groups-rings-fields

[next obsidian]

In the end it’s all the same

It’s just aesthetics at this point

[noble nexus]

there's p much no reason to ever define a map on a tensor product without using the universal property tbh

if you define it for simple tensors and extend linearly you need to show its well defined

which basically comes down to proving it comes from a bilinear map anyway

[copper kestrel]

how does one even consider starting this? i guess find the order of all subfields of E of the form Z_p, then find a field of prime characteristic p and by a theorem we have that there exists a subfield isomorphic to Z_p?

[quiet pelican]

The additive group of the prime subfield of E is a subgroup of the additive group of E
Then ||apply Lagrange||

[copper kestrel]

and then somehow find the dimension

[copper kestrel]

yeye

[south patrol]

Z_p isn't a field

[copper kestrel]

so we have the subfield of order 2,4,8,16,32 i think

[copper kestrel]

wha

[copper kestrel]

and then find which one has prime characteristic i guess

i think?

[quiet pelican]

*Has prime order

[copper kestrel]

huh

but thats only the subgroup with 2 elements

[rocky cloak]

Can you think of such a field? What characteristic does it have?

[copper kestrel]

Z/2

i think thats a field at least but maybe not

[rocky cloak]

Right, so since the only subfield of prime order is Z/2, that is the prime subfield

It cannot be Z/3, Z/5 or any other prime field

[noble flower]

Could someone help me understand this part of the text on K-Algebra, for example what are they getting at with the canonical map

Is a K-Vector space a vector space with field K?

they hadn't defined that but I assume its that

but what is a K-algebra structure

[copper kestrel]

hm

i can see that actually yeah

[rocky cloak]

A ring structure + a K-vector structure as described in the first paragraph

[noble flower]

hm whats a K-vector structure?

like the ring A follows the rules of a vector space with field K?

[rocky cloak]

A function
KxA -> A
that satisfies the axioms of a vector space

Together with addition etc

[noble flower]

is KxA the binary operation x : K X A -> A?

or something along the lines of that

oh

I misread

I just repeated what u wrote lol ok I see

[rocky cloak]

Like given an element lambda of K and a of A then
lambda a
is another element of A

That's scalar multiplication, which is the main part of a vector space structure

[copper kestrel]

how would i determine the dimension of E over its prime field?

i dont exactly know what that means

i know what dimension means

[delicate orchid]

a vector space of dimension n over a finite field of order p has order p^n

[copper kestrel]

oh!

i think i remember hearing that

[delicate orchid]

I mean it's easy to see, any vector space of finite dimension over a field k is just a product of ks

[copper kestrel]

hm

[south patrol]

a basis of V gives a linear map k^n -> V sending ith standard basis vector to ith thing in ur basis

[copper kestrel]

right

[south patrol]

And this is a bijection

[copper kestrel]

so then i have to find the dimension of E in general and then have 2^dim E?

no wait thats dumb

[coral steeple]

Isn't the correct statement "any non-zero, non-unit factor of the image of f must be associate to x^2+1"?

Er, sorry, that's not right. Is the reasoning here that any quadratic factor of the image does not reduce (since x^2+1 has no roots in F3), meaning that the only possible factorisation of the image into irreducibles is into irreducible quadratics?

That's not quite right either

I can get myself to the same conclusion, but by more tediousness than the author seems to be using here

[rocky cloak]

F3[x] is a unique factorization domain. So the only possible factorizations up to associates is f*1 and (x^2 + 1)^2

Any factor must be associate to some product of its irreducible factors.

The only way to make a quadratic polynomial out of multiplying together quadratic polynomials is to just pick one

[coral steeple]

Right, so we arrive back at my first statement I think: any irreducible factor (i.e. nonunit factor) must be associate to x^2+1

[rocky cloak]

I mean that is also a true statement.

[coral steeple]

But not quite how the author is doing it?

[rocky cloak]

I guess I'm not sure what you're asking anymore.

Do you feel there is something wrong with what's in the red box?

You mean that it's emphasising the wrong thing relative to what you need it for?

[coral steeple]

Yeah pretty much

[coral steeple]

I thought that you meant my statement was not relevant to the argument here

[rocky cloak]

Yeah sure, I guess it's sort of taken for granted that you can check the polynomial has no roots

Either by doing it mod 3 or using the rational root theorem

[rocky cloak]

I guess it's not really true that this means f would be the square of a quadratic either. Not sure what argument they have in mind there

[coral steeple]

Well suppose (x^2+1)^2=fg in F3[x], neither f nor g units. Then f and g factor into irreducibles. Since x^2+1 is irreducible, uniqueness of this factorisation show f and g are both associate to x^2+1

[rocky cloak]

Yeah in F3[x]

[coral steeple]

And lifting that back to Z[x] ought to show that the original polynomial is associate to a square of a quadratic, but since the polynomial is monic you're done

AHh I see what you're saying

[rocky cloak]

Why would it ought to show that?

[coral steeple]

Is your point that I could write the original polynomial over Z as something other than a square, and that would reduce mod 3 to a square?

[rocky cloak]

Sure, like
(x^2 + 1)(x^2 - 2)
or whatever

[coral steeple]

Right

I'm confused-- is there a problem just with the boxed sentence or is the whole solution iffy? This seems to be a problem with it to me

[rocky cloak]

I'm more concerned with what's after the box

[coral steeple]

Yeah, the lifting

The best statement the author could make is we would have a factorisation in Z[x] into a product of two quadratics, I think

[rocky cloak]

I mean I'm sure you can throw some more computations at it and it would be fine. But I think they may have cut some corners

[coral steeple]

I've never seen an argument like this before

[rocky cloak]

Anyway, you can prove this quite cleanly with Galois theory. Much less computation required

[coral steeple]

I mean I know that reduction mod p gives a criterion for checking irreducibility, but the author does this "the factorisation must be a square" trick a few times in course notes and assignment solutions

[coral steeple]

We have not gotten to the Galois theory part of the course unfortunately

[coral steeple]

Do you mean that you think there is some way to think about reducing mod 3 and showing that the polynomial is irreducible that way, or just that it is possible to verify that fact computationally?

[rocky cloak]

Like expanding (x^2 + ax + b)(x^2 + (a+3m)x + b + 3n)

[coral steeple]

AHh

Why don't you have a 1+3k coefficient on the x^2 terms?

[rocky cloak]

Well, I know the factors would need to be monic. But you can add that in if you want

The first equation would then just read
1 = 1+3k

[coral steeple]

Ok yeah

Is there some (non-Galois-theoretic?) underlying reason why the author does not seem to need to check they're not losing information about the factorisation while reducing? It happens quite a bit (twice in this image)

[rocky cloak]

Where are you getting these notes? Are they solution sets to your homework or from a book or something else?

[coral steeple]

The first image are course notes, the second are solutions

[rocky cloak]

I might ask your professor / TA / whoever makes these solutions then

(I might = you should xD)

[coral steeple]

Good idea

Thanks for the help! @rocky cloak

[ivory anvil]

do matrix row operations form a group?

[frigid epoch]

yus

all are represented by invertible matrices

[ivory anvil]

right

[azure cairn]

invertible matrix row ops do, that's the general linear group

wait sorry i misread

[azure iron]

i think usually we restrict the notion of elementary row ops to invertible ones

[twilit wraith]

this is a problem on my chapter on rational canonical form

and honestly i really just dont know

[rotund dragon]

i would try this for an RCF with a single block first

then try to generalize to an RCF with multiple blocks

[twilit wraith]

i mean for a single block RCF you just get that the constant term is the top right element times some power of -1

but idk how its related to the determinant of A

[azure cairn]

is that jordan normal form

ok nvm i googled it and it's not

i could help if the field were algebraically closed

well wait

[rotund dragon]

the determinant of a block diagonal matrix is the product of the determinants of the blocks

[twilit wraith]

Yes so I get the product of the top right elements

[rotund dragon]

so the thing you just mentioned can be used to say something in particular about the constant term of A

because of how the characteristic polynomial interacts w/ each of the companion matrices making up the RCF

[twilit wraith]

Is it just that it annihilates them

I get that part but I still dont even remotely see where the determinant comes in

Maybe im forgetting a vital property of the determinant

[azure cairn]

so if i understand rational canonical form correctly, we write $A = PRP^{-1}$. note $A \sim R$ and so the char poly, trace, and determinant of $A$ is the same as $R$. write $R$ in block-diagonal fashion, like $R = \diag\{C_1, \dots, C_n\}$, where
$$C_i = \mqty[0 & 0 & \dots & 0 & -c_{i, 0} \\ 1 & 0 & \dots & 0 & -c_{i, 1} \\ 0 & 1 & \dots & -c_{i, 2} \\ \hdots & \hdots & \ddots & \hdots & \hdots \\ 0 & 0 & \dots & 1 & -c_{i, m_i - 1}]$$

note $\tr(C_i) = -c_{i, m_i - 1}$. do cofactor expansion about the first row of $C_i$ and you get $\det(C_i) = (-1)^{m_i} c_{i, 0}$.

[cloud walrus]

Altanis

[azure cairn]

then you can relate the determinants and trace of each block diagonal (or companion matrix i think it's called for rcf?) matrix in R to R itself

by uh $\tr(R) = \sum_i \tr(C_i)$ and $\det(R) = \prod_i \det(C_i)$.

[cloud walrus]

Altanis

[twilit wraith]

Oh wait does the rcf of a matrix have the same trace and determinant

[rotund dragon]

yeah

[twilit wraith]

Oh I see it now

[rotund dragon]

more generally that's true up to similarity

[twilit wraith]

Oh I should've known that man

Det is multiplicative

Thanks a bunch yall

[wicked patio]

oh yeah det and trace are both determinable from eigenvalues, which similarity preserves

also, trace is preserved under cyclic permutation, i.e. Tr(ABC) = Tr(BCA) = Tr(CAB)

[wicked patio]

So you can say Tr(PRP^-1) = Tr(RP^-1P) = Tr(R)

[twilit wraith]

ah nice

ugh i shouldve known that the whole thing bottlenecking me from solving this was just that i didnt know properties of det and tr

[crystal vale]

How do I prove vi)?

[quiet pelican]

Prove X_f is isomorphic to Spec(A_f) is probably the nicest way
But icr if that book does localisations by now

[crystal vale]

No

[feral slate]

Asks how to prove something
Is shown a way to prove something
no

Absolute gigachad

[quiet pelican]

I mean it’s a no to the implicit question “does the book do localisations by now”

[feral slate]

Oh,that would make sense

I just thought it was funny out of context

[fading acorn]

oh

sorry for ghost ping lol saw vi as iv

[fading acorn]

honestly this is the most reasonable thing

[crystal vale]

Any other idea?

[fading acorn]

you could probably copypaste the proof of (v) for an analogue to (vi) because these are just essentially the same way lol

you have to fix some parts

for more hint: Spec A is just D(1), or D(u) for any unit u

[wraith cargo]

this would require knowing that X_f is the spectrum of some ring

and a priori they do not know that

I assume

[twilit wraith]

this one should be quick bc i think i know what to do but i just wanna make sure

its easy to see that $T^3 + T^2 - \text{id} = 0$ and so $p(x) = x^3 + x^2 - 1$ annihilates $T$

[cloud walrus]

hiidostuff

[twilit wraith]

furthermore $p$ is irreducible so its the minimal polynomial

[cloud walrus]

hiidostuff

[twilit wraith]

and even more, the invariant factors of $T$ must just be some repeated list of $p$ if anything because the invariant factors have to divide $p$

[cloud walrus]

hiidostuff

[twilit wraith]

is this right?

[crystal vale]

What is D?

[wraith cargo]

since the X_f are a basis of open sets of X, for any X_f it's enough to consider an open covering of the form X_g with X_g \subseteq X_f. Now use the definition of X_f and mimic the proof of v)

[fading acorn]

hmm cant we just take principal open cover X_f\subseteq\bigcup D(f_i), and taking complement gives \bigcap V(f_i)\subseteq V(f) then we could apply the same argument no?

[wraith cargo]

this is how I'd do it

[crystal vale]

But I am considering X_f \subset \bigcup X_g_i

[fading acorn]

take complement and then you can write f^n=\sum a_ig_i for some i's (and this is finitely sum)

then use X_f^n=X_f

[crystal vale]

I get V( {g_i} ) \subset V(f), how do I show f^n is finite sum of a_ig_i?

[fading acorn]

you can show V(a)\subseteq V(b) if and only if rad(b)\subseteq rad(a) by fixing (iv)

[wraith cargo]

yeah this was my idea but I thought you were going for something else

[crystal vale]

Yes

I got it, thank you @fading acorn

[crystal vale]

For iii) my guess is for Hausdorff topological space the maximal irreducible components are singletons but I am not sure, I am saying because I can say for finite sets which has cardinality more than one, that set cannot be irreducible

[azure cairn]

Is there a name for the theorem that all subgroups of (Z, +) are of form Za?

[rocky cloak]

Don't think it has a name, but a subgroup of a cyclic group is cyclic would be the relevant theorem

[candid patrol]

euclide xd

[rocky cloak]

If X is the union of two closed sets. What does that say about the complement of those two sets? Can you relate that to the Hausdorff condition?

[azure cairn]

Ah ok

[crystal vale]

okay i can say both are clopen sets

[copper kestrel]

how would someone approach part a of 24? i know part b is showing uniqueness but im confused for part a

sorry about no cropping i am lazy to crop

[rocky cloak]

clopen is a bit strong maybe.

But they are open, because complement of closed is open

[copper kestrel]

i know this is linear algebra but it is in my abs alg book so i thought i'd send it here

[rocky cloak]

Start by using the definition of a basis

[crystal vale]

you can write any elements in terms of linear combination of basis vectors, and you know where you basis elements maps

[copper kestrel]

right but like what does "completely determined" mean?

[rocky cloak]

That if you know phi(beta_i) then you know all of phi (i.e. phi(v) for all v)

[copper kestrel]

hm

[rocky cloak]

Part a is the "at most one" to part b's "exactly one"

[crystal vale]

yes but i can't see the connection it with Hausdorff

[copper kestrel]

i dont really understand what im trying to prove then, like how do you prove that if you have one thing then you know the entire linear transformation? like what does that mean in a statement form

[rocky cloak]

Alright, so before you see the connection you would need to answer the question.

If you have X the union of two closed sets, what does that tell you about the complements?

[rocky cloak]

It means you're supposed to determine phi(v) given the vectors phi(beta_i)

[crystal vale]

they are open too

wait

[copper kestrel]

ohhhhhhhhh

[crystal vale]

i am wrong here

[rocky cloak]

Right, so the statement sort of contains two bits of information.

The sets are closed and X is the union of them

[crystal vale]

if they are disjoint then they are open

[copper kestrel]

so like v = some linear combination and then apply phi to both sides and use linearity to get phi(v) in terms of phi(beta)

[rocky cloak]

So far you've only used the first bit of information to conclude anything

[crystal vale]

but are they disjoint?

yes maximal one are disjoint

[rocky cloak]

Are you asking if the closed sets are disjoint?

[crystal vale]

i mean if X is the union of two closed sets then i don't think they are open too, if they disjoint then i can say they are open

but here maximal are disjoint so i can X is the disjoint union of closed sets

[rocky cloak]

I'm not really sure what you're saying or what you mean is maximal.

What I'm trying to say is that being Hausdorff is usually framed as something about intersecting open sets, while being irreducible is framed as taking unions of closed sets.

So we somehow have to turn unions of closed sets into intersections of open sets. So we need a way to turn a closed set into an open set, and the complement is a way to do that

So if
X = F1 u F2
for closed sets F1 and F2, and their complements are U1 and U2, what is the relationship between U1 and U2?

[crystal vale]

are F1 and F2 disjoint?

[rocky cloak]

No

The definition of irreducible is only about unions of closed sets, nothing about disjointness

[crystal vale]

I am using this definition

[rocky cloak]

Well, then it's much easier.

You don't have to translate from closed sets to open sets, it's just immediate.

Just imagine a Hausdorff space with more than one point and notice that it's not irreducible

All I was saying before is that
X = F1 u F2
is equivalent to
Ø = U1 n U2

So if you can find two open sets that don't intersect you're done

[crystal vale]

Yss

[tardy hedge]

NotKnow

[crystal vale]

Yes De Morgan Law

[crystal vale]

Kiand123

[rocky cloak]

So you see it?

[crystal vale]

Yes, so if a set Y is which has cardinality more than 1 then still it is Hausdorff then it shows there exists two disjoint open sets, hence it cannot be irreducible therefore singletons are the irreducible components in Hausdorff space

Thank you

[crystal vale]

For iv) part, i observe that V(p) is irreducible for all prime ideal p, but for maximal I have to show p has to be minimal polynomial.

What I am thinking is somehow I can prove irreducible subset are the form of V(q), where q is a prime ideal then I am done.

I can only say irreducible subset are of the form V(I)

Somehow if I can prove...

[rocky cloak]

So you may as well assume I is a radical ideal and use that

V(I) = spec A/I

Then figure out when spec R is irreducible

[crystal vale]

When the nilradical of R is prime ideal of R

[rocky cloak]

Yeah, and what does that say about I if the nilradical of A/I is prime

[crystal vale]

Here are you identity V(I) with spec A/I

[crystal vale]

If I consider I as radical ideal then I is prime ideal

Thank you @rocky cloak

[empty estuary]

How do we remember all the elements of it

[rocky cloak]

What is your question?

[empty estuary]

How do I remember its matrix form?

It's slightly confusing

I can remember only identity

[rocky cloak]

Well i is the one with i's on the diagonal, so that makes sense.

j is a 90 degree rotation so somewhat natural thing that would square to -1 and k = ij

[empty estuary]

I did not get it properly 😌

I'm new over discord and abstract algebra

Is this how we rotate?

[rocky cloak]

,rotate

[empty estuary]

How do you get J? By rotation. Pls explain more

By multiple?

[rocky cloak]

I'm not sure I understand what you're asking. In your book they write that j is given by
[0 1]
[-1 0]
which is the 2x2 matrix for a 90 degree counter clockwise rotation

[empty estuary]

Is it not counter clock wise matrix?
[0 -1]
[1 0]

[rocky cloak]

Yeah, clockwise sorry

[empty estuary]

Please check my K is different than book

@rocky cloak

[rocky cloak]

Well your i is also different

[quiet pelican]

i has i, -i on the diagonal

[empty estuary]

ohh damn

[quiet pelican]

Not i, i

[empty estuary]

How did they assume i?

diagonal should be same no?

I'm totally confused about how they are constructing matrices

[rocky cloak]

The real span of 1, i, j and k is called the quaternions.

Then the typical way to construct matrices would be to pick some (right) C-basis, and just write down the matrices for left multiplication by 1, i, j, k.

I guess the basis that would be the most natural to me would be something like {1, j}, but that would give
[0 -1]
[1 0]
for j, so maybe they have some different basis or are thinking about right multiplication or something else

If you pick the basis {1, -j} you get their matrices at least

[ripe harbor]

#help-28 message also fits here

[supple ice]

we do not know algebraic independence of pi and e

[crystal vale]

What does it mean by Lattice?

I am not sure

But what is the definition

[noble nexus]

https://en.wikipedia.org/wiki/Lattice_(order)

[ripe harbor]

So the pick for L was suboptimal..?

[supple ice]

yeah choose an other L

[ripe harbor]

In what way should you do that? Do you have some intuition behind it

[supple ice]

it’s much better to use an algebraic field

[ripe harbor]

Thats a field L so that L/Q is a finite field extension?

[supple ice]

by algebraic field here I just meant a subfield of C generated by algebraic numbers

[ripe harbor]

Ah

Reason being so that we know the algebraic dependencies?

[wraith cargo]

Shouldn't something like Q(pi+i) work

[supple ice]

I think yes

[rocky cloak]

I guess the easiest example might be something like ||Q adjoin a complex third root of 2||

[supple ice]

yeah exactly the same example I had in my mind

[rocky cloak]

That is at least the smallest example

[tulip otter]

how to draw the spectrum of a ring

[rocky cloak]

Depends on the ring I guess

[tulip otter]

right yea. Well lets say Spec(Z) as a start. So Spec(Z)={(p)| p prime}U{(0)}

[noble nexus]

typically you draw each point by drawing it's closure

so the generic point you draw as a large line "containing" all the other points

because it's closure contains the other pts

[velvet hull]

more than one way to present it, in the specific case of spec(Z) you can alternatively visualize it is a tree with infinitely many branches coming out of it

[tulip otter]

what does generic point mean

[noble nexus]

The prime 0 in an integral domain I think b/c the closure is everything

Idk if it has a precise definition

[fading acorn]

closure of the point {P} is dense in your space

[tulip otter]

whats considered an easy example to visualize?

[fading acorn]

just read Eisenbud's geometry of schemes ch2 there are many examples lol

[tulip otter]

well the thing is that I am doing A&M commutative algebra

[fading acorn]

oh that one exercise?

asking how to interpret Spec Z, Spec R, Spec C, Spec Z[x] maybe?

[tulip otter]

an exercise says to draw pictures of the spectra of Z, R, R[x], C[x] and Z[x]

[tulip otter]

yea

[fading acorn]

oh lol

[fading acorn]

so that's why you are not active these days

[tulip otter]

actually I am not active these days because I am not really studying these days, mostly wasting time

[velvet hull]

C[x] is pretty easy

why are you trying to visualise spec anyway

[tulip otter]

ok so for example maybe Spec(C[x]) is easy because you can associate each prime ideal to a complex number

[noble nexus]

well visualizing spec is really nice

[tulip otter]

since the prime ideals are (x-a) and (0) for any complex a

[noble nexus]

why wouldn't you try to visualize see it

[tulip otter]

so then this should be the complex plane plus some point at infinity?

[noble nexus]

it's not a point at infinity

[fading acorn]

its not point at infinity

[tulip otter]

just because its an exercise in A&M

[velvet hull]

oh yeah right i remember that exercise actually

[tulip otter]

well I didnt know what to do with (0) so i said what came to mind

[noble nexus]

it's funny I remember in that one Borcherds video on spec he actually mentioned that the first time everyone learns about spec C[x] they immediately think "oh it's just C with a point at infinity"

b/c you see C plus an extra point and neuron activation

[tulip otter]

yea everyone gets tricked by that lol

but what to do with (0) tho 🥀

[fading acorn]

the real C with point at infinity is ProjC[x,y]

[noble nexus]

it's just the whole line

but a single point

it's very weird to think about it this way but tbh imo it's the best way to visualize non Hausdorff spaces

also another quirk with C[x] is that it's actually better to think about it as a line than a plane

which I find hilarious

"the complex line"

[rocky cloak]

(0) should be close to all the other points since it's closure is everything.

So it's kind of a "generic" point, existing everywhere at once

[tulip otter]

so it traces the same curve/shape traced by all other points of the spectrum?

[noble nexus]

idk what you mean by traces the same curve

Sometimes ppl draw a line with some cloudy fuzzy point behind it to represent the generic pt

[wraith cargo]

Q[x]....

[tulip otter]

I am not sure what that means too tbh, I am just not getting how to draw Spec(C[x]) or Spec(R) for any commutative ring R in general

[fading acorn]

yeah smth like this:

[wraith cargo]

For general R you cant really draw it!

[fading acorn]

tfw R with arbitrary krull dimension

[wraith cargo]

You'd have a lot of thick and generic points

Since nilpotents are basically infinitesimal thickenings and generic points abound!

Like this is what Spec(Z[x]) looks like

[rocky cloak]

So Spec(C[x]) is just C plus an extra point.

So you would just draw it however you want to draw C.

[thorn jay]

the joy of realising why those circles are drawn by the curve [(x^2+1)] there relating with factorisation of primes in Z[i] is great

[thorn jay]

"the complex plane is best visualised as a 4-dimensional euclidian space"

[ripe harbor]

Right, how do you justify it though?

[wraith cargo]

Show that pi-i isn't in the field :)

[ripe harbor]

Can we say all elements are of the form r + s(pi + i) with r, s in Q?

[wraith cargo]

God I wish

No it'll be polynomials evaluated in pi+i

So you can have arbitrarily high powers of pi+i

[ripe harbor]

Oh, right

So we need to show that $\pi - i \neq a_0 + a_1(\pi + i) + a_2(\pi + i)^2 + \dots + a_n(\pi + i)^n$ with $a_i \in \mathbb Q$ for all $n \in \mathbb N$

[cloud walrus]

ILikeMathematics

[noble flower]

does anyone know what this restriction "s |_{Q_1}" notation means

its at the very start of chapter 2

and I searched through chapter 1 and couldnt find it

I mean I think they defined it in the parenthesis

[noble nexus]

just restrict the source and target maps

no?

[noble flower]

what does restrict mean here

[noble nexus]

like as a function

idk how this books axiomatizes quivers but there's kinda only one interpretation that makes sense

[noble flower]

hm ok I see

thanks

[tulip otter]

tysm everyone for helping me earlier and sorry for disappearing suddenly but i had to go. I will keep everything yall told me in mind and try to look more into this later on to figure out how things work more clearly

[crystal vale]

I got it, thank you

[tulip otter]

have a great day/night!

[crystal vale]

I am a little bit not sure about this P(X) here, can anyone explain what we are doing here ?

K[t1,...,tN]/I(X) is the quotient ring, are we restricting our domain K^N to X?

[trail cave]

Let $\pi\in R = \mathbb{Z}[\sqrt{2}]$. Suppose $N(\pi)=p$ for p a rational prime. Show $\pi$ is prime in R.

The norm here is $N(a)=|a^2-2b^2|$. I've gotten to the point that I can say, for $\pi | xy$, x and y in R, $p | N(x)$ or $p | N(y)$. How do I 'get back' from the norms here to say that $\pi$ itself must divide x or y?

[cloud walrus]

Bottlecap Desu~(Bottlecap Gang)

[crystal vale]

I think for one direction i don't need \phi has to be regular mapping

[thorn jay]

well, if ϕ isnt regular then the composition η ∘ ϕ isnt necessarily polynomial

[crystal vale]

Oh my bad

Does every ideal of A×B looks like I×J, where I is an ideal of A and J is an ideal of B ?

[thorn jay]

yes for rings

[crystal vale]

For vii), let K1 be the prime ideal in B then π_2(K1) is prime ideal in K, so π_2(K1) = 0 and also π_1(K1) = I is a prime ideal in A/p, so K1 = I ×{0}.

But A/p has only one prime ideal which is {0}.

So K1 = {0} × {0}

[thorn jay]

idk why i said "for rings" thats a kinda unnecessary addendum lol

[thorn jay]

no, moving prime ideals through rings is via quotients and preimages

Spec(B) should be the disjoint union of Spec(A/p) and Spec(K)

[crystal vale]

Sorry i don't get it

[thorn jay]

if I is a prime ideal then π1(I) isnt necessarily a prime ideal

it needs the condition ker π1 ⊂ I

[crystal vale]

Ah my bad

[crystal vale]

How?

[crystal vale]

How?

[thorn jay]

if q is a prime of B, then it is of the form I × J where I is an ideal of A/p and J an ideal of K
=> B/q = (A/p)/I × K/J
but an integral domain cannot be a product of two nontrivial rings (as those necessarily introduce zero divisors), so either J = K or I = A/p

[thorn jay]

if I is an ideal of R, then I = (1, 0)I + (0,1)I

[crystal vale]

Yes but how do I find K1 and k2 such that I = K1 × K2

[thorn jay]

think

[crystal vale]

Let me think

[cursive spindle]

Noggin

[crystal vale]

I got it, thank you

[noble flower]

is alpha^-1 in the set of arrows?

since s is a function from the set of arrows to the set of points

for every alpha in the set of arrows, there must be a alpha^-1 in the set of arrows right?

if so, then wouldnt every point be a predecessor of each other

[rotund dragon]

only if there exists at least one path between that pair to begin with

[noble flower]

hm yeah

apparently alpha^-1 isnt in the set of arrows tho

[rotund dragon]

its not yeah. the goal is primarily notational convenience here

[crystal vale]

does Cayley Hamilton theorem holds for modules?

[thorn jay]

its not a well-defined question for non-free modules but besides that i dont know

[novel star]

matsumura, commutative ring theory, theorem 2.1

if $M$ is a finite module over a (commutative) ring $A$ and $\text{End}(M)$ is considered as an $A$-algebra, every element $\phi$ of $\text{End}(M)$ satisfies a relation $a_0+a_1\phi+a_2\phi^2+\ldots +a_n\phi^n=0,$ with $a_i\in A$

[desert verge]

finite module meaning M has finitely many elements?

[cloud walrus]

qchs

[novel star]

finitely generated

[desert verge]

yes.. that makes more sense

[desert verge]

wait does the polynomial relation it satisfies have anything to do with some kind of characteristic polynomial of phi?

[novel star]

yes it is precisely that

you build a matrix wrt some finite generating set

and take char poly of that

[desert verge]

does fg imply free

im fucked

[thorn jay]

consider abelian groups

many a fg abelian group that isnt free

[novel star]

no

[desert verge]

oh yes

Z/5Z as a Z-module

[novel star]

really any proper fg ideal

but somewhat surprisingly, projective->free in local rings

matsumura isnt clear in the proof tho

read kaplansky's paper instead

[orchid crystal]

what is this group about?

[desert verge]

groups, rings, fields

see the channel description in the header

[thorn jay]

oh a dutch person

[elfin wraith]

https://tenor.com/view/culture-austin-powers-dutch-gif-10931343

[mystic ether]

This is probably obvious but my brain just isn't parsing it for some reason. Suppose we have a group action of a group G on a set X. Suppose H is a subgroup and F is a subset of X such that the H orbits intersect F in a unique point. Show the same property holds for the cosets of H

that is, suppose Hx cap F is a unique point for all x. Then show (gHx) cap F is still a unique point

I don't think that (gHx) cap F = g(Hx cap F)... rather I think probably (gHx) cap F = g(Hx cap g^{-1}F)

so an equivalent version of this would be to show that the G translates of F also intersect the orbits of H in a unique point. But I'm also still unsure how to show this

[mystic ether]

it's clear that the H translates of F intersect the orbits of H in a unique point, and it's also clear that this property holds if H is a normal subgroup of G, but if H is a general subgroup I can't see it

[mystic ether]

Apparently it’s not necessarily true

[tall igloo]

i agree. for example take H = {e, (1 2)} in \Sigma_3 the symmetric group on 3 elements acting on the 3 vertices of an equilateral triangle via symmetries. if I label the points 1, 2, 3, then pick F = {1, 3} for example, the assumption is correct. however, if i pick g = (2, 3), then (gH)(3) = g({3}) = {2} which is not in F

i dont know what the most general thing you can say is, but you can get away with it if F is G-invariant instead of some random subset

or as you said if H is normal also works

[mystic ether]

If F is G invariant the statement is somewhat trivial, since then it is also H invariant and the assumption implies H acts as the identity on it

I’m trying to say something about the measure of SL(2, R)/SL(2, Z) by studying a fundamental domain for SL(2, Z) acting on the upper half space (namely that it’s finite)

But I really have no idea how

[tall igloo]

gl! my ability to be useful became even more trivial than the previous statement when you said the word "measure" lol

[mystic ether]

Fair lol

[azure cairn]

Is there a symbol denoting smth js a subgroup of another

Like how $U \le V$ is sometimes used to show U is a subspace of V

[cloud walrus]

Altanis

[azure cairn]

Same question for subrings

[frigid epoch]

that notation is also used for groups, with a variant for normal subgroups

$A \lhd B, A \unlhd B$

[cloud walrus]

PKThoron

[twilit wraith]

The \leq is for any algebraic structure

[marble hinge]

How to determine the minimal number of generators that can generate a particular group? Is that an NP-complete problem computationally? Is there a name for this characteristic of a group?

[azure cairn]

Oh cool

[rocky cloak]

It's called rank, you can look up the rank problem.

(beware of the related, but distinct concept of rank of module/abelian group)

[marble hinge]

Nice, Wikipedia gives a summary in https://en.wikipedia.org/wiki/Rank_of_a_group

The rank problem is decidable for finite groups

And for many groups it’s even algorithmically undecidable! So afaiu it means that there can’t be any algorithm at all to give us the answer

[azure cairn]

Is it more common to denote the group identity as 0/1 or e

[noble nexus]

0 for abelian group, 1 and e are pretty split

I'm team 1 myself

[karmic moat]

i'm team id

[marble hinge]

I am in process of switching from e to 1

[rocky cloak]

If using + for the operation, use 0 for the identity
If using \cdot or juxtaposition, use 1
If using some other symbol, maybe rethink your life choices

[quiet pelican]

What about $\circ$

[cloud walrus]

micoi the group things

[noble nexus]

you don't really write circ for groups tbh

if the operation is composition just use juxtaposition

[rocky cloak]

I mean if you're working with a specific group where the operation is composition go ahead

[noble nexus]

I only find myself using circ for like real valued functions or smth so it's not confused with pointwise product

[quiet pelican]

Very occasionally (I’m mostly thinking of very early intro group theory, where people aren’t necessarily comfortable with “juxtaposition can be used to imply different operations when we have two groups floating around”)

[karmic moat]

what about * or whatever they use for joining two paths

should i also rethink my life choices if i use * for concatenation

[rocky cloak]

x, *, \cdot and juxtaposition are all the same symbol in my book

But you might reflect on your life just in case

[karmic moat]

Okay I'll go for a walk later and reflect on my life

Thank you wise jagr2808

[ripe harbor]

Regarding earlier

This seems to get messy

I'd still like to use this method because Q(pi + i) seems like the most natural pick

[rocky cloak]

If you're able to turn this into a polynomial relation for pi you can use that pi is transcendental.

[ripe harbor]

Oh yeah that should be doable, only looking at the real parts of both sides we would have something like (after subtracting pi): 0 = tilde a1 + tilde a2 pi + tilde a3 pi^2 + ... + an pi^n

And since pi is transcendental, this doesn't have a solution, so pi - i != (...) and we are done

Does that look right?

Thanks!

[ripe harbor]

You or someone else suggested picking a third root of 2 instead of pi + i yesterday

How is that motivated/how would you think of that? [the objective was finding a field that is not closed under conjugation]

[rocky cloak]

Well, Q(cuberoot(2)) is a well known example of an extension that isn't normal. Surgically that is to say it doesn't contain all the cuberoots of 2, just one of them.

But of course the conjugate of a complex cuberoot of 2 is another cuberoot

[ripe harbor]

Ah, ok. Thanks

[errant quartz]

For (b), does it not suffice as a proof to say that any basis of L has this property?

[quiet pelican]

Why wouldn’t it?

[errant quartz]

I don't know, but the book gives two fairly convoluted proofs

So I'm wondering whether there's a reason the book uses different proofs

[next obsidian]

I think the main content is showing they’re algebraic

[errant quartz]

That follows from (a), no?

[next obsidian]

Okay but how do you prove (a) lol

[errant quartz]

dim(L)=n, take n+1 powers of any of the elements and they must be linearly dependent

[next obsidian]

Yeah

So that’s the main content

[errant quartz]

There's a nuance in this that I didn't catch at first, which is that these n+1 powers might not be distinct

Or at least it takes work to show that they are distinct

I think I will come back to that later since the book doesn't mention it and I have to do homework

[quiet pelican]

If they’re not distinct, you immediately have a linear dependence

Between the pair that are the same

[errant quartz]

Oh, true

[ripe harbor]

Why are the elements of K(A) of the form a0 + a1A + a2A^2 + ... + anA^n?

My definition of K(A) is it's the smallest field containing the field K and the set A

So shouldn't it also go backwards since inverses are allowed, A^(-1), A^(-2), ...?

[azure cairn]

ok since everyone likes 1 (including my textbook author) seems ill use 1

i like e though

[ripe harbor]

as an alternative, pick a in L \ K. Then [K(a) : K] >= 2. Repeat (now with L \ K(a) and so on) until you can't anymore, then K(a1, ..., an) = L. Because the degree keeps increasing by at minimum a factor of 2 this will happen

[rocky cloak]

Yes, in general elements of K(A) looks like f(A)/g(A) where f and g are polynomials with coefficients in K and g(A) is non-zero.

However if A is algebraic these can all be written as f(A) where f is a polynomial of degree less than the degree of A

As an example take K=Q and A=sqrt(2).

Then sqrt(2)^-1 = sqrt(2)/2

[azure cairn]

is this what ur referring to?

[ripe harbor]

Oh. We didn't deal with how elements of K(A) look like at all in class, do you think there is a way to justify pi - i is not in Q(pi + i) without that?

[azure cairn]

this is called the structure theorem for cyclic groups right?

[ripe harbor]

Probably not, right? We do need to represent elements somehow

[azure cairn]

or well the point in (c)

[rocky cloak]

That name makes sense at least

[rocky cloak]

I mean if you have
pi - i = f(pi + i)/g(p+i)

You can multiply by g to simplify. But if you don't know what Q(pi + i) is at all then I guess it will be tricky

[ripe harbor]

I guess I could say this just trivially follows from field axioms

Well, rather, K[A] c K(A) and so Quot(K[A]) c K(A) but also the other inclusion holds trivially and so K(A) = Quot(K[A])

[noble flower]

Could someone help me understand what the difference is between module and algebra

My book definse an algebra as: Let K be a field. A K-Algebra is a ring A with an identity and A is a vector space over the field K compatible with multiplication of the ring

Module is defined as: Let A be a K-Algebra. A right A-module is a pair (M, *) where M is a K-vector space and *: MxA -> M (m,a) -> ma, is a binary operation satisfying distributivity, associativity, x1 = x

well is there differene between A-module and module?

is a module always derived from some algebra?

My understanding of an algebra is a vector field that is also a ring where multiplication between the ring and field is defined an associative

thought im not sure if I understand exactly what a module is

I think I'll have to look at another source for modules, I think this book assumes some understanding of module already

[rocky cloak]

A module is usually defined over some ring. You would say R-module to specify which ring when it's not clear. And R could of course be an algebra

[noble nexus]

an R-algebra is just a R-module with a multiplication

[rocky cloak]

In general you have M be an abelian group instead of a vector space

[noble nexus]

book defining modules as being over an algebra instead of over a ring is a bit nonstandard I guess

[rocky cloak]

It's not uncommon to focus on modules over algebras, so if that's all you're gonna deal with you might as well I guess 🤷‍♀️

But yeah, not the standard definition

[glad osprey]

Your understanding is correct if you mean vector space instead of vector field. And a module over a ring is basically just like a vector space, except the scalars are in a ring instead of a field

[noble flower]

hmm I see, yeah I think I was getting confused because the definition online were different from the book

thanks for the help

[glad osprey]

yeah, just forget the definition in the book. Defining a module over an algebra is just confusing if you're just using the ring structure of the algebra anyways. And they've made it extra confusing by mandating that M should be a K-vector space instead of just an abelian group

[rocky cloak]

I mean they are using the algebra structure when they require M to be a vector space. And that does actually simplify it.

Then you can think of a module as an algebra map A -> End(V) for a vector space V.

Instead of having to think about how the algebra structure induces a vector space structure on M (since you absolutely will be using the vector space structure of M alot)

[azure cairn]

Is it correct to consider a homomorpbjsm as a map for which applying the map and applying the group operation is commutative?

Is there a commutative diagram for this lol

[glad osprey]

hmm yeah, good point

[glad osprey]

yes, but you'll get lynched if you post it in this channel 💀

[rapid cave]

This is exactly a commutative diagram

[slim kayak]

Sure.
[\begin{tikzcd}[cramped,sep=scriptsize]
{G \times G} && {H \times H} \
\
G && H
\arrow["{\phi \times \phi}", from=1-1, to=1-3]
\arrow["m"', from=1-1, to=3-1]
\arrow["m", from=1-3, to=3-3]
\arrow["\phi", from=3-1, to=3-3]
\end{tikzcd}]

[cloud walrus]

Kerr

[rocky cloak]

A commutative diagram is just a nice way to write an equation. So yes.

[rocky cloak]

The usual terminology is that the homomorphism preserves the group operation

[knotty badger]

i like to think of it as saying that applying the homomorphism is orthogonal to/independent of applying the group operation

in the commutative diagram they are literally orthogonal in a geometric sense

but they're also "conceptually orthogonal" in that they're independent processes which don't affect each other

[fading acorn]

group homomorphism is a functor between groups seen as groupoids with single object

[tall igloo]

[ripe harbor]

I want to show that $$\mathbb Q(\sqrt 2 + \sqrt 3 + \sqrt 5) = \mathbb Q(\sqrt 2, \sqrt 3, \sqrt 5).$$ I showed before that $$\mathbb Q(\sqrt a + \sqrt b) = \mathbb Q(\sqrt a, \sqrt b)$$ and I thought about using that somehow. $$\mathbb Q(\sqrt 2, \sqrt 3, \sqrt 5) = \mathbb Q(\sqrt 2 +\sqrt 3)(\sqrt 5)$$ but this doesn't seem to help, we still need to show that $\sqrt 5 \in \mathbb Q(\sqrt 2 + \sqrt 3 + \sqrt 5)$

[cloud walrus]

ILikeMathematics

[velvet hull]

How much Galois theory do you know? You can instead show that the degree of the minimal polynomial of sqrt2+sqrt3+sqrt5 over Q is 8, and the degree of the extension on the RHS is 8 (because it is a triple tower of quadratic extensions) as well so they must coincide

[azure cairn]

Nice

Ty all!

[knotty badger]

this same pattern works for all algebraic structures

you have "internal" operations like the group operation, or ring multiplication, or scalar multiplication

and "external" operations via applying homomorphisms

homomorphisms are defined to be orthogonal to/independent of the internal operations

[azure cairn]

Hm interesting. Do formal texts use this terminology?

[ripe harbor]

Haven't covered any Galois theory at this point.
Hm, that's good, so both are degree 8 field extensions - but how does that help

[knotty badger]

not at the moment, but something close to it

[velvet hull]

Both are degree 8, one is contained in the other, so they’re the same

[knotty badger]

there's a categorification of orthogonality/independence called naturality

[ripe harbor]

Oh

[knotty badger]

essentially the motivation for the invention of category theory

[ripe harbor]

This is just linear algebra, right

Thanks!

[velvet hull]

If you don’t have Galois theory then I don’t see a way of doing this without brute force crunching the powers of sqrt2+sqrt3+sqrt5, which does not look fun

[ripe harbor]

Only thing left is finding that degree 8 minimal polynomial of sqrt(2) + sqrt(3) + sqrt(5) lol

[velvet hull]

Yeah that’s kinda hard to do without some Galois theory

[elder stump]

(Or wolfram)

[ripe harbor]

The alternative is probably messing around with 1/(sqrt(2) + sqrt(3) + sqrt(5)) just like how I showed Q(sqrt(a) + sqrt(b)) = Q(sqrt(a), sqrt(b))

[elder stump]

I looked at that just now. Doesn't seem to get you anywhere good

[velvet hull]

That’s secretly the same as computing the minimal polynomial, lol

[quiet pelican]

You can get it via a generalisation of that way
But I have no clue how to explain the generalisation without Galois theory
(I could explain the solution, but it would look like I’m pulling things out of thin air that happen to magically work)

At which point just use your favourite computer algebra system to get the min polynomial

[ripe harbor]

Haha

Thanks!

[ripe harbor]

How does the Galois theory way go? Haven't had that at this point but it's still interesting

[quiet pelican]

Actually here’s a neat way to do it, which needs a fair bit of algebra but not more than the two square root case
Show that Q(sqrt(2) + sqrt(3) + sqrt(5)) has degree 2 over Q(sqrt(2) + sqrt(3)) = Q(sqrt(2), sqrt(3))

Ie, show that sqrt(5) is not in Q(sqrt(2), sqrt(3))

[velvet hull]

The main result here that my idea uses is showing that the minimal polynomial is exactly the product of (x ±sqrt2 ±sqrt3 ±sqrt5) where you have exactly 8 terms ranging over every possible sign permutation

[quiet pelican]

Which isn’t that bad if you know a Q-basis for Q(sqrt(2), sqrt(3))

[velvet hull]

It’s not… too hard to show that this is a polynomial in Q, but showing that this thing is irreducible does need a bunch of nontrivial theory

[quiet pelican]

Actually no

Because my method needs you to already know that sqrt(2), sqrt(3) are in Q(…)

[ripe harbor]

Why? It seems to me like it works out

[rocky cloak]

To show that E has degree 2 over F you need to actually show that E contains F.

[ripe harbor]

oh

[rocky cloak]

Here E would be Q(sq2 + sq3 + sq5) and F = Q(sq2 + sq3)

[ripe harbor]

I saw a way where they take (sq 2 + sq 3 + sq 5)^3, ^5 and ^7, compute all that, then set up a coefficient matrix and write A * (vector with all the roots) = (a, a^3, a^5, a^7), then show that A is invertible and so the roots are expressible in sq 2 + sq 3 + sq 5

But thats not better than CAS-ing the minpoly

[quiet pelican]

That’s how I’d so it without Galois theory

[rocky cloak]

I think you can do it not too bad:

As sq5 is not in E= Q(sq2, sq3) it must be that sq2+sq3+sq5 and sq2+sq3-sq5 share the same minimal polynomial over E. Then they do the same over Q.

Repeat replacing sq5 with the others and boom

[ripe harbor]

Why would you take exactly ^3, ^5 and ^7? And not ^2, ^3, ^4?

[quiet pelican]

I’d do the latter
And I don’t see the methods as meaningfully different

[ripe harbor]

Ah

Maybe the coefficient matrix is not invertible or something in that case

Thanks!

[ripe harbor]

share the same minimal polynomial over E.
Why? Is the justification Galois theory?

[rocky cloak]

No, just observing that in general the for something like K(sq(t)) the minimal polynomial of a + bsq(t) is
x^2 - 2ax + a^2 - tb^2

So doesn't depend on the sign of b

[ripe harbor]

ah

[rocky cloak]

Or I guess said more simply: the minimal polynomial is
(x - a + bsqt)(x - a - bsqt)

I guess the one thing missing is that none of the
±sq2 ± sq3 ± sq5
are equal. But that shouldn't be too hard

[crystal coyote]

What do you think about Fraliegh's first course in abstract algebra book?

[crystal vale]

i think it is good for first course

[azure cairn]

linear maps over vector spaces using + as the group operation are homomorphisms right

[thorn jay]

yes

[low mural]

i lowk don't know exactly how to start this. i know how to find the order of A_20 but my brain isn't fully working rn

[chilly radish]

They order of A_n is always n!/2, but that's not really relevant here. Use disjoint cycle decomposition to build a permutation of order 165=3*5*11. Argue the permutation you found is even and thus lies in A_20

[low mural]

ooo ok i'll try it

thanks!!

[azure cairn]

is this the idea in linalg behind how the induced map $\tilde{T}: V/\ker T \to \im T$ is able to equate cosets in $V/\ker T$ by shifting by some zero-element under addition (i.e. $T(v + w)$ for $w \in \ker T$ is simply $Tv$)?

[cloud walrus]

Altanis

[azure cairn]

cuz this part isn't making a lot of sense to me yet

[tough raven]

Oh you're comparing group theory with linear algebra.

Then yes, they are the same idea, one the incarnation of it in groups the other in vector spaces.

[azure cairn]

ok thatg'll be a useful analogy

quotients were a little rushed imo in the linalg book i read

[cursive spindle]

many such cases

[azure cairn]

so artin defines $\bZ n$ to be the set of all integers divisible by $n$. for example, $\bZ (2) = {\cdots, -4, -2, 0, 2, 4, \cdots}$. but by this definition, isn't $2 \bZ = \bZ 2$??

[cloud walrus]

Altanis

[azure cairn]

why did artin not just write it as nZ initially

[azure cairn]

(Zn is a subgroup of additive Z)

[cursive spindle]

right this property is special

[azure cairn]

how so

should i read more before asking

[cursive spindle]

yes you should

it should come up at some point

[azure cairn]

okay

[cursive spindle]

but keyword is normality

[azure cairn]

this isnt even the section on cosets yet

[azure cairn]

like normal subgroups?

[cursive spindle]

yes

[azure cairn]

i skimmed future pages and saw it

so i guess normal subgroups are in my future!!

[cursive spindle]

yeah normal subgroups are useful and it's what you need to make sense of quotients

[proud vigil]

just wanna add that in general aH != Ha, and they're just writing this as right cosets instead of left because here they happen to be the same

[azure cairn]

ah ok so that's why

OH

[azure cairn]

the group needs to be abelian so that aH = Ha right

[proud vigil]

the cool thing is no

[azure cairn]

cuz ah = ha iff the composition commutes

what

[proud vigil]

aH and Ha are both sets

[azure cairn]

well if the group is abelian then aH = Ha right

[proud vigil]

yes abelian is sufficient

[azure cairn]

and there may be more conditions for aH = Ha

yeah

[cursive spindle]

play with S_3

[azure cairn]

artin explained S_3 really weirdly

or maybe im dumb idk

it was like

lemme get a ss brb

i understand symmetric groups but i dont fully understand what this arrangement of S_3 is meant to convey..?

[proud vigil]

do you have a chapter on free groups

[azure cairn]

i havent come across those yet no

idk if artin has that

[azure cairn]

the symmetric group is, by def, the group of all bijective maps f: T -> T (which are permutations)

so the symmetric group would just be all functions that would permute over some set T

[cursive spindle]

for now you can play around with cycle notation or the weird matrix-like notation

[azure cairn]

S_3 is the group that describes how permutations operate on sets of size 3

[cursive spindle]

{1,(12),(13),(23),(123),(132)}

[dull ginkgo]

Proving the group presentation of S_n and A_n was miserable

[kind temple]

which ones

[dull ginkgo]

Arbitrary n for like the (1, i) and (i, i + 1) cycle presentations respectively

[azure cairn]

for (132), this cycle says 1st element -> 3rd element, 3rd element -> 2nd element, 2nd element -> 1st element, yes?

and thats what the function is

OH

[tidal torrent]

iirc doesnt A_n only have elements of an even # of 2-cycle transpositions

[azure cairn]

i get S_3 now

[cursive spindle]

correct

[azure cairn]

it's the set of 6 permutations that uniquely shuffles any set of size 3

[dull ginkgo]

Yes

[tidal torrent]

S_3 is of order 6 that is non-abelian and non-cyclic i think

ill check rn

[dull ginkgo]

# of cycles is a homomorphism to Z/2 and it’s the kernel iirc

What

[cursive spindle]

another way to write these things is like I said with the matrix like notation. You order the numbers in increasing order in the first row
1 2 3
X X X
then in second column you right the corresponding number where it gets maps to

so (12) would be
1 2 3
2 1 3
for example

[tidal torrent]

using array notation you would have to go from right to left i think

[azure cairn]

so for (132) would be
1 2 3
3 2 1

[vapid vale]

people usually omit the numbers in increasig order

[azure cairn]

i see

[vapid vale]

you just write 321 or 164523 or whatever

[tidal torrent]

yeah (123) is equal to (231)

[vapid vale]

this is preferred notation in like algebraic combinatorics/rep theory

[cursive spindle]

yeah worth mentioning for future uses of group theory, cycle notation (esp transpositions) are super helpful

[tidal torrent]

cycle notation is fun i guess

wasnt it cauchy who came up with this

[cursive spindle]

one thing you can do is take an arbitrary element in S_n and see how to decompose it into adjacent transpositions

there is a nice geometry way to do these things and it comes up in algebraic combinatorics/knot theory

[tidal torrent]

i actually dont know how to compose permutations without using array notation

whenever its written in cycle notation i always have to put it into array notation

[kind temple]

(1 2 3) (1 2) is
1 |-> 2 |-> 3
3 |-> 3 |-> 1
2 |-> 1 |-> 2

so the permutation is (1 3)

[tidal torrent]

oh it is the same idea

makes sense

[azure cairn]

(123)(12)
1 2 3
3 2 1

what did i do wrong

[cursive spindle]

i think that's right

[azure cairn]

so (1 2 3)(1 2) = (3 2 1)?

[cursive spindle]

no read it correctly

[azure cairn]

oh

ummm

oh

OH

(1 3)

yea

[cursive spindle]

yup

[azure cairn]

cuz 1->3 and 3->1

and 2 is fixed

waow

ty for teaching me This

[tidal torrent]

2 maps to itself yeah

if you were to compose the two permutations

[tidal torrent]

if you were to go from left to right of a permutation group of order 2 (S_2?)

that would still be the same as going right to left since the group is abelian right

[tidal torrent]

yep was right

unlike Z6

[ivory anvil]

how much structure does something have to have to have a notion of irrational numbers?

[restive idol]

Do irrational numbers make sense algebraically in general?

[ivory anvil]

is that a question for me?

[azure cairn]

i think you need all the rationals and ordering

and rationals imply you need a field so

[coral shale]

i dont think u need a field

[azure cairn]

wait idk if ordering is needed

let me think about this a bit more

[restive idol]

Nono, I’m curious for myself!

[restive idol]

Why do you need ordering?

[azure cairn]

ok so first i think you would need a field because the def of an irrational number is something inexpressible as a/b, a,b\in Z

[coral shale]

you can easily reword that without division.

[azure cairn]

you would need some notion of division no?

[ivory anvil]

i dont see why you'd need distributivity and stuff for that

it seems like the "integers" could be anything?

[tidal torrent]

i dont think it would be necessary to define a well ordering for all sets

could you say WLOG without needing to include zermelos theorem (no not the game theory one)

by the definition of a rational number isnt the set of rationals is closed under multiplication technically

[restive idol]

I would feel uncomfortable about that. Wouldn’t we like integers to behave similar to how Z does? One big reason people care about it in the first place is to do number theory

So, we should surely give it a ring structure at least

[azure cairn]

an integral domain seems like the lowest structure to start with then?

[restive idol]

Yes I think so too!

[coral shale]

maybe we can venture into Rings of Fractions to avoid fields and still come up with the relevant notion of irrationality
idk

[azure cairn]

an integral domain would have to generate a field of fractions/rationals to determine what is/isnt rational right?

im thinking if you take an integral domain R, take F to be the field of fractions R generates, then any set E for which R and F are subsets of, E \ F is the sett of irrationals

[coral shale]

if u start with a not domain. can we start with one

[restive idol]

I think this set E should be obtained analytically?

[azure cairn]

take E to be the extension field or smth idk

[coral shale]

do we really have to start with an integral domain

[azure cairn]

ive never heard of this

[tidal torrent]

idk why i thought you were supposed to start off with a euclidean domain

h

[cursive spindle]

Related but you can work with algebraic numbers and these have a lot of structure

[azure cairn]

algebraic numbers are numbers that are roots to rational-coefficient polynomials yea?

like opposite of transcendental

[restive idol]

This would give more (maybe less?) than what I’m willing to call irrationals though?

[tidal torrent]

does this have to do with the constructible numbers

[cursive spindle]

Yes

[tidal torrent]

yeah all numbers that are algebraic are solutions of any nth degree polynomial equation

[azure cairn]

id assume less cuz it omits transcendentals

[cursive spindle]

It has more than that yes but no transcendentals

[azure cairn]

oh yeah and it gives u rationals too

youd need to filter those out somehow

but a bigger problem is no transcendentals imo

[cursive spindle]

I mean if you like masochism go ahead and deal with transcendentals

Another related thing which I think is nice to study in the context of transcendentals is periods

[coral shale]

e + pi is rational

[tidal torrent]

is e*pi rational

[coral shale]

if u dream hard enough

[coral shale]

this is a joke btw

[tidal torrent]

ik

ln(-1) is rational too

[cursive spindle]

Periods include all algebraic numbers and some transcendentals (pi, ln(2), zeta(3) etc) too

[ivory anvil]

yes, (e*pi*pi)/(pi)

no need to thank me

[tidal torrent]

i once thought that pi is algebraic because of x^2-pi=0

[ivory anvil]

what is ln(-1)?

[tidal torrent]

ipi i think

[ivory anvil]

right

[tidal torrent]

what a zeta3

i dnot remember

[restive idol]

Riemann zeta function I believe?

[tidal torrent]

oh no

i will stay away from the r word

r words keep the algebraists away

[restive idol]

It’s not? I have no idea then!

[restive idol]

Is it similar to a swear word for you 😂?

[tidal torrent]

yes

i despise it

[coral shale]

why is this channel in such close proximity to #recreational-math

bs

[hollow dust]

it ranges from that to basic algebraic geometry

at some points

makes sense cuz groups rings and fields well everything got those basically 🤣

[coral shale]

where are the groups in my non-associative algebra

[thorn jay]

symmetry groups

or displacement/lmult groups if youre working with (left)quasigroups/loops

[crude kiln]

this guy has access to negative ascii characters

[somber goblet]

just use the 8th bit to encode sign

[ripe harbor]

I want to prove that if E / (E n F) is finite then so is F(E) / F. Any ideas? I thought about picking a finite subset M so that (E n F)(M) = E and then writing F(E) = F(E n F)(M) but that doesn't seem very helpful

[rocky cloak]

Why wouldnt that be helpful?

And Im assuming you mean E/(E n F) is a finite extension, not that E\(E n F) is a finite set

[ripe harbor]

Yes, sorry

[ripe harbor]

Why can we follow F(E) / F is a finite extension out of F(E n F)(M) / E n F being finite?

[rocky cloak]

What is F(E n F) equal to?

[ripe harbor]

Is it F(E)?

[rocky cloak]

Not quite no

What are the elements of E n F?

[ripe harbor]

{x | x in E and x in F}

[rocky cloak]

So they are elements of F...

[ripe harbor]

Yes

Oh

That's just F

[rocky cloak]

So that gives you F(E) = F(M)

[ripe harbor]

So from that we get F(M) / E n F is finite

[rocky cloak]

No, thats not necceraly true

so thats not what you get

[ripe harbor]

Oh wait, we have F(E) = F(E n F)(M) = F(M)

[rocky cloak]

thats right

So F(E)/F = F(M)/F

[ripe harbor]

now we just want to show F(M) / F is finite

Well that's clear because of how we picked M

M is a finite subset

Thanks!

[ripe harbor]

What does it mean to 'characterize' algebraic field extensions with finite field extensions? All finite field extensions are algebraic. Now let L/K be an algebraic field extension. Then there exists M c L with L = K(M). If such an M can be finite, then L/K is also a finite field extension, otherwise not. Is this right?

[limber tapir]

I think it means that you can think of any algebraic field extension as being, in some sense, the "union" of all the finite subextensions. I think more formally its the colimit of all the finite subextensions. This boils down to every element x in L/K of the extension being contained in the finite extension K(x)

Often you can then extend results that are "preserved by colimits" from the finite case to the case of arbitrary algebraic extensions

[ripe harbor]

I haven't had colimits yet. So what you are saying is, given $L/K$ an algebraic field extension, let $$P \coloneqq {U \subseteq L \mid \text{$U/K$ finite field extension}},$$ then $$L = \bigcup_{U \in P} U?$$

[cloud walrus]

ILikeMathematics

[rapid cave]

U/K not L/U

L/U includes U=L, and so its trivial

[ripe harbor]

Right

[rapid cave]

You can edit your previous message fyi

[ripe harbor]

I did, the TeX bot doesnt recognize it anymore though since the "edit time" has passed

[rapid cave]

Yeah this is correct

[ripe harbor]

Hm yeah it makes sense, in the end it's just L having some infinite K-basis and we are unioning a bunch of subspaces together of finite basis, since all linear combinations of the infinite basis are finite atleast one of these subspaces will include them

Thanks

[ripe harbor]

Let $\alpha, \beta \in \mathbb C$ and $m, n \in \mathbb N$ with $\gcd(m, n) = 1$ and $\alpha^m = 2$, $\beta^n = 3$. Show that $$\mathbb Q(\alpha, \beta) = \mathbb Q(\alpha \cdot \beta)$$ and find the minimal polynomial of $\alpha \cdot \beta$.

[cloud walrus]

ILikeMathematics

[ripe harbor]

We know that $(\alpha \cdot \beta)^{m} = 2 \beta^m$ and $(\alpha \cdot \beta)^n = 3 \alpha ^n$, so $\alpha^n$ and $\beta^m \in \mathbb Q(\alpha \cdot \beta)$. Maybe now WLOG let $n \leq m$ and consider $(\alpha + \beta)^n$?

[cloud walrus]

ILikeMathematics

[rapid cave]

Show that x^n-2 and x^m-3 are irreducible over Q. This shows that Q(a,b) has degree nm over Q. Show the same for Q(ab).

[ripe harbor]

Oh, that's good. We have

[Q(a, b) : Q] = [Q(a, b) : Q(a)] : [Q(a) : Q] = nm

since the two polynomials you gave are the minimal polynomials. And then we need to find a minpoly for alpha * beta, which should be

X^(mn) - 2^n 3^m

Thanks!

[rapid cave]

you should work out the details on why these are polynomials are irreducible

[ripe harbor]

Where exactly did we use that gcd(m, n) = 1?

[rapid cave]

[Q(a,b):Q(a)] might be less then n otherwise

Q(a,b) has two subfields Q(a), Q(b), which are of degrees m,n respectively. so [Q(a,b):Q] is divisable by both m,n and hence is divisable by nm.

[ripe harbor]

Wait, this line

[Q(a, b) : Q] = [Q(a, b) : Q(a)] * [Q(a) : Q] = nm
isnt quite correct as it stands, is it? I mean the intermediate step, we know that [Q(a) : Q] = m but we dont know yet if [Q(a, b) : Q(a)] = n, do we? We just know [Q(b) : Q] = n

So yes, its your justification, n and m both divide the LHS

But how do we know that it's not larger than nm

[rapid cave]

Q(a,b) is the compositum of Q(a) and Q(b), so you can create a Q-spanning set of size nm from a Q-basis of Q(a) and a Q-basis of Q(b)

if a_1,...,a_m is a Q-basis of Q(a) and b_1,....,b_n is a Q-basis of Q(b) then {a_i * b_j} where 1 <= i <= m, 1 <= j <= n spans Q(a,b) over Q

[ripe harbor]

I haven't done much with the compositum yet, is there a way to see this without it too?

[ripe harbor]

Intuitively I see why this is true but we haven't had it in class

[rapid cave]

how did you define Q(a,b) then?

[ripe harbor]

Smallest field/intersection of all fields containing Q and the set {a, b}

[rapid cave]

ok

The set of all Q-linear combinations of a_i * b_j is a field like that, so Q(a,b) is contained in it.

[ripe harbor]

oh, yes

So the basis of it has <= nm elements

Ok, yeah that's it. Thanks!

[rapid cave]

@ripe harbor also, returning to your previous argument,
The minimal polynomial of b over Q(a) divides the minimal polynomial of b over Q and this means that [Q(a,b):Q(a)] <= [Q(b):Q].

[azure cairn]

in my textbook, will this ever be connected to linear maps

i recognize this as the following being equivalent:
$$T(u) = T(v)$$
$$u - v \in \ker T$$
$$u \in v + \ker T$$
$$u + \ker T = v + \ker T$$

[cloud walrus]

Altanis

[azure cairn]

for a linear map T and vectors u, v

and using the vector space as an additive group, $a^{-1}b = -a + b = b - a$

[cloud walrus]

Altanis

[delicate orchid]

This is just the first isomorphism theorem for two different structures

You’ll see it’s true for rings and modules and algebras and so on and so forth

[azure cairn]

OK

butt. my understanding is right that this is the same thing in linear algebra, where u consider vector spaces as groups under + and linear maps as homomorphisms?

[delicate orchid]

Yup thats what I said

[azure cairn]

and all abelian groups are normal cuz $gag^{-1} = gg^{-1}a = a$, so in particular the abelian subgroup of additive integers $\bZ n = n \bZ$

[cloud walrus]

Altanis

[azure cairn]

i like this interpretation a lot more now that i've worked with group isomorphisms. since isomorphisms (which are homomorphisms) are "orthogonal" to the law of composition, if you have an isomorphism phi: G->G', you can simply do a law of composition in either G or G' and uniquely identify it with that same composition in the other group by mapping through phi or phi^-1

although i guess ive seen this in linalg by like doing an operation on one set of coordinates and applying a change of basis, which is an isomorphism

[azure cairn]

oh shit i just realized conjugation resembles change of basis lmao

[cursive spindle]

Yupie

[knotty badger]

Yay

[azure cairn]

i understand the if and only if is not required for the symmetric condition (Since you can just swap the roles of a and b and thatll be effectively the converse which is true), but is it not sloppy to write if and not iff?

[knotty badger]

Are you more convinced about homomorphisms are “orthogonal” to the group operation in the sense I outlined?

[knotty badger]

No I think it’s fine

[azure cairn]

well yes it's a true statement no?

they are independent operations

[knotty badger]

Mhm

[knotty badger]

One way I like to frame it is that

Homomorphisms are conceptually orthogonal to the group operation

In that they’re independent operations

But they’re also geometrically orthogonal in the commutative diagram

The two maps are at right angles to each other

So in a sense, the cat theory point of view merges the conceptual and geometric sides of orthogonality

[azure cairn]

ye

[proud vigil]

it's usually better to have weaker premises / definitions and stronger conclusions when they don't affect the argument

regardless, simplicity / conciseness is always better, iff doesn't convey anything meaningful here that if doesn't already convey

maybe as an analogy, to show that a nonempty subset of a vector space is a subspace, you only need to show closure under addition and scalar multiplication, even though this is equivalent to closure under all linear combinations, and the former is often simpler to show

[azure cairn]

How'd you know I know linear algebra

But yes I get it

This is a definition though, not a thm. I feel like it can just be said that the if case is sufficient

But I'm being pedantic

[proud vigil]

equivalence relations have useful properties, showing that something is an equivalence relation lets you invoke theorems about them

[ripe harbor]

divides the minimal polynomial of b over Q
Why?

[supple pecan]

for any cyclic group G with generator a, is it true that a^-1 is always a generator too?

will those be the only 2 generators? or are there weird groups where theres many or something

[supple ice]

yee

[untold belfry]

If the group has prime order, then every nontrivial element generates it.

In general, the order of a^n is |a|/gcd(n,|a|).

In G is finite and |a^n| = |a|, <a^n> = <a> by a cardinality argument.

If G is infinite, G is the integers, and the only two single generator sets are 1 and -1. You can determine what other subsets generate using Bezout's identity.

Returning to the finite case, a^n generates <a> iff |a| and n are coprime.

[marble hinge]

There may be any any number of generators

I.e. there are groups that are generated by exactly N generators, where you choose N

For example, a free group on N generators

[rapid cave]

All polynomials that have b as a root form an ideal of Q(a)[x].
The minimal polynomial of b over Q is such polynomial so its in that ideal.

[ripe harbor]

In cycle notation, how do we do composition? I.e. (123) means we map 1 -> 2, 2 -> 3, 3 -> 1, right?
Now how do we quickly determine (123)(132)?

Is this 1 -> 3 -> 1, 3 -> 2 -> 3, 2 -> 1 -> 2 and so it is e = (1)?

i.e. from right to left?

[rapid cave]

yeah

[ripe harbor]

Thanks!

[wraith cargo]

the idea is that (132) and (123) are bijections from {1,2,3} to itself so the notation reflects this i.e. you compose them as functions

[ripe harbor]

Yeah, thanks

[marble hinge]

For (123)(132) you can also notice that one is the inverse of the other

Hence the product is e

[ripe harbor]

How do you notice that?

[marble hinge]

The inverse is a cycle written backwards

[ripe harbor]

oh, so (321)

[marble hinge]

(123)’s inverse is (321)

And you can also cyclically rotate it, so can start with 1

And so (321) = (132)

[ripe harbor]

Thanks! Also, is there a general method for determining all subgroups of a group? Like for S3, we have {e, d, d^2}, {e, s1}, {e, s2}, {e, s3} and the trivial one, {e}. How do we know those are all?

[marble hinge]

No general method, afaik. There are many techniques and properties that one can use.

The first one is the Lagrange theorem

Which you probably know

[ripe harbor]

Yeah, |G| = |H| * [G : H]

[marble hinge]

Yeah, so this lets you deduce that there are definitely no subgroups of orders 4,5, so no need to check that anymore

[ripe harbor]

Ah, right

[marble hinge]

You can also draw a subgroup lattice

[ripe harbor]

Is that the subgroup diagram?

[marble hinge]

Because a subgroup of a subgroup is a subgroup - it enables you to see the structure of subgroups better

[dense hearth]

Is there mathematicians here and how do I find them

[supple ice]

yeah but they’re mostly a conjectural subset no explicit examples are known

[dense hearth]

What you trying to say lol

I asked about graduate students in mathematics field (not applied)

[quiet pelican]

Why?

[marble hinge]

So we know that non-trivial subgroups of S_3 can only be of order 2 or 3. And we know that identity is there, so there remains a free slot for one or two elements. And the subgroup must also be closed. And this should be enough to persuade us that there are only subgroups that you listed i.e. various 2-cycles and e, and two 3-cycles and e

[supple ice]

my bad I was answering in maximal generality graduate students in pure math should be easier to detect at least up to isomorphism

[dense hearth]

I'd like to have mutuals online to know how they made it, their secret to succeed..etc

[dense hearth]

Alright I'm first year student

[marble hinge]

This is the one for S_3:

[supple ice]

where

[dense hearth]

Wym by where, country or continent or u mean what field

[ripe harbor]

Right, thanks! Probably pretty helpful when there are more than 1 layers

[supple ice]

all of that

[marble hinge]

Yeah, D&F promise that there are some “hair raising examples” of such diagrams even for groups of relatively small orders :)

[dense hearth]

North Africa, Algeria, Maths field

[supple ice]

okay

[dense hearth]

You?

[supple ice]

I'm french

[marble hinge]

I couldn’t find this book easily

[dense hearth]

You study in France

[supple ice]

but I think it's not the best place to talk about that

[dense hearth]

Alright

[candid patrol]

loser 🤣

[supple ice]

why lol

[supple ice]

you too french

[candid patrol]

oui oui baguette

[supple ice]

gros golmon va

[candid patrol]

he said i'm so good in maths

[supple ice]

lmao hell nah

[supple pecan]

i see thank you, my course is introductory so idk what some of these mean but i get the general picture

[supple pecan]

i see thank you so much

[marble hinge]

find an element g that takes your sigma "out of the subgroup"

i.e. find a $g \in G$ such that $g \sigma g^{-1} \notin H$

[cloud walrus]

dying_sphynx

[marble hinge]

for example || sigma=(12), g=(13), then (13)(12)(13) = (23), so not in H, so H is not normal ||

not sure if it's "the quickest" way, likely not, but it's quick enough :)

also note that for this "closure with respect to conjugation" normality check, one only needs to check generators. So if you know generators h of H and g of G, then you can only check those pairs for belonging to H, i.e. check that ghg^{-1} \in H just for generators

the same, but more verbosely, from D&F: