off the top of my head how to prove hilbert basis theorem is to show that leading coefficient map is an ideal-mapping

[tough raven]

This is the special case that F is a field, so we can just use the fact that F[X] is a PID.

[dull ginkgo]

The spoiler'd text is what my proof would be for the field case which is significantly easier

[tough raven]

You have used the fact that every ideal is principal… it's the same proof…

[dull ginkgo]

yes I know

[tough raven]

Or close enough.

Cool

[dull ginkgo]

i'm just saying idk how hilbert basis is generally proven

would it be "cheating" to basically just look up a table of quadratic residues here

[tough raven]

Depending on the context, maybe?

[dull ginkgo]

idk how else I'd prove it here without trying to find the quadratic residues by hand

[tough raven]

if this is meant to motivate techniques for computing quadratic residues, say.

[dull ginkgo]

this is the whole problem

[tough raven]

Well, p = 2 isn't about quadratic residues

Unless it is…

Wait these aren't necessarily field extensions

[dull ginkgo]

no just to shortcut the isomorphism between rings

that's basically checking if it splits in one field, and not the other, there is no isomorphism

[tough raven]

I don't think it's one line to get to the point of comparing quadratic residues. I would guess you weren't intended to look them up, but if you want to, why not?

[dull ginkgo]

i want to do it the way the author intended

[kind temple]

if i'm remembering this correctly, at least for p = 2, we have that x^2 - 2 = x^2, and x^2 - 3 = x^2 - 1, so R_1 is modding out by an irreducible while R_2 is modding out by something that splits

[south patrol]

it shouldn't be bad by quadratic reciprocity and the formula for (2/p)

[dull ginkgo]

that's a question down the line

[kind temple]

i think i could be horribly misremembering though

[south patrol]

Actually lol what I just said was much too much generality

[dull ginkgo]

p = 2:
X^2 - 2 = X^2
X^2 - 3 = X^2 + 1 = (X + 1)^2,

[south patrol]

You can just compute yeah when they are quadratic residues and if you find that tedious ig just google but should be very quick since p <= 11

[dull ginkgo]

Actually i wonder if the isomorphism comes from this

X <-> X - 1?

wait I wonder if R[X]/(f(X)) ~= R[X]/(f(X + a))

More general question

Let R be a ring, a be an automorphism of R, and J be an ideal. Then is R/J ~= R/a(J)

you'd think a*(x + J) = a(x) + a(J) literally causes no issue

[eager willow]

well yeah, what's the kernel of the composition R -> R -> R / a(J) where the first map is a?

[dull ginkgo]

cool that proves for p = 2 the iso literally is x -> x + 1

[eager willow]

frobenius moment

[dull ginkgo]

thank you characteristic 2 frobenius endomorphism

the only time you don't make me sick

For p = 5:

[cobalt heath]

How is x -> x+1 an iso

[eager willow]

The inverse is x -> x - 1

it's a homomorphism because it is uniquely defined on F[x] and descends to the quotient

[dull ginkgo]

well, x in F_2[X]/(X^2) vs F_2[X]/((X + 1)^2)

[cobalt heath]

Oh, sorry, I was missing the context

Yeah, it's iso on F[X]

[dull ginkgo]

Anyway for p = 5, 2 * 3 = 6 = 1 (mod 5)

so X^2 - 3 = X^2 - 2^-1 basically

[eager willow]

yeah that's the ring element x, not like f(x) = x+1 forall x

[dull ginkgo]

So if X^2 - 2 splits, so does X^2 - 3

and the automorphism should come from there i'd assume?

[cobalt heath]

Uh, why?

[eager willow]

this is characteristic 2

it's an automorphism

[dull ginkgo]

har 5

*char 5

this time

[eager willow]

pwned 😦

[dull ginkgo]

2 * 3 = 6 = 1 mod 5 so 3 and 2 are inverses mod 5

So if 2 is a quadratic residue, so is 3

[cobalt heath]

Hm I don't think X^2 - a splits iff X^2 - a^(-1) splits

[cobalt heath]

Oh

Quadratic residue is so hard

[dull ginkgo]

if (X^2 - a) = (X - b)(X + b) then (X^2 - a^-1) = (X - b^-1)(X + b^-1)

so uh ye

[cobalt heath]

Yeah that works

Idk what I was thinking

[dull ginkgo]

If it's irreducible then both rings are iso to F_25

[cobalt heath]

I guess p=11 case is where it is not isomorphic

[dull ginkgo]

oh my god

5^2 = 25 = 2*11 + 3

that was easy

uhhh for 2

[cobalt heath]

Time to memorize quadratic residue formulas

[dull ginkgo]

i don't think 2 is a quadratic residue mod 11

wait

yeah it's not a quadratic residue

so there's no iso

[cobalt heath]

Ye

[south patrol]

yeah (2/p) is 1 if p = +/-1 mod 8 and 0 otherwise

well okay excepting (2/2) lol

[cobalt heath]

Do you memorize this stuff?

[delicate bloom]

that with (-1/p) and QR

[dull ginkgo]

The iso explicitly between the case of p = 5 is a + bu -> a + 3bu i think

sending u to 1/u = 3u

[cobalt heath]

I guess so

[south patrol]

Well like

Not intentionally

It's just very common and I've had various number theory courses where i've needed it

[dull ginkgo]

"OH BOY TIME TO PULL OUT A BUNCH OF CRITERIA"
2 elements

[south patrol]

(Z/(2))[x] lol

notation moment

[dull ginkgo]

horror

[south patrol]

wel the second has a typo anyway lel

[dull ginkgo]

we like to ignore the typos because there's no errata

[cobalt heath]

Wow, you manage to memorize by osmosis

Nowadays I often think if I should not be mathematician

[south patrol]

No I mean usually i find this stuff p hard to remember lol

But doing enough diophantine equations has drilled it in ig

It's also something you can easily check by hand once you know it's a mod 8 thing

[cobalt heath]

Ah, are you going to research adjacent to NT?

[south patrol]

like 2 is a square mod 7 (3^2) and mod 17 (6^2) but not mod 5 or mod 3

[dull ginkgo]

eugh

[south patrol]

No but I've enjoyed what I've done thus far

Would be very cool if i could incorporate it somehow but i'm not very good at it rn lol

[dull ginkgo]

Is the easiest way to show that it's a field of eight elements as follows:

every element of that quotient has the (remainder) form ax^2 + bx + c. If the quotient is not mono, then we have ax^2 + bx + c = 0, thus contradiction of irreducibility :p

[cobalt heath]

Hmmm

[cobalt heath]

Are there people who accel at these stuffs, even more than you?

[dull ginkgo]

GOOD ENOUGH

[delicate bloom]

just plug in 0 and 1, no root -> no linear factor so irreducible cuz cubic

[dull ginkgo]

i just did that lol

so it's irreducible, thus {x^2, x, 1} form basis of finitely generated module thus 2^3 = 8 tldr

[south patrol]

Definitely

Well there are people for whom number theory is the main interest

[cobalt heath]

Field extension is a vector space shenanigans

[dull ginkgo]

I just smacked down "integral" and |F_2|^|{1, x, x^2}| = 8

[cobalt heath]

Woah.
Hmm, may I ask what is your main interest?

[dull ginkgo]

technically integral element schenanigans

[south patrol]

Well

both

:chad:

[south patrol]

Ig homotopy theory

i would like to incorporate some AG into it

[cobalt heath]

Homotopy..theory? Sounds incredibly difficult

[south patrol]

hbu

[crystal turtle]

[dull ginkgo]

Oh boy time to try to find irreducible quadratic and cubic mod 5 please kill me

actually I already have 2 is a quadratic non-residue so x^2 - 2 is irreducible, for the cubic I can use cardano form

[south patrol]

uh think like some bits of algebraic topology focused on homotopy-y stuff, and "higher" category theory

[crystal turtle]

it's not that hard since you just need to find ones without roots

[south patrol]

but hoenstly idk fully what i will do for phd yet lel

[crystal turtle]

homotopical algebra and higher category theory

[cobalt heath]

Impressive, I envy you for being able to do these stuffs

[dull ginkgo]

X^3 - aX - b irreducible mod 5 is gonna be a bit sketchy.

the ponder

[dull ginkgo]

i mean lmao yeah that's what I meant

[cobalt heath]

My English is lacking, what does this mean?

[south patrol]

"how about you?"

[cobalt heath]

Oh

[chilly ocean]

"Huge Big Ultraproducts"

[barren sierra]

what year are you in?

[cobalt heath]

I wish I had (rather, allowed to have) some interest in some stuffs, but I fear I have to decide from asking for advisor to multiple profs and see who accepts me..

[south patrol]

masters, starting le phd in sept/oct heh

hbu

[barren sierra]

oh cool

yea starting PhD in August myself

am currently senior

[cobalt heath]

Sorry about this random rant that might be meddling you..

[south patrol]

nice

[south patrol]

Wdym sorry? Like you're keeping options open for nor?

[cobalt heath]

It was kind of a rant. Also yeah, I am keeping my options open.

It feels like if I learn more, I would be locking myself to that branch.

[delicate bloom]

here's a trick, use the fact that the multiplicative group of F_q has q-1 elements. Now throw a cyclotomic polynomial at it

[dull ginkgo]

Oh shit... should've done that

I found X^2 + 2 and X^3 + X + 1 lmao

[delicate bloom]

idk in practice that might not be so direct haha

[dull ginkgo]

Two options here:

1. Verify by hand using x^3 - x = x(x + 1)(x - 1)
2. Use CRT to show Z_6 ~= Z_2 x Z_3, of which x^3 - x = 0 for any x in Z_2 OR Z_3

which is the least psycho

[delicate bloom]

2

[cobalt heath]

Just show by obviousness

I guess this is equivalent to showing x^3 = x (mod 6) always

[dull ginkgo]

....

HASN'T INTRODUCED DIRECT SUMS

OKIE DOKIE BOSSMAN

flip flops between the notations

like yeah they're equivalent for like, finite products/sums but

fuck man,

[cobalt heath]

Didn't they?

Ah, with groups you just use products

[dull ginkgo]

I don't think so, at least not a difference between direct products and sums

for neither groups nor rings

definitely not as products/coproducts though idk the coproduct for general rings and it is probably horrifying

[cobalt heath]

Hmmm.. dunno if there exists coproduct of rings

[dull ginkgo]

no one's grading this so I'm going to use funny a + v vector quaternion bullshit

makes it more intuitive.

wait any quaternion with no real and purely imaginary vector of norm 1 squares to -1

proof by: sphere

[cobalt heath]

Wdym

Ah, you mean squares to -1

[celest furnace]

That’s my boy

Hungry. Angry. Efficient at solving problems.

[dull ginkgo]

actually kinda stuck

[cobalt heath]

Casually invoke sth like Gauss lemma

[dull ginkgo]

for what, to show the latter poly is irreducible?

[cobalt heath]

Idk

But also, for me this kind of seems obvious - so that idk proof

[dull ginkgo]

You can use rational root theorem on the latter polynomial to show it's irreducible

it's just to show that it isn't principal

[glass zealot]

cant you just show that ax^3 + bx^2 + cx + d can't be a generator of the ideal

[dull ginkgo]

Actually

[glass zealot]

by comparing coeffients with 3 a lot should just vanish

[dull ginkgo]

a(x)b(x) = 3, which can only happen if a(x) and b(x) is constant due to degree rules for domains

[glass zealot]

not a cool way but surely works

[celest furnace]

Yeah this is it

The rest is just showing that the polynomial isn’t a multiple of 3 and that is clear

Same idea as the classic (2, x)

[glass zealot]

over Z you also get that one must be a unit, which doesn't leave much possibility

[dull ginkgo]

So summed up:

[south patrol]

lol

option 3

0^3 = 0

1^3 = 1

2^3 = 8 = 2

3^3 = 27 = 3

4^3 = 64 = 4

5^3 = 125 = 5

[celest furnace]

My maam knows her cubes

[dull ginkgo]

Assume (f(x)) = (3, p(x)) where p(x) is that cubic

1. There must be some a(x) : a(x)f(x) = 3, but deg(3) = 0, and deg(af) = deg(a) + deg(f) = 0 so deg(f) = 0, i.e f(x) = c for some c, and c | 3 so c = 1 or c = 3. f(x) cannot equal 1 as it wouldn't be a proper ideal, so c = 3:
2. There must be some a(x) : c*a(x) = p(x), but p(x) has leading coefficient 1, contradicting c = 3

[celest furnace]

You might need to show it’s not proper

[dull ginkgo]

i mean (1) is always the whole ring as for any other ideal generated by a unit

[celest furnace]

Yeah why is the ideal in the question proper

[dull ginkgo]

(u) for a unit is R because (ru^-1) u = r must be in that ideal for any r in the ring

[celest furnace]

I agree

[glass zealot]

why is <3, p(x)> not the whole ring

[celest furnace]

Why is (3, p(x)) not just the whole thing

[dull ginkgo]

if it is then we have a contradiction, if it isn't then we still have a contradiction with the leading coefficient

oh wait not just (3)

well here's the thing, we have to say that p(x) =/= 3q(x) for some q

otherwise (3) = (3, p(x))

wait

[celest furnace]

That’s only necessary not sufficient

[dull ginkgo]

for any general p or just the p in question (of which 3 doesn't divide every coefficient)

[celest furnace]

Just the p in the question

[dull ginkgo]

3 doesn't divide every coefficient, thus there doesn't exist a q(x) such that 3q(x) = p(x)

[celest furnace]

Yep that works

I was stumbling before

You showed the only possible generators are (3) and (1)

Wait what

How does this show that for every a(x), b(x) a(x) * 3 + b(x) (x^3+x^2-2x-1) is not just 1

[glass zealot]

isn't the easiest way to show that 2 is not contained

[dull ginkgo]

I am confused

[celest furnace]

Like you claimed in your proof that the ideal was proper

why does this imply b(x) is just 0?

[dull ginkgo]

yeah because 1 isn't in the ideal and that's easily checked through degree rules.

[celest furnace]

im actually not seeing it right now lmao

how do you show that 1 isn't in the ideal by degree rules

[dull ginkgo]

actually I realized you can't easily, you're right hmm

[celest furnace]

what if there was cancellation among the terms?

[glass zealot]

because X^i isn't invertible for i>0

[celest furnace]

what if a was picked so the x^i's cancel?

[dull ginkgo]

I wonder if you could observe that the leading terms have to cancel and the constant terms need to sum to 1

[celest furnace]

yeah something like that should work

see but its not exactly immediately trivial right

[dull ginkgo]

yeah

[glass zealot]

oh yeah im stupid

[glass zealot]

btw here what about -1 and -3

[dull ginkgo]

same thing, I should've put that down

[frank cosmos]

im getting tripped up on some basic notation in noncommutative vs commutative rings

in a noncommutative ring R, does (a) mean Ra or RaR?

what does it mean for a noncommutative ring to be principal? that all ideals are of the form Ra, or all left ideals are of the form Ra

[crystal vale]

In a non-commutative ring two sided ideal generated by {a}, (a) = { na +ra + as+ xay, n in Z and r,s,x,y in R } ( we don't assume R has unity)

[crystal vale]

Let N and K be submodules of M with I = Ann(N) and J = Ann(K). Show that I+ J is a subset of Ann( N and K) and give an example such that it will be strictly inclusion.

So I have shown that part and for strictly inclusion if I take M= R^2( R is a set of Real numbers ) over R. Let N is submodule generated by (0,1) and K is submodule generated by (1,0). Then I = {0} and J = {0} and Ann( N and K ) = R^2, right?

[rocky cloak]

This can be contextual. But I'd say that if the word "ideal" generally refers to a left ideal, then (a) is probably a left ideal. And if "ideal" generally refers to a two-sided ideal, then (a) should be a two sided ideal.

[rocky cloak]

A principal ring is one such that every left ideal and every right ideal is principal.

You might also talk about a left principal ring, where you require every left ideal to be principal, and similar for a right principal ring.

In particular every (two sided) ideal will also be of the form Ra

[frank cosmos]

ah i see thanks

[boreal inlet]

Let's say H is a proper subgroup of G, such that G/H is abelian. Also, for every non-identity element in H, Centralizer of x in G is H. Then show that H is the commutator subgroup of G.

For this, I can see that the commuator subgroup [G, G] must be inside H, as otherwise G/H cannot be abelian

I couldn't prove the other direction

Oh wait

nvm we done

[delicate orchid]

I don’t buy this. Let G be abelian, then for all x in G C_G(x) = G so we have to have H = G

But clearly the commutator subgroup is trivial

And every quotient of an abelian group is abelian

[boreal inlet]

For a fixed $y \in G \setminus H$, $f: H \to [G,G], f(x) = xyx^{-1}y^{-1}$. This is well defined, and injective.

[cloud walrus]

Riku

[delicate orchid]

Actually this holds for any group that isn’t centreless. Each central element will force your H to be the entire group

[boreal inlet]

And the rest is trivial because G/H can only be abelian iff [G, G] is a subgroup of H

That is, H is a group such that it has cardinality less than equal to the commutator subgroup, and also the fact that G' is also a subgroup of H

[rocky cloak]

Why is this injective? And are your groups always finite?

[rocky cloak]

I guess the fact that H is a proper subgroup is actually relevant

[delicate orchid]

Goofy phrasing

[delicate orchid]

Ok this group is centreless by assumption, then G \cong Inn(G) hence x -> c_x(y^-1)y := x^-1y^-1xy is indeed injective

[rocky cloak]

Hmm, okay I see why it's injective. Like if

x y x^- y^- = z y z^- y^-
Then since H is normal and abelian you would have
xz^- y = y xz^-
Which contradicts y not in H, unless xz^- is the identity.

But unless H is finite, there seems to be a missing step

[crystal vale]

Let M be R- module and M is a direct sum of N + N' and K is a subset of N and also submodule of M. Then I want to show that if N is the direct summand of M then N/K is a direct summand of M/K.

So my guess is M/K is a direct sum of N/K and N'/ intersection of N' and K

Is my guess correct if not any hint?

[delicate orchid]

N’ intersect K is empty because N’ intersect N is empty

[crystal vale]

My bad 🥲

[delicate orchid]

My method would be much more direct, use the fact that m = n+n’ uniquely for all m in M (by the properties of the direct sum) and then use this to write each m+K as n+(something)+K uniquely

Alternatively, bash the coproduct and quotient diagrams together

[crystal vale]

But I want to find that something

[hidden haven]

You mean {0}

[hidden haven]

Yes your guess is correct, once you realize that the second summand in your guess is N' itself

[crystal vale]

So M/K is a direct sum of N/K and N', right?

[delicate orchid]

Yeah, it would’ve been clear to see that the “(something)” in my post was just n’

[delicate orchid]

Both initial objects. I win - you are nitpicking and biased

[next obsidian]

Moldi is always right

[crystal vale]

Okay thank you

[languid trellis]

i stand with wew 😔 ✊

[boreal inlet]

Oh yeah sorry, group is finite

As for injectivity

let for x, x' in H, we have f(x) = f(x')

Then

Then xyx^{-1}y^{-1} = x' y x'^{-1}y^{-1}

From this we get

(x'^{-1}x)y = y(x'^{-1}x)

The first elements belong to H as H is a subgroup

So, y belongs to the centralizer of (x'^{-1}x)

So, y should be in H

That is a contradiction

Hence, the element x'^{-1}x must be identity

And hence x = x'

So the map is injective

@rocky cloak

[winged void]

Small question what is the intuition behind the inverse of a modulo

Someone asked me and I did not know how to answer

[dull ginkgo]

The way I understand it is as a byproduct of Bézout lemma

[winged void]

That’s true you mean xa+yb =1

[dull ginkgo]

Yes

We know the x and y exist

Observe when you take the equation mod b

[dull ginkgo]

assuming a and b are coprime

[winged void]

Sure

[dull ginkgo]

TLDR if a and b are coprime, then a has an inverse mod b, as there exists (x,y) coprime such that xa + yb = 1, so there exists an x such that xa = 1 modulo b

[winged void]

But what about the y

It could be as well ya = module b

Or not

[dull ginkgo]

If you want you can say the y shows that there is an inverse of b mod a

ax + by = 1 implies y is b’s inverse mod a, and x is a’s inverse mod b

[winged void]

Ah I see

[dull ginkgo]

Extended Euclidean algorithm is a way to calculate these lol

extended Euclidean algorithm is determining a word in SL(2,Z)

[dire siren]

it's just like the notion of inverse in any other group

[dull ginkgo]

Gives a bit of insight behind why coprime numbers mod n form a multiplicative subgroup

[winged void]

true

[dull ginkgo]

The converse is a bit more annoying but not too difficult

[languid trellis]

we also immedaitely get that in Z/nZ, |U(Z/nZ)| = \phi(n), which gives us euler totient theorem immediately : )

[round jay]

I'm doing question 4.9, stuck on showing that a_d*g(x) = 0. Any ideas guys?

I'm unsure how to use the fact that g is minimal, since all that allows you to do is conclude that if h has degree bigger than g then fh is nonzero, but i don't see how that'd be useful at all.

[dull ginkgo]

I’ll take a gander

[dull ginkgo]

Here’s an idea

So we know that x in our poly ring commutes with everything and is not a zero divisor, right

let f(x)g(x) = 0 right, so the product of the leading coefficients must equal 0

Now let’s imagine that g(x) = ax^N + g*(x)

so f(x)(ax^N + g*(x)) = af(x)x^N + f(x)g*(x) = 0. That implies af(x) ≠ 0, as if it did, then f(x)g*(x) = 0, contradicting g’s minimality.

(I’m typing out my thought process here, i haven’t done this exercise before)

[round jay]

What is g*?

Nvm

[dull ginkgo]

The sum of terms lower than the leading term

[round jay]

Wait, you commuted a and f right?

[dull ginkgo]

May have by accident

[round jay]

How do we know they commute

[dull ginkgo]

I meant f(x)a ≠ 0

Same conclusion I just wrote it wrong because I’m used to considering “working terms” on the left lmao

[round jay]

Oh ok

[dull ginkgo]

Tbh I don’t like this approach

Let me try a different route from that same starting point

[round jay]

Sure

I tried that same approach but didn't get far

[dull ginkgo]

maybe

Minimality instead from g(x) = g’(x)x + a_0

[willow cipher]

aluffi right? have u looked at the errata?

[round jay]

Omg lol

This dang book

Thank you

[willow cipher]

yeah i have the same version its such a pain. no worries

[dull ginkgo]

It doesn’t seem like commutativity is necessary

[round jay]

Wait I feel like with commutativity we can conclude g is 0

The coefficients of xg(x) are the same as the coefficients of g

If xg(x) is not 0, then f(x)xg(x) can't be 0, a contradiction since xg(x) has higher degree than g(x). Thus, xg(x) is 0

Thus g(x) is 0

Thus f can't be a 0 divisor.

And so the question makes no sense.

Wait I'm tripping so hard

Lol

I think what I just said is wrong

I confused minimal with maximal

[dull ginkgo]

f(x) = ax^N + f*(x)
g(x) = bx^M + g*(x)
0 = f(x)g(x) = abx^(N+M) +f*(x)bx^M + ag*(x)x^N + f*(x)g*(x)

not hard to see that ab = 0, and f*(x)g*(x) = 0

[round jay]

Yeah

[dull ginkgo]

Thus f*(x)bx^M = -ag*(x)x^N

[round jay]

Well all those terms are 0 right?

Since their coefficients are the coefficients of f(x)g(x)

[dull ginkgo]

I am thinking

[chilly ocean]

let $f(X) = \sum_{i = 0}^n a_i X^i$ and $g(X) = \sum_{j = 0}^m b_jX^j$.
where $g$ is a polynomial of minimal degree such that $fg = 0$.
Then the coefficients of $fg$ must be zero so that $a_nb_m = 0$.
thus $\operatorname{deg}(a_ng) < \operatorname{deg}(g)$.
Since $f(a_ng) = a_nfg = 0$ we have that $a_ng = 0$ by minimality of $g$.

[cloud walrus]

Kleptomancer

[chilly ocean]

So we do use commutativity

[round jay]

Nice

Wish I would have seen that

[chilly ocean]

sorry

[round jay]

No problem, thank you for the help

[chilly ocean]

sure thing ^_^

[wraith swan]

What does it mean for a group with two generators x, y two only have the relations x^2 = e and y^3 = e?
I guess this means that xy != y^2? And that only xyxy^2 = e? But that is a relation as well, so...

[dull ginkgo]

There's other relations that can be kind of deduced from those,

Like (x)^2y^3 = e ofc,

Only two relations in this context means that every other relation on that group can be derived from the x^2 = y^3 = e ones

[wraith swan]

Right, but if I say xyxy^2 = e, then I add the restriction that they are commutative

[dull ginkgo]

The word problem is horrifying and scary

[dull ginkgo]

it's an open problem to know if a relation is derived or independent of given ones apparently

and is why you should cry yourself to sleep at night if you're a group theorist

[delicate orchid]

given a word in x and y, the only cancellation you can do is when you have two xs in a row, or three ys

not counting the obvious "xx^-1 = e" as a relation because this is a group

[dull ginkgo]

relation on the monoid ecks dee

[wraith swan]

Right, that is what I thought but then my book asks to prove that that group is isomorphic to C_2 * C_3. But that would imply the group has to be abelian, right?

[dull ginkgo]

is this a free product

In which case yes it's isomorphic to the free product

Wait

[wraith swan]

My book hasn't mentioned it yet

[dull ginkgo]

C_2 and C_3 are cyclics right?

[wraith swan]

Yeah

[dull ginkgo]

or are those the dihe

okay

Uhhhh yeah that's a free product :thecosmoshumswithatunesoVIOLENT

[wraith swan]

* is the coproduct, direct sum

[dull ginkgo]

just x^2 = y^3 = e?

[wraith swan]

yeah, it even says only

[dull ginkgo]

Like every element in that group with the relations is just pairs of xy^n multiplied together

which is the free product, you can't really "swap" the positions of x and y here

[wraith swan]

Oh wait, I guess the * is not the direct sum but the coproduct in Grp

[dull ginkgo]

Like if you had a third relation, that would allow you too, namely the dihedral relation, you'd have the dihedral group of order 6

Free Product is the coproduct in Grp

[wraith swan]

ah ok

[dull ginkgo]

https://en.wikipedia.org/wiki/Free_product#:~:text=In mathematics%2C specifically group theory,new group G ∗ H.

[wraith swan]

So C_2 * C_3 in Ab is a different structure than it in Grp?

[dull ginkgo]

well, the coproduct would be different

[wraith swan]

ok

[dull ginkgo]

In Grp it'd be the free product, in Ab it would be the direct sum (direct product)

[wraith swan]

Oh shit, I think I am messing up notations. The direct sum is \oplus not *....

[dull ginkgo]

in Ab we'd want to also allow the elements to commute, so we'd have to add "commutativity relations" to the free product

which makes it into a direct sum

[wraith swan]

Ok I think I am starting to understand the difference between the free product and the direct sum

[dull ginkgo]

yeppers

[wraith swan]

So are products also different in Grp and Ab?

[dull ginkgo]

no

Just cartesian product,

[wraith swan]

Ok 🙂

[dull ginkgo]

The justification for the difference too also can be interpreted as like, it's universal property

of which is how I remember to not mix up which is a (co)product

[wraith swan]

Well the book only has it defined with the universal property

[dull ginkgo]

We can map "out of both A and B" from the coproduct

So like, if we have a map from A into X, and B into X

[wraith swan]

Right for Abelian groups, since it's isomorphic to the product

[dull ginkgo]

We can map any pair (a,b) into X

just as a+b

which I realized looks a whole fucking lot like the tensor product in a weird way in retrospect

anyway

Here's how I interpreted the free product from the universal property

[wraith swan]

This was a pretty small chapter in the book, with some exercise that really let you do some thinking... I think the next few chapters will explain more

[dull ginkgo]

ah ok

Tldr:
If we have a map from A into X, f
and a map from B into X, g

Every element of the free product is some alternating word like a_1 b_1 a_2 b_2 a_3 b_3... right
well we can just have a map from the free product that sends a_1 b_1 a_2 b_2 a_3 b_3... to f(a_1) g(b_1) f(a_2) g(b_2) f(a_3) g(b_3)...

In a way it is the "most general" group that can be built from A and B, and that map out comes rather naturally by just sending the pieces of the word to their respective images from their subgroup (MUCH LIKE A DISJOINT UNION)

[wraith swan]

Right, thank you 😄

Self study is a little difficult, because no one is checking my exercises and can't ask a lecturer...

[dull ginkgo]

I'm also self studying dw

[delicate orchid]

this is kind of a tricky example to introduce free products with

like it's easy enough to find an order 2 and order 3 generator of PSL(2,Z) but showing that they're free generators is scary

[dull ginkgo]

what?

Isn’t it like the simplest ecks dee

[dull ginkgo]

Holy fuck no

[delicate orchid]

sure, if you know the ping-pong lemma

or the rep given by their action on the upper half plane of C

[dull ginkgo]

the funny part is this is literally extended Euclidean algorithm ignoring sign changes to an extent

[delicate orchid]

"literally [...] to an extent"
drivel

commit.

[dull ginkgo]

You can state SL(2,Z)’s relators using commutators

And it’s abelianization comes instantly

[delicate orchid]

PSL(2,Z) is not abelian

[dull ginkgo]

Ye I know

[delicate orchid]

babble

[wraith swan]

What does SL, PSL mean?

[dull ginkgo]

But SL(2,Z) is more easy to find relations for

[delicate orchid]

it is obvious once you know that SL(2,Z) is C_4 \ast C_3 that PSL(2,Z) is C_2 \ast C_3

[dull ginkgo]

Special linear group, projective special linear group

[dull ginkgo]

Isn’t it amalgamated

[wraith swan]

Ok, I guess the book will talk about those later...

[dull ginkgo]

What book?

[delicate orchid]

maybe. I don't really care

[dull ginkgo]

Understandable completely

[wraith swan]

Aluffi Chapter 0

[delicate orchid]

you cannot convince me that any of this is obvious regardless. Like hell you independantly would have discovered the ping pong lemma on your own

[dull ginkgo]

Fuckin Jacobson introduced wreath products for a single problem and then drops like this fucking nuke in the Sylow section

[wraith swan]

Right, SL is mentioned in the next 3 chapters

[delicate orchid]

wreath products are easy

[dull ginkgo]

They are like 3 group actions together wtf you on about

[delicate orchid]

this is touching on modular forms which is analytical number theory therefore it is hard

[dull ginkgo]

or at least how Jacobson presents it

huge emphasis on the last clause there

[delicate orchid]

there's one group action in a wreath product lol

[dull ginkgo]

also how he wrote it incorrectly

[wraith swan]

Well thank you for helping Tube. And have a good night you two

[dull ginkgo]

did the queen posthumously award you a gaslighting license

[delicate orchid]

$\phi H \rightarrow S_n \Rightarrow G \wr_{\phi} H := G^n \rtimes_{\phi} H$

[dull ginkgo]

good night fam

[cloud walrus]

Wew Lads Tbh

[delicate orchid]

it's literally a semidirect product

[dull ginkgo]

ye I know lol

[delicate orchid]

a NICE semidirect product at that

also wtf why didn't latex render my _n properly

[dull ginkgo]

THE ANOMALY

[delicate orchid]

el machina di diabolo...

[dull ginkgo]

I mistyped, it's two group actions. But the one is just condensed into the $H \rightarrow \mathrm{Aut}(\text{whatever the fuck})$ map

[cloud walrus]

THE TUBE

[delicate orchid]

sure u can think about that as another group action if you want

[dull ginkgo]

just need to consider the semidrect $G^\text{whatever the fuck} \rtimes_{\phi} H$

[cloud walrus]

THE TUBE

[dull ginkgo]

where G^{whatever the fuck} is given the whole image group shit

[delicate orchid]

apparently the cohomology of these things is very nice

[dull ginkgo]

Group cohomology?

I forget if that works for nonabelians

[delicate orchid]

of course it does? it's easy for abelian

well, finitely generated anyway

[dull ginkgo]

i will get there when I get there in Jacobson II

half that goddamn book is (co)homology shit

[delicate orchid]

taking cohomology commutes with direct sums so you can just structure theorem it

[dull ginkgo]

[agile burrow]

hm, i never did cohomology of wreath products

[prisma ibex]

spectral sequence time

[delicate orchid]

Yeah but unlike my previous excursion with the LHS spectral sequence this one tends to collapse on E_2

It does for S_n \wr S_m iirc

[agile burrow]

ok nice so it fits into a neat SES

[delicate orchid]

Yur

Cohomology of S_n with coefficients in F_1

[dull ginkgo]

open #groups-rings-fields
"F_1"
leave #groups-rings-fields

[south patrol]

Good

[dull ginkgo]

I need to show $(3, x^3 - x^2 + 2x - 1)$ is proper in $\mathbb{Z}[x]$

[cloud walrus]

THE TUBE

[lapis latch]

My general advice would be to reduce mod 3

[dull ginkgo]

In general if we have an ideal generated by $a, b$ is the image of this ideal under an epimorphism $\varphi$ is the image the ideal generated by $\varphi(a), \varphi(b)$

[cloud walrus]

THE TUBE

[lapis latch]

Yes

[dull ginkgo]

Is this provable through preimage of ideal being ideal

wait

[lapis latch]

Although I’d say surjective. Just apply the map to the defn of (a,b)

[dull ginkgo]

the smallest ideal containing a and b

phi((a,b)) must contain (phi(a),phi(b)) but it's the reverse that seems screwy

wait

preimage

[lapis latch]

the R-linear combination formulation would be my go to

[dull ginkgo]

oh wait shit $(a,b) = (a) + (b)$ for commutative rings right

[cloud walrus]

THE TUBE

[lapis latch]

Yeah

[dull ginkgo]

lmao

10 can prove 9

Interesting

it's F_27 is it not?

I'd take it that Z[x]/I is a field of 27 elements

Let J = (3, p(x)) where p(x) = x^3 - x^2 + 2x - 1
so J = (3) + (p(x)) contains (3)

Thus s: Z -> Z/(3)Z is an epimorphism, and J is sent to (p*(x)) , where p*(x) = x^3 - x^2 - x - 1 mod 3
So s(J) = (0) + (p*(x)) = (p*(x))

By third isomorphism: (Z/(3))/(J/(3)) = Z_3/(p*(x))~= Z/J

Observing p*(x) = x^3 - x^2 - x - 1. p* is a cubic thus it is either irreducible or has a linear factor (i.e root).
For any x in Z_3 (of char 3): p*(x) = (x^3 - x) - (x^2 + 1) = -(x^2 + 1), which is irreducible mod 3, so so is p*
Thus the quotient is isomorphic to the field F_27

[lapis latch]

Very nice

[dull ginkgo]

so it's a domain from that lol

then that kind of circles back to 9 showing the ideal is proper (as F_27 is not trivial) so 1 does not generate J

Assume (f(x)) = J
Then there must exist some g(x) :
g(x)f(x) = 3, which by degree inequality implies deg(f) = 0, so f(x) = c for some c, and also that c | 3, so c = 1 or 3 (sign is wlog).

As shown c ≠ 1 as the ideal is proper, so c = 3. But that implies there exists a h(x) :
3h(x) = p(x), but then 3 must divide every coefficient of p(x) which obviously doesn’t happen for any of them so we have a contradiction

All rings in this chapter are implied to be commutative

what does "holds without restriction" mean here

[dull ginkgo]

wait shit wrong chat

Actually I'll do it here because I want to know if this is an acceptable proof

I want to prove that the product of all nonzero elements in a finite field $\mathbb{F}_q$ is $-1$. The simplest case is $q = 2$, where the only nonzero element is $-1 = 1$, so it is trivial. Otherwise, we know $\mathbb{F}_q^\times$ is cyclic of order $q - 1$. Let $r$ be the generator.\

$w = \prod{n = 1}^{q - 1}{an} = \prod{k = 1}^{q - 1}{r^k} = r^{\sum_{k = 1}^{q - 1}{k}} = r^{\frac{q(q-1)}{2}}$ \\

However, from here we have two cases. If q is a power of 2, then $\frac{q(q-1)}{2} = m(q-1)$ so $w = (r^{q-1})^m = 1$. This is a nonissue as $1 = -1$ for characterstic $2$ fields. \
For the remaining odd prime powers, we have that $w \neq 1$ as $\frac{q(q-1)}{2}$ cannot divide $q-1$. However, we know that $w^2 = 1$, so $w$ is involute, of which there is only one involute element for cyclic groups of even order. $(-1)^2 = 1$, so $w = -1$

[cloud walrus]

THE TUBE

[dull ginkgo]

I didn't feel like using Vieta's formulas lmao

wait I'm a fucking idiot, the constant map is an endomorphism from F_q[x] to F_q, so it'd send X^{q-1} - 1 = \prod(x - a_n) to \prod(-a_n) = -1

you just need to account for the sign, which that is a characterstic argument

[weary spade]

prove that if H is a subgroup of G such that 2 |H| = |G|, then H is normal

[dull ginkgo]

If 2|H| = |G|, then the index of H in G is 2... what are the cosets?

[weary spade]

H and gH

[dull ginkgo]

okie

there's two left cosets and two right cosets, right

H and gH are left cosets, H and Hg are right cosets... so any element not in H must be in gH and Hg

what can you say from there

[weary spade]

gH = Hg?

[dull ginkgo]

yep

Since any element is either in H or gH, likewise H or Hg. so if x isn't in H, it must be in both of them. Thus gH = Hg

so H is normal

Even if H and G are infinite, if [H : G] = 2, then H is normal in G

[weary spade]

how does gH = Hg imply H is normal though

sorry if that's a dumb question i'm kind of new to this

[dull ginkgo]

So, lets say x is in gH and Hg

then there is some h_1 such that x = gh_1

and some h_2 such that x = h_2g = gh_2... so h_2 = gh_1g^-1

:3

[weary spade]

h2 g = g h1, then h2 = g h1 g^-1, and this holds for all x, therefore H is normal

holy shit it makes sense!

[dull ginkgo]

yeppers

[weary spade]

but quick question what's an index

i've been skipping class haha

[dull ginkgo]

Number of cosets of H in G

[weary spade]

oh ok that's simple

[dull ginkgo]

or by Legrange's theorem, |G|/|H|

All cosets are the same size, so they split up |G| into the index-amount

[weary spade]

yes ofc

ty

[dull ginkgo]

you're welcome

[weary spade]

my final is gonna kill me lmao. for some reason this intro to algebra course has to cover every topic under the sun. CRT, FTA, knots, groups, fields, rings, quotient rings, ideals, euler's theorem, permutations, the list goes on and on

[dull ginkgo]

My textbook is like that

"basic algebra"

smaller than D&F and goes into like, crazy deep shit

[weary spade]

CRT is kinda fun tho ngl

[dull ginkgo]

Group theoretic or ring theoretic

But yeah ring theory CRT is fun

You can prove some suprising stuff with it

[weary spade]

we're just working over integers for CRT lmao. does CRT even apply to different number systems?

[dull ginkgo]

It's actually a strong statement for rings (even noncommutative)

[weary spade]

don't you need to be able to divide though? you can't do that in every ring

[dull ginkgo]

It has to do with the ideals,

I have come to an interesting observation

for simplicity let Q(a,b) be the legendre symbol

[void cosmos]

a nice problem that I managed to solve, try it for fun:
Can a sequence {a1, a2, ... , an} of elements of
a group of order n be chosen so that none of the partial products
is equal to the identity?

the partial products being any product(a_i ) i = p to q where p and q are any positive integers

the length of the sequence (which may, of
course, contain repetitions) is the same as the order of the group.

from halmos's problem book

[dull ginkgo]

Is this group abelian or non abelian?

I assume it doesn't need to be abelian

[void cosmos]

just a group no assumptions

here is a screenshot so that everything is as clear as possible

[dull ginkgo]

THE RAGE I FEEL TO THE 2 LEFT INVERSE ONE

FUCK

infinitely many left inverse HATRED

anyway

[weary spade]

just a hunch but i think this depends on n

might be wrong tho

[void cosmos]

nope

but good

i think it's intuitive to think about its parity

but you can try it out

like pick out a group of order 4 or 3 or 5 or whatever

[dull ginkgo]

Let m be the lcm of the orders in G, i.e min m such that g^m = e for each m in G

m | n

I wonder if you can do some sneakery with that

[void cosmos]

try it out with a real example

i gotta go study some cloud computing but ping me if you have a solution

have ufn

[void cosmos]

yeah think about that

[dull ginkgo]

no it's a problem in Jacobson

I did it before

[void cosmos]

well ig that's how i did this problem somehow

anyways enjoy

[dull ginkgo]

Partition G threeways, into involutions, and the remaining split by inverse

Real quick

If I let $\mathbb{Z}/n\mathbb{Z} = \mathbb{Z}_n$, then is the legendre symbol just the natural projection of $\mathbb{Z}_n^\times$ to $\frac{\mathbb{Z}_n^\times}{(\mathbb{Z}_n^\times)^2} \cong \mathbb{Z}_2$ for odd primes

[cloud walrus]

THE TUBE

[weary spade]

Part 1:
a^(p-1) = 1 mod p by FLT
(a^(p-1)/2 + 1)(a^(p-1)/2 - 1) = 0 mod p
If a is a QR, then a = x^2.
(x^(p-1) + 1)(x^(p-1) - 1) = 0
x^(p-1) is never -1 by FLT, so x^(p-1) = 1 mod p, so a^(p-1)/2 = 1 mod p

[dull ginkgo]

this was my observation

of which it is literally just that

[old hollow]

Let $\alpha = \sqrt[3]{2 - \sqrt2}$, and let $\beta = \sqrt{2 - \sqrt2}$. Is $\beta \in \mathbb{Q}(\alpha)$? Give proof.

[cloud walrus]

a.b.s._.0.

[old hollow]

I feel like the answer is "no", but I'm unsure how to prove it

[tough raven]

Hmm

Q(α), Q(β) both contain sqrt(2) and hence include Q(sqrt(2)).
Can you figure out their degrees over Q(sqrt(2))?

[tough raven]

More precisely, the formulas for α, β strongly suggest that [Q(α) : Q(sqrt(2))] = 3 and [Q(β) : Q(sqrt(2))] = 2.
What would it take to prove this?

[celest furnace]

Wow great proof sir!

[tough raven]

In the meantime,
If A is a subring of an integral domain B such that B is integral over A and the inclusion of A in B is a ring epimorphism, is it necessarily true that A = B?

[old hollow]

Ok, that makes sense, I can prove both of those statements

But what's the view towards showing β in Q(a) though?

edit: OHH is it like, the degrees 2 and 3 are coprime, so the combined extension is exactly degree 6?

[celest furnace]

Inclusion is surjective?

I mean that’s all you would need to show they are equal right

[tough raven]

2 does not divide 3, so Q(β) cannot be a subfield of Q(α).

(Assuming that the degrees are in fact 3 and 2.)

[tough raven]

Yes, I want to know if that holds.

[celest furnace]

Is that not what epimorphism is?

[old hollow]

ohhh thank you

[mighty kiln]

Epimorphism means right-cancellative

[celest furnace]

For sets that is just surjective right?

[mighty kiln]

Yea

[celest furnace]

So it does

[mighty kiln]

No it is not

[tough raven]

I mean

An integral extension cannot be a localisation.

maybe

Do you mean that it is an epimorphism because it is sandwiched between the ring and its field of fractions?

Thanks!

[mighty kiln]

Wait does it

[tough raven]

Uhhh

IDK

Let A inside B inside F = S^{-1} A ("inside" means "is a subring of"; we are assuming that S consists of cancellative elements).
Then any x in B is equal to a/s, a in A, s in S (1/s need not be in B though).
So we want to say f(x) = f(a)/f(s).

OK but s is cancellative, so if f(s) is also guaranteed to be cancellative we can say f(a) = f(x)f(s) which "fixes" f(x).

[dull ginkgo]

An inclusion is a monomorphism of A into B. So if it’s epi, it’s a bimorphism

Oh shit

I hate rings I hate rings I hate rings

[tough raven]

Well, the question is precisely about the difference between bimorphism and isomorphism in this case.

[mighty kiln]

Oh hm

[dull ginkgo]

B is integral over A do you mean that every element of B is a root of a monic poly of A

[mighty kiln]

If R → S is epi that implies Aut(S/R) is trivial right

Oh how about Q ⊂ Q[2^1/3]

No that fails

[dull ginkgo]

I’m pretty sure

bimorphisms of modules are iso?

[mighty kiln]

What's a bimorphism

[dull ginkgo]

Epimorphism and monomorphism

[mighty kiln]

Yea

[dull ginkgo]

Not necessarily always an isomorphism

[mighty kiln]

Modules yes but rings no

[chilly ocean]

monomorphism + epimorphism doesn't imply isomorphism in rings?

[dull ginkgo]

The inclusion of the integers into the rationals is actually epi

[mighty kiln]

Or any localization of an ID

[dull ginkgo]

ore localization trauma

[chilly ocean]

what? you mean mono?

[mighty kiln]

It is epi and mono

It is epi because for any f: Q → R, f(n) and f(m) determine f(n/m) for any nonzero integers n,m

So the composition Z → Q → R determines Q → R

Is there an injective epimorphism from an ID that isn't localization

[dull ginkgo]

I feel like there is a sort of correspondence with the universal property of the localization in a way?

[chilly ocean]

ty i was thinking epi = suryective not categorical epi

i hate rings

[rocky cloak]

Think this should give you an example
https://mathoverflow.net/a/161/157483

[mighty kiln]

Ohh

So that's like

[rocky cloak]

Are you asking about the converse here? Because the automorphism group of Q[2^1/3] is trivial, but the map is not epi

[mighty kiln]

Yea I realized that

[rocky cloak]

But the original statement is true

Even works in any category.

[mighty kiln]

For reference this would be
C[t²-1, t³-t] ⊂ localization of C[t] by t-1

[rocky cloak]

How about A = Fp(t^p) and B = Fp(t) ?

Or do you need it to strictly be an integral domain, and not a field?

[rocky cloak]

Hmmm, wait.

If C = B[z]/z^2, then t -> t and t -> t+z, might still agree on A.

Darn

[cobalt heath]

Was Z -> Q an epimorphism

My head not work

[rocky cloak]

It is yes

[mighty kiln]

Any localization of an ID

[cobalt heath]

What is the easiest way to see this fast?

[rocky cloak]

Like this

[tough raven]

It's pure curiosity so I need nothing.

[tough raven]

Niceee.

[rocky cloak]

Maybe not so nice though

[tough raven]

Rings that are not fields strike again 😔.

[cobalt heath]

Wait, so Z -> Q is both epi and mono

..how does it even work

[rocky cloak]

Well, it works just like that

[cobalt heath]

My mind says that if epi and mono then it is iso
Gotta tame my mind somehow

[tough raven]

Just as you said. Gotta live with it.

[rocky cloak]

Maybe examples from topology are more familiar. Continuous bijection does not mean homeomorphism

And morally commutative rings are basically types of topological spaces

[cobalt heath]

Hm. Is a continuous bijection both epi and mono?

[tough raven]

In any concrete category, injections are mono and surjections are epi.

[cobalt heath]

Oh. Hmm

[rocky cloak]

And in Top you even have the converse

[tough raven]

(But monos need not be injections, or more commonly, epis need not be surjections.)

[mighty kiln]

Aren't dense subsets epi but not surjective

[rocky cloak]

Not in Top, but in something like the category of Hausdorff spaces, sure

[mighty kiln]

Wait what

[rocky cloak]

The easy way to see it is that if X has the trivial topology, then any function to X is continuous. So being epi in Top is the same as being epi in Set

[mighty kiln]

Oh

[rocky cloak]

But if you do something to make limits unique (like restricting to Hausdorff spaces) then you're cooking

[tough raven]

Hmm, so suppose B/A is integral mono and epi.
Pick x in B satisfying f(x) = x^n + … + a0 = 0, where f is chosen with n minimal.
Let C be the “universal splitting ring” for f over A, e.g. C := A[T1, …, Tn]/(Σ Ti = -a_{n-1}, …, T1 … Tn = (-1)^n a0).

Then hopefully there “should” be an A-morphism B -> B (×) C mapping x to any Ti, forcing n = 1?

Not that it would actually just be this simple (unless A is a field, I think, so that {f | f(x) = 0} is a principal ideal).

[rocky cloak]

Apearantly, true when A is Noetherian, but not in general
https://mathoverflow.net/a/21718/157483

[tough raven]

Huh.

[dull ginkgo]

.

[cobalt heath]

Damn rings

[rocky cloak]

I guess this example is not an integral domain. Forgot you also required that

[tough raven]

Fair enough.

[mighty kiln]

Wait does this one not work

[rocky cloak]

Idk, is that an integral extension?

[mighty kiln]

Oh right

[cobalt heath]

How do you say that A -> B is an "integral extension"?

When the map is not inclusion.