groups-rings-fields

[south patrol]

I wasn't sure if they meant bilinear as in "maps to k" or maps into any vector space, hence what I wrote, but then I realise the two are kinda obviously equivalent anyway lol as any nonzero vector space has k as a direct summand lol

[tardy hedge]

Thanks

When i have the time i really want to just go through like all the exercises in dummit and foote for this. It has always bugged me that i havent really understood it properly

[rocky cloak]

Well, if it's over a field, then m(x)n is only 0 if m or n is 0 anyway

[frigid epoch]

There are a few kernels of the concept you might need:
\
Bilinear maps from $M \times N$ are linear maps from $M \otimes N$.
\
In fact, there's a bilinear map $h: M \times N \to M \otimes N$ that induces a bijection of maps $h^*: \rm{Bil}(M \times N, P) \to \rm{Hom}(M \otimes N, P$ for all $P$, which is natural in everything ever because category theory.
\
$M \otimes N$ consists of linear combinations of pure tensors $m \otimes n$. If we're in Vect$_k$, then you can write down a cool basis if you have bases for M and N, and the dimension of the tensor product will be the product of dimensions (as opposed to the sum, like it would be for direct sums or [cartesian] products).
\
The construction of $M \otimes N$ will take every $m \otimes n$ as a basis element and generously quotient out relations that you want. It's highly worth going through this, especially if you don't have too much experience with generator-quotient constructions.
\
For noncommutative rings and modules over them it gets a bit icky I believe.

[cloud walrus]

PKThoron

[frigid epoch]

Christ my texxing skills 😄

[arctic trail]

Suppose we have a non-commutative ring R, is it true that if I = I^2 is an idempotent left ideal which is finitely generated, then I is a direct summand of R?

I know this is true for commutative rings, the proof is rather intuitive

we can use transform I^2 = I and a set of finite generators x = [x1,...,xn] to construct a matrix A over I^(nxn) such that Ax = x
hence (1-A)x = 0, so multiplying by the adjoint we get that det(1-A)x = 0

[south patrol]

This is true. Idk why I assumed field lol

[rocky cloak]

Consider the ring of upper triangular matrices
[k, k; 0, k]
and let I be the ideal
[k, k; 0, 0]

Then I is idempotent, but is not a direct summand as a left module.

Somewhat interestingly I is actually a direct summand as right module (I is a twosided ideal).

[vivid kestrel]

sorry if thats inappropriate but thanks for everyone who helped me with various questions the last week, passed my exam and couldn't have done it without so much free patience and help

[thorn jay]

yooo congrats dude!!

[tough raven]

Consider the R-bilinear map (m, n) ↦ m (⨯) n. Clearly if it takes (m, n) to 0, m (⨯) n = 0.

[tough raven]

For vector spaces over a field, yes. For modules over a (commutative) ring, no. You need maps into all modules for it to be true. This isn't because of the tensor product in any way, but just because modules may not be "determined" by linear maps from M to R.

For example, any ℤ-linear map f: ℚ → ℤ is 0 (simply because f(1) is divisible by every positive integer n, hence 0), so if R = ℤ and M (⨯) N is isomorphic to ℚ (for example, because M = N = ℚ), m (⨯) n will be killed by all bilinear maps to ℤ, even though it need not be 0.

[south patrol]

Indeed agreed

Ngl I kinda assumed we were limiting to maps into the ground ring as I thought otherwise it was much easier

[tough raven]

bump

Should I ask in #linear-algebra?

[arctic trail]

can you write this in TeX

it's hard to understand in plaintext

[tough raven]

Let $V$ be a vector space over a field $K$ and $k$ a positive integer.
For $\omega \colon V^{k-1} \to K$ alternating multilinear, define
\begin{equation*}
  \iota_{\omega} \colon \bigwedge^k V \to V \colon v_1 \wedge \dots \wedge v_k \mapsto \begin{multlined}[t] \omega(v_1, \dots, v_{k-1}) v_k + \dots \\ + (-1)^{k-1} \omega(v_2, \dots, v_k) v_1. \end{multlined}
\end{equation*}

For any subspace $W \subset V$, for $\eta \in \bigwedge^k W$ we have $\iota_{\omega}(\eta) \in W$ for any $\omega$.
Is the converse true: if $\eta \in \bigwedge^k V$ and
\begin{equation*}
  W \coloneq \{ \iota_{\omega}(\eta) \,\vert\, \omega \text{ alternating $(k-1)$-multilinear on $V$} \},
\end{equation*}
must $\eta$ lie in $\bigwedge^k W$?

[cloud walrus]

Raghuram

[rocky cloak]

I guess pick a basis
e1, e2, ..., em, ..., en
Such that e1, ..., em is a basis for W.

Then a basis for Wedge^k V is e(i1)^...^e(ik) (i1 < i2 < ... ik)

Say eta is not in Wedga^k W. Then eta is supported on
e(i1)^e(i2)^...^e(j)^...^e(ik) where e(j) is not in W. Then pick w such that w(e1)^w(e2)^...^w(e(ik)) = 1 and 0 for on other basis vectors. Then
i_w(eta) = ± e(j)

(sign depending on j)

[arctic trail]

\begin{align*}
\iota_{\omega} \colon \bigwedge^k V &\to V\v_1 \wedge \dots \wedge v_k &\mapsto \omega(v_1, \dots, v_{k-1}) v_k - \omega(v_1, \dots, v_{k-2},v_k)v_{k-1}+ \cdots + (-1)^{k-1}\omega(v_2, \dots, v_k) v_1
\end{align*}

[cloud walrus]

Trivial Lemma

[tardy hedge]

To show M->N is a R-bimodule homomorphism, we need to check that its a homomorphism through the left R-action and then also a hom through the right R-action?

[tardy hedge]

Is the tensor product A(x)A taken over F? It needs to be taken over F for A(x)A to be an F-algebra right?

[nova cliff]

yeah I'm pretty sure one should tensor over F

[tardy hedge]

Also, i have no idea what an A-bilinear splitting means.

[nova cliff]

not totally sure either

I would guess that you need A-linearity on the left and on the right

namely, Delta(abc) = a Delta(b) c, where A tensor A is an A-bimodule in the obvious way

while there's a pause, maybe I'll ask about this lang exercise
for (a), there must be a typo here because the dimension of the matrices does not agree with the degree of the characteristic polynomial, no? I guess I should replace GL_2(C) with GL_6(C)?

[tough raven]

It means that Delta is A-bilinear, as @ dfoiler said.

[tough raven]

Yeah, probably.

Maybe M_6(C)..

[tardy hedge]

What does the “splitting” mean?

[tough raven]

A short exact sequence 0 -> M -> N -> P -> 0 is said to split if there is a map P -> M such that P -> M -> P is the identity; the map P -> M is called a splitting.
Now the multiplication map mu: A (x) A -> A is surjective and A-bilinear, so if we call the kernel I, there is a short exact sequence 0 -> I -> A (x) A -> A -> 0 of A-bimodules. By definition, A is separable if and only if this short exact sequence splits and in this case, Delta is the splitting.

[frigid epoch]

fwiw in a splitting sequence, $N \cong M \oplus P$

[cloud walrus]

PKThoron

[tardy hedge]

Did u mean a map P->N?

[frigid epoch]

yeah

as an exercise, you can write down the iso

[tough raven]

Let $R$ be a commutative ring, $V$ an $R$-module, and $v$ a linearly independent element of $V$.
For every $k$, there is an $R$-linear map $\varphi_v \colon \bigwedge^k V \to \bigwedge^{k+1} V \colon v_1 \wedge \dots v_k \mapsto v \wedge v_1 \wedge \dots \wedge v_k$.
Is the chain complex of $R$-modules
\begin{equation*}
  0 \to R = \bigwedge^0 V \xrightarrow{\varphi_0} \bigwedge^1 V \xrightarrow{\varphi_1} \dots
\end{equation*}
exact?

It is not too difficult to check that, denoting $W \coloneq V/Rv$, $\ker(\bigwedge^k V \to \bigwedge^k W) = \operatorname{im}(\varphi_{k-1})$, so this comes down to showing that the map induced by $\varphi_k$, also denoted by $\varphi_k \colon \bigwedge^k W \to \bigwedge^{k+1} V \colon w_1 \wedge \dots \wedge w_k \to v \wedge w_1 \wedge \dots \wedge w_k$ (which is well-defined because it vanishes if any $w_i \in Rv$) is injective.

It isn't really necessary to assume that Rv admits a complementary submodule instead of taking W = V/Rv to formulate the question, but maybe it helps prove this. 🤷

No, wait: I know how to do this when it splits, and I'm curious if it's true when it doesn't.

[cloud walrus]

Raghuram

Raghuram

[rocky cloak]

What happens if R=Z and V=Q?

Then the sequences just looks like
0 -> Z -> Q -> 0
right?

[tough raven]

Right.

[rocky cloak]

Even just V = Z, but picking the element to be 2 instead of 1 should cause trouble

[white oxide]

How can I show that dim R[x, y]/(x^2 + y^2) is infinite

Well I guess 1, x, x^2, x^3, x^4... can still somehow be considered basis elements in some way

Oh yeah I guess I could just assume that there is a finite basis p_1(x, y), ..., p_N(x, y), but then x^M cannot be a linear combination of these elements where M is the maximum of the degrees which appear in the p_i

[minor fulcrum]

is it known how many subgroups there are of S_n?

[tough raven]

I know how many normal subgroups there are.

More seriously, at least as many as there as isomorphism classes of groups of order n, so a lot (but perhaps compared to n!, only a lot, not a lot).

[vast stump]

some info on this https://oeis.org/A005432

[minor fulcrum]

oh I should have checked oeis, good call

n^2 is way slower than I'd expect

[candid patrol]

Same for An 😎

[cinder fox]

If $R$ is a ring and we have $3k=0$ for $k\in R$, can we conclude that $k=0$?

[cloud walrus]

VirtualCode

[cinder fox]

Or rather, when can we conclude that k=0?

[crystal vale]

No

Take ring Z/3Z then for every k in Z/3Z, 3k = 0, @cinder fox

[crystal vale]

If the ring has char 3, then k ≠ 0, but if the ring does not char 3 and if it is an integral domain then we can say k = 0.

[thorn jay]

Also in Z/nZ where n is not divisible by 3

In general thats a nonsufficient requirement for the characteristic

[south patrol]

Interesting

[thorn jay]

n^2 compared to n! Is so slow,
Also gives a very weak upper bound to the number of groups of order n lmao

[rotund aurora]

the bound is not n^2, it's c^n^2

[frigid epoch]

what have you taken before that's non-advanced?

I think abstract algebra is pretty legit IF you have experience with axioms and proofs

[rotund aurora]

if you like it, yes

[frigid epoch]

groups, rings and fields are all not too difficult to work with, but they do leave the comfort of the real numbers that you experience in early analysis/calculus and linear algebra

dunno if this applies to every algebra course but:
mine was motivated by the question of: can polynomials of degree 5 or higher be solved?

oh it's fine then

go ahead and take algebra imo

yeah that's true, although not 100% true

rings will very much look like the integers

and fields will very much look like the rationals, reals and complex numbers

groups leave common intuition quite a bit, but at the same time it's not too hard to get an intuition for them (especially since many groups turn out to be groups of numbers after all)

also this server has frequent abstract algebra lectures tho i'm not sure where they're at right now

here

I didn't learn algebra by a book, just by a great lecture so I can't help there

[lusty marlin]

Dummit and Foote

[frigid epoch]

yeah abstract algebra

groups, rings, fields

[glad osprey]

Fraleigh or Gallian are good options if you want something a bit more beginner friendly

[thorn jay]

Highschool algebra isn't even classified as "real" math, because you're not really formally proving anything.
But yes, if you do any math beyond highschool, then "algebra" will mean "abstract algebra".

In that sense highschool algebra isnt algebra, and just referred to as "algebraic manipulation"

[tardy hedge]

When you have A and B R-algebras, the tensor product A(X)B over R is also an R-algebra. How is R mapped into A(X)B?

[coral spindle]

r |-> r (x) 1 = 1 (x) r.

[tardy hedge]

that is using the fact that A and B are R-algebras , for the multiplication there right

[coral spindle]

I guess so?

The important part is that it's a tensor product over R I think.

[tardy hedge]

Ok

what is it that makes the multipliation on simple tensors (a(x)b)(c(x)d) = (ac) (x) (bd) well defined here?

Ohh normally you dont even have multiplication in M or N for some M(x)N because usually they are just modules

[south patrol]

Well this can be defined via the universal property like

(A (x) B) (x) (A (x) B) = (A (x) A) (x) (B (x) B) -> A (x) B

[minor fulcrum]

yeah I misread, it's it's log(#subgroups) = n^2

that makes more sense

[thorn jay]

Yeah lmao it does

[frigid epoch]

are simple groups exactly the groups with no abelian quotients?

it must be, right?

given that [G,G] is always normal and G/[G,G] is always abelian

so the fundamental group of the poincaré homology sphere is simple

[thorn jay]

[G, G] can be simply G

[frigid epoch]

yeah, ik, by "no abelian quotients" I kinda swept the trivial group under the rug

but if G = [G,G], is it simple?

[thorn jay]

Dont think so

[bitter rover]

No

Take the Cartesian product of two non-Abelian groups

[thorn jay]

You mean two perfect groups

[bitter rover]

Yes, yeah, realized that was too much.

[thorn jay]

But yes, for example the direct product of A_5 with A_5 has trivial abelianisation but isnt simple

[bitter rover]

But like A_5 × A_5 will do, being the smallest perfect group

yeah

[thorn jay]

Lmao

Great choice

[frigid epoch]

ah ok

[bitter rover]

A_5 always in the chamber

[thorn jay]

So based

[frigid epoch]

https://en.wikipedia.org/wiki/Binary_icosahedral_group

apparently this is the funda group of the poincaré homology sphere

[rocky cloak]

So Cp are abelian simple groups. And you can have non-simple groups without abelian quotients

[empty perch]

Hey folks, just a quick reminder that the Abstract Algebra Lecture series is meeting regularly again ^_^
We will meet to discuss inner product spaces in about 15 minutes over in the ⁠⁠#1055201711679082516. More information can be found in our thread: ⁠⁠#1317307081535000606

[tardy hedge]

Why in the construction of tensor product we want the relation (m,rn) = (mr, n) to be satisfied

[frigid epoch]

because for any bilinear map, f(m,rn) = r·f(m,n) = f(rm,n)

[tardy hedge]

Thanks

[fast pendant]

What intuition should I have for the alternating group? I know its definition, I know its properties, but I don't get what it means for a permutation to be even.

[thorn jay]

Any permutation can be decomposed into a series of swaps

It turns out that the parity of the number of these swaps (called transpositions) does not depend on the specific decomposition

This means that one can divide permutations into those which you get after an even number of swaps, and those which you get after an odd number of swaps, and there's something fundamentally different about them too

If you only track the parity of the permutation, then it behaves exactly like how it would for numbers

This means that the even permutations form a subgroup, which is even normal

[fast pendant]

Yeah I know how it's defined

if this makes sense I understand the what but not the why

[thorn jay]

Let P is a permutation function on a set S. For a pair (i,j) of elements in S such that i < j , if P(i) > P(j), then the permutation is said to invert the order of (i,j). The number of such pairs is known as the parity of the permutation. If permutation inverts even number of such pairs it is an even permutation else it is an odd permutation.

https://math.stackexchange.com/questions/361822/odd-even-permutations

[surreal swan]

I am loving Dummit and Foote Abstract Algebra book
Going through it slow since employed
And every now and then I get stuck on a problem for a couple of weeks
But they are always actually achievable tbh, and really force you to develop different lines of thought

[thorn jay]

So, how you can think of this is as follows: if you have two rows of n dots and connect them according to the permutation, then the number of pairwise crossings of lines determines if the permutation is even

[fast pendant]

Yeah

I can identify even permutations, I don't think that's helping much. Thank you though

[thorn jay]

I've seen it critically used in one place at least

May not be useful as intuition what they ate

Are

But

http://www.math.ubc.ca/~cass/courses/m308-02b/projects/grant/fifteen.html

[rocky cloak]

One perspective could be to think about permutation matrices.

Even permutations are exactly those with determinant 1 and odd are those with determinant -1.

So even permutations are orientation preservering, while odd ones are orientation reversing

[swift tundra]

This is also kind of like associativity of multiplication

In the sense that if we take R tensor R, this condition is literally associativity

[fast pendant]

Thanks - found this, which is what I was looking for I think

Orientation preservation is a good way of thinking about it

[tardy hedge]

Anyone notice that chat gpt is actually pretty good at helping with math now

If u disagree i feel like u kinda in denial lowkey

Sorry i wanted to say it here cuz my algebra homies r here

[elfin wraith]

I can’t say it’s something I regularly try

But I imagine it massively depends on what you define as help and what specific things you ask about

[thorn jay]

Its great at reproducing what it sees on the internet into logically sounding sentences

[elfin wraith]

That is the general idea

[tardy hedge]

Try it, its a lot better these days than when i was playing around with it for math like a year ago

Make sure to press the “Reason” button

[fast pendant]

Question

I thought that a (commutative) ring with no proper ideals must be a field. What's a local ring? Is that for the non-commutative case and chatgpt is being dumb

[chilly ocean]

I’m gonna test it on a question I’ve tested it on before and see if it answers any better

[chilly ocean]

It still got it wrong

Question was “is there a fiber bundle p:E to B where the fiber of E is not T_4(normal Hausdorf), but E itself is?”

[fast pendant]

It's much better since I last used it

It took me a long time to find a nitpick, that it didn't explain well where the open cover came from and that it doesn't need contradiction

Definitely a huge improvement since a year ago when it couldn't do basic vector algebra

[glad osprey]

Is that chatgpt? A simple commutative ring is always a field, so I guess it's technically correct that it's either a field or a local ring 🙃 and a simple ring is local, so it's doubly correct

Chatgpt be like: an even number is either divisible by 2 or by 1

[rocky cloak]

Simple rings are not in general local.

The only simple local rings are fields

And this doesn't seem like it's pretty good at helping with math

[frigid epoch]

What is a simple ring?

[rocky cloak]

A ring without any nontrivial quotient rings

Or equivalently one without any non-zero proper two-sided ideals

[frigid epoch]

Oh dear, that sounds strict

Any examples outside of commutative fields?

[rocky cloak]

nxn-matrix rings over division rings

[frigid epoch]

Ah

Is that all of them?

[rocky cloak]

Those are all the Noetherian ones

[frigid epoch]

Sick

What about non-noetherian ones?

[rocky cloak]

Another could be, let V be a countable dimensional vector space, let R = End(V) and let I be the ideal of R of endomorphisms with finite dimensional image.

Then R/I is simple

[frigid epoch]

I really wish they did more noncom algebra in uni

[frigid epoch]

"Cocountable endonorphisms"

[rocky cloak]

Then there's the weyl algebra (ring of differential operators on the polynomial ring)

That's all the examples I know

Well, you can modify the End(V) example for other infinite cardinals, but other than that

[glad osprey]

Oh, I see. I was thinking that if (0) is the only ideal then it is maximal, but I guess that's only true in the commutative case

[rocky cloak]

A ring is local if any of the following equivalent conditions are satisfied
It has a unique maximal left ideal
It has a unique maximal right ideal
The set of non-units forms an ideal
If you mod out the Jacobson radical you get a field

You can have only one two-sided ideal while still having plenty of maximal left ideals

For example the 2x2-matrix ring over an infinite field has infinitely many maximal left ideals

[glad osprey]

aha, that makes sense

[tardy hedge]

If you take a polynomial (say over Q), and then in its splitting field factor it completely, can that polynomial still be considered a polynomial over Q (even tho some linear factors x-a might not have a in Q) since if you multiply all the split factors together you’d get back what you started with?

[frigid epoch]

I suppose yeah, in the same way that $\sqrt{2} \cdot \sqrt{2} \in \mathbb{Q}$

[cloud walrus]

PKThoron

[proud python]

Is a real polynomial that splits over C still a polynomial in R?

[tardy hedge]

Yea i guess so^

[gleaming sierra]

okay so im learning about galois fields ahead because of some essay, and i don't have much previous knowledge on algebraic structures so i may sound stupid ☠️ but can someone explain to me why we even need irreducible polynomials to construct a galois field? from what ive seen the elements of any gf(p^k) are all just all polynomials up to degree k-1 with coefficients in gf(p), i dont really get the role of the irreducible polynomials.

[rocky cloak]

I mean, the polynomial never changes

[glad osprey]

Not sure if this is what you're asking, but for a field F and a polynomial p(x) in F[x], F[x]/(p(x)) is a field if and only if p(x) is irreducible. So to construct GF(p^k) we must quotient by an irreducible polynomial in GF(p)

[gleaming sierra]

yeah that's pretty clear, however the elements of GF(p^k) are the same no matter what irr. polynomial we use and we can find them by simply listing all polynomials up to degree k-1 with coeff. in GF(p) so i kind of feel like i am missing something

[glad osprey]

Yeah, in some sense the elements of GF(p^k) are the same, because it's always a k-dimensional vector space over GF(p). But the irreducible polynomial tells you exactly how to multiply elements of GF(p^k). Like, the elements of GF(p^k) are not just polynomials of degree less than k, because that's not closed under multiplication; the elements are cosets of the irreducible polynomial p(x)

[gleaming sierra]

OHHHHH you're right

i forgot about multiplication being mod p(x), thank you a lot!!

[tough raven]

You need to choose a polynomial of degree k to tell you how to multiply polynomials of degree up to k-1. You'll only end up with a field if the polynomial you choose is irreducible.

For example, think about the construction of ℂ from ℝ. The elements of ℂ are polynomials in i of degree up to 1, but when you multiply two of them you'll get terms with i^2. To reduce this back to degree up to 1, you need to pick a relation i^2 = ai + b to use to reduce it. The reduction will end up being polynomial long division by the polynomial in i, i^2 - ai - b, so you will have constructed the ring ℝ[i]/(i^2 - ai - b). The abstract algebra reveals that this is a field iff the quadratic polynomial is irreducible.

Of course, for ℂ and ℝ, the standard choice is a = 0, b = -1, so that the polynomial is i^2 + 1.

(Example: if you used i^2 = 1 to reduce, because i^2 - 1 = (i+1)(i-1) is not irreducible, (i+1)(i-1) would reduce to 0 even though neither factor reduces to 0, so the cancellative law would fail in the ring you came up with.)

[gleaming sierra]

thanks for the explanation!!

[tardy hedge]

W explanation

[native magnet]

Aight I am smoll baby in this world, but I must understand. I feel like I partly kinda sorta understand fields. I am currently grappling with groups. What is a center of a group???????

[velvet hull]

the center of a group is the set of all elements of the group that commute with everything in the group

you can check that it is always a subgroup, and in fact a normal subgroup

even better, if you know some more group theory, then G/Z(G) is isomorphic to Inn(G)

[native magnet]

So is the center a subgroup?

[velvet hull]

convince yourself that the things I said are true

[native magnet]

Haha. That is wild

I believe you, but is the center a single element???

[prisma ibex]

no

it's always a subgroup yes

if the group is Abelian then the center is the whole group

[native magnet]

Huhhh

Okay Abelian is it can add

[prisma ibex]

uhhh

[native magnet]

Baby 👶

[prisma ibex]

Abelian means the group operation is commutative

[native magnet]

Okay cool, so it can move around

[prisma ibex]

ab=ba yeah

[native magnet]

Okay cool

[prisma ibex]

the center is just like, the largest subgroup where you can do this

sometimes that's the whole group, sometimes it's the opposite

[velvet hull]

well, the largest subgroup that commutes with everything else* but yeah
its a common enough misconception that I had to point it out, this does not rule out there being bigger commutative subgroups

[prisma ibex]

yes

[native magnet]

So if it is commutative the whole group is the subgroup

[prisma ibex]

yeah trivially

[native magnet]

So what's the purpose of the center? Is it more useful in the non-Abelian?

[prisma ibex]

well one reason to care is this is sort of the universal way to extract an Abelian group from any group

[native magnet]

Ohhh

so if you know the center of any group, then you can commute the items in the center

[prisma ibex]

I mean sure

one result you prove early on in group theory is if G/Z(G) is cyclic then G is Abelian

so that puts some constraints on how exactly the center can sit inside the group and how things can commute

[native magnet]

I dunno what the cyclic means

[prisma ibex]

if we're talking about finite groups think Z/nZ (integers mod n)

[native magnet]

I got finite groups

[velvet hull]

something something 5/8ths
when i first came across that result it seemed like a deep structure theorem but no the proof is really boring

[prisma ibex]

the only cyclic group that is not finite is Z itself

cyclic just means generated by a single element

[native magnet]

Like identity?

[prisma ibex]

no?

[native magnet]

What's generated by a single element mean?

[prisma ibex]

look up what a group presentation is

[native magnet]

Aight

[velvet hull]

lmao that might be a bit too much

[prisma ibex]

generators and relations is not that much lol

[chilly ocean]

{g} generates a group G if every element h of G is equal to g^n for some integer n, where g^0 is the identity and g^(n+1)=g(g^n), g^(n-1)=g^-1(g^n), and g^-1 is the inverse of g.

[native magnet]

gimme a moment to process that

[native magnet]

Okay so like 1 is a generator is Z or R because all values can be made by 1*anything

And in the presentation, Z = <a|> just tells you what the generator is and then relations are the rules applied to the generator

[sly crescent]

Is there a name for a subgroup G of H such that N_H(G)/C_H(G)=Out(G)?

[chilly ocean]

1 is not a generator of R, not all real numbers are formed by repeated addition and subtraction of 1.

[frigid epoch]

If we're looking at them as additive groups, then only the integers can be generated by 1

You don't have access to scalar multiplication with real nums

[native magnet]

What about multiplicative groups of R?

Oh wait sorry

Just read the last message

[frigid epoch]

The multiplicative group of R lacks the element 0

It's a totally different group

[native magnet]

Oh yeah I've seen that

It's like R* or something

[frigid epoch]

Yep

Well it's actually not totally different, there's an easy bijection between R and (R+)*!

(Aka just the positive invertible reals)

[hard hearth]

let $G = G_{0} \supset G_{1} \supset \cdots \supset G_{m}$ be an abelian tower. let $i \in {0\cdots m-1}$. I will insert a group $G '{i}$ between $G{i}$ and $G_{i+1}$ such that $G_{i} '/G_{i+1}$ becomes cyclic. Take any non identity element $xG_{i+1} \in G {i}/G{i+1}$ and consider the cyclic subgroup generated by $xG_{i+1}$. this corresponds to a normal subgroup $G '{i}$ containing $G{i+1}$. Insert $G '{i}$ in between $G {i}$ and $G{i+1}$. doing this for all $i$, we have a refinement $G = G{0} \supset G '{1} \supset G{1} \cdots \supset G '{m} \supset G{m}$. note that the size of each quotient has decreased. adding more quotients if necessary, we bring down the size of each quotient to be low enough to apply induction on the size of the quotients to get a cyclic refinement. is this proof good enough for the first part? the second part follows anyways

[cloud walrus]

fastrack_and_backtrack

[hard hearth]

hmm, no i feel like something is really off here the proof isnt sitting right

i could have just refined the tower so that the size of each quotient went below a certain threshold and still applied induction. the generating a cyclic subgroup part seems like a waste

[rocky cloak]

The proof looks great, not sure what puts you off.

The reason to generate cyclic groups is so that you get cyclic groups, which is what the proposition says

[hard hearth]

i wrote down what came to my mind immediately, but later on my concern was with the fact that $G_{i+1}/G '{i}$ may not be cyclic, so all im doing is refining the tower till all the quotients go below a certain size and then applying induction. so i could have just added in any random intermediate normal subgroup $G '{i}$ instead of one whose quotient $ G '{i}/G{i+1}$ is cyclic

[cloud walrus]

fastrack_and_backtrack

[hard hearth]

that is what i thought too, maybe im missing something

[rocky cloak]

I mean you can, but then the proof becomes more complicated right. Because then you need an extra step that to show that the result is cyclic

But yeah, you can argue that
inductively refining gives you a refinement where the quotients are simple, and simple abelian groups are cyclic

[hard hearth]

nonetheless, the proof given in the book is much neater cleaner 😅

wait a second how do they make the assumption that G is abelian?

[rocky cloak]

You can just consider Gi+1/Gi seperately

This is also just exactly the same proof you wrote earlier

[hard hearth]

ah of course. with this information its pretty much the same proof

its still more compactly written

[rocky cloak]

Well just because they've hidden some of the details in "it clearly suffices"

[blazing crag]

Please help. My homework problem ask me to prove that the prime-power decomposition of group T (subgroup containing all number of order p^n and the identity) is isomorphic the the direct product of cylic factor of order of some power of prime p

[thorn jay]

That seems interesting, where does it come up?

[lusty marlin]

What is Out(G)?

[thorn jay]

Aut G / Inn G

Inn G being the group of conjugation actions on G

[thorn jay]

Well, first of all, N_H(G) / C_H(G) comes from the homomorphism
f : N_H(G) -> Aut G
sending an element in the normaliser to the automorphism corresponding to conjugation by that element.
G < N_H(G)
So
Inn G < im f < Aut G
So if Out G ≈ im f there's also not really some semidirect product you can make from it so I'm confused where you got it from tbh

[woeful wolf]

Hey Im unable to understand how to prove this

Let $G$ be a group and $H$ a subgroup. Then
$(G:H)(H:1)=(G:1)$

where number of cosets of H in G is the first term and the order of H is the second.

I saw a proof that proves that G is the disjoint union. But Im not able to think of it intuitively, as in Im not able to see how this would work

[cloud walrus]

TensorBlaze

[coral spindle]

So [G:H] is just the order of G/H (the set of cosets of H in G). So provide a bijection G/H x H/1 <-> G/1 and you're done.

Hint: first choose a set of representatives for G/H as this will make defining the bijection easier.

[storm kiln]

For $p$ a prime in the P.I.D. $R$ and $N$ an $R$-module prove that the $p$-primary component of $N$ is a submodule of $N$ and prove that $N$ is the direct sum of its $p$-primary components (there need not be finitely many of them).

[cloud walrus]

Jelle

[storm kiln]

I don't really understand the question, shouldn't there only be finitely many p-primary components, since they are based on the factorization of a, where (a) = Ann(N)? Or do they mean that the other p-primary components are just 0 if the prime doesn't appear in the factorization of a (i.e. has exponent 0)?

they also only defined p-primary components on torsion modules, but I suppose that non-torsion modules have Ann(N) = 0, which kinda makes sense

(D&F btw)

[rocky cloak]

Ann(N) can be (0).

Even if N is torsion

[storm kiln]

Oh, true. Unless N is finitely generated right?

[rocky cloak]

That's right

[storm kiln]

How are p-primary components defined when Ann(N) = (0) though?

oh, wait they did specify non-zero annihilator in the definition of p-primary components

but not in the exercise

[rocky cloak]

I would define the p-primary component of N to be all elements n such that the radical of Ann(n) is p.

Not that you need to include 0-primary components for the exercise to be true in that case

To take an example if
N = R/p (+) R/p^2 (+) R/p^3 (+) ...
Then N is p-primary with annihilator 0

I don't know how they define it in D&F

This MSE seems to suggest that N was meant to be a torsion module in the exercise

https://math.stackexchange.com/q/2175376/306319

[storm kiln]

This is where they defined it, but the theorem seems the same as the exercise though

[rocky cloak]

Maybe the exercise is supposed to keep N torsion, but drop the assumption that the annihilator is nonzero....

[storm kiln]

Hmm, I guess I'll just skip it. I understand their proof of the theorem at least

[tardy hedge]

Section 12.1 of d/f was so ahh

[storm kiln]

Yeah, I gave up trying to completely 'understand' some of the proofs. I can read them and I can follow the steps, but can't really picture the full proof in my head

[tardy hedge]

Ong same

Did that section last term

[woeful wolf]

i did it but i remain unsatisfied

not to sound like a dunce or anything, but i feel i want to apply what i learnt about cosets a little more

ig thats a personal opinion

[rocky cloak]

What have you learned about cosets then?

[woeful wolf]

well cosests are disjoint, so i wanted to use that in my proof. i learnt that for $z \in H,\ zH=H$, and its applicability in proving this. I realised that when we form a union of two representations with a coset in the subgroup H, denoted K, we can form an equation of two different subgroup-cosets and the representations of K and H. I wanted to see how zH = H is applied here (for H and K). I feel i may be getting a little lost, and that the implicit argument is actually bijection.

[cloud walrus]

TensorBlaze

[woeful wolf]

so something along the lines of

Let G be a group and H, K its subgroup, and $K\subset H$. Our representation for H is $x_i$, and for K is $y_i$. it seems trivial to say that

[\bigcup_{i,j} x_iy_i K = G]

Using this I wanted to see if I could form an equation between two of these and prove that it is a disjoint union

[cloud walrus]

TensorBlaze

[coral spindle]

Yes. Use the property of representatives.

I.e., write down what it means for x_i to be representatives for the cosets of H and likewise for y_i and see what you get

Apply it to that situation

[woeful wolf]

okay cool ill try that

[delicate orchid]

Number 15... iterated cosets... the last thing you want in your coset representatives is some subgroup's transversal. But as it turns out... that might be what you geeEEttt..

[thorn jay]

Get out of my head get out of my head get out of my head get out of my head

[coral spindle]

Marry me

[delicate orchid]

maybe later original gangster

[thorn jay]

Is there a name for / anything interesting about ideals whose radical is prime?

[tropic obsidian]

it is true that any module is the direct limit of its finitely generated sub modules?

wait nm

it is true

[thorn jay]

Yeah, because every element generates a cyclic submodule, so the direct limit should contain every element, i.e. be the whole submodule

[rocky cloak]

It's close to being a primary ideal at least

[thorn jay]

It is

But, that property is, i think, not retained under preimages

Because that is true iff nil R is prime and S < R => nil S is prime

Which is not true, im pretty sure

[rocky cloak]

The radical being prime you mean?

If s^n is in f^-1(I), then f(s)^n is in I, so rad(f^-1(I)) = f^-1(rad(I)), and primes are preserved under preimage

[thorn jay]

Ah okay, i was thinking too generally

I've seen somewhere that the nilradical being prime is equivalent to the spectrum with the zariski topology being irreducible

But i dont reckon there's much more to it

[south patrol]

Indeed minimal primes correspond to irreducible components

And it is more or less immediate that nilradical prime <=> there exists a minimum prime (I.e. unique minimal prime)

[thorn jay]

Oh yeah that was it

[south patrol]

But also like quotienting out by the nilradical induces a homeomorphism

So you can reduce to reduced things ig

[rocky cloak]

To deduce the irreducible simply reduce to the reduced case

[south patrol]

(The generalisation is that a scheme is "integral" iff it is irreducible and reduced)

[thorn jay]

I see

That makes sense

Damn, imagine schemes for arbitrary coherent systems on a variety of algebras V

[white oxide]

Would this be something like a minimum degree division argument, suppose that there were some element g in k[x]/(f) with deg g >= n, let g be an element with minimal degree such that g is not in (f), then we can divide by f using the division algorithm and show that the remainder polynomial is actually of minimal degree not in (f)

[white oxide]

Nvm that doesn't work

I don't really see why this is true - it suffices to show that $x^r \in (f)$ for all $r \geq n$. Now, $x^r = f(x) g(x) + r(x)$ . If $r = 0$ then we are done, otherwise $\deg r < f(x)$. But I don't know where to go from here

[cloud walrus]

okeyokay

[thorn jay]

You dont need to show that every element of k[x] is reachable

Every element in k[x]/(f) can be written as some polynomial of order less than n + (f)

Because that is essentially what quotienting by (f) does; it reduces the order of the polynomial until it is less than that of f in some way

[white oxide]

I see

Maybe something like let $g \in k[x] / (f)$, if $g \in (f)$, we are done, otherwise $g = fq + r$ with $\deg r < \deg f = n$. Since $g + (f) = r + (f)$ and $r$ is a polynomial of degree less than $n$, we are done (at least for the spanning part)

[cloud walrus]

okeyokay

[thorn jay]

This is for g in k[x], not g in k[x]/(f), but yes

[white oxide]

Oh yeah

[thorn jay]

Then, for it being a basis, we can suppose that
(a_0 + a_1 x + ... + a_n-1 x^n-1) + (f) = 0
For some combination of coefficients a_i
Call the polynomial
a_0 + a_1 x + ... + a_n-1 x^n-1
p, for convenience. What does this say about p? Why is this only possible if p=0, looking at the orders of p and f?

[white oxide]

Yea, because a polynomial of higher degree cannot divide a polynomial of lower degree (unless that polynomial is zero)

[thorn jay]

Exactly

So p must be zero hence basis etc

[white oxide]

got it, thank you!

[gilded fulcrum]

How do i find all ring homomorphisms from Z to Z?

[coral shale]

think where certain specific elements must go

[gilded fulcrum]

Ok lets say f is a ring homomorphism. I know f(0) = 0 and f(1) = 1. Both the zero map and the identity map preserve these properties. How do i prove that these are all ring homomorphisms?

[wraith cargo]

these aren't all of them

[thorn jay]

Zero map doesnt

Zero map has f(1) = 0 ≠ 1

What element(s) is Z generated by?

[gilded fulcrum]

1?

[south patrol]

You can also think of all the group homomorphisms

[thorn jay]

So if you know where 1 goes, youve uniquely determined the map

[south patrol]

Then you get an extra constraint from being a ring map lol

[thorn jay]

Generating sets my beloved and most abused

[coral shale]

i call hacks. describing the infinite with the finite

belittling infinity

[south patrol]

This is the main game of maths right lol

[thorn jay]

I am playing jod

[south patrol]

Reduce the infinite to the finite

And solve the finite by passing to the infinite

[thorn jay]

Not finitely generated?
W-we'll just use a filtration its all good

[south patrol]

Lol

kappa generated for some kappa less than the cardinality hopefully

E.g. Q oh wait

[thorn jay]

Hey at least Q is nicely described by a universal property

[south patrol]

Everything has a nice universal property

[coral shale]

everything ur willing to talk about

[south patrol]

For example if A is a ring then A is universal for maps into A

[thorn jay]

Yeah yeah but thats after you've constructed A

[south patrol]

I take this example from Kevin Buzzard

[thorn jay]

Cheater

:(

[thorn jay]

What is big math hiding from us

[south patrol]

Lol

Tbf like as a derived person, universal maps are the only things

[thorn jay]

Every map is universal

In some way

[south patrol]

Universal for maps in the mapping space to that map

[thorn jay]

Yes

Totally

Agreed

[lusty marlin]

It is a nice exercise to think about unity-preserving ring homomorphisms from ℤ to any ring with unity R.

[crystal vale]

If I say G is closed under an associative product, then that operation is closure?

[crystal vale]

I showed that if G is a finite set closed under an associative product and that both cancellation laws hold in G. Then G is a group.

Now I have to construct a counterexample when G is infinite set.

Any hint?

[patent harbor]

think formal words

[next obsidian]

Indubitably. Perchance. Forthcoming.

[crystal vale]

I don't get it

[velvet hull]

what are the cancellation laws? more importantly do they say that the inverse must exist, or only if they exist

[crystal vale]

If ab = ac then b = c and if ca = ba then c = b

[velvet hull]

oh okay, so this guarantees that multiplication must be injective

however, injectivity only implies bijectivity when the set is finite

because when you consider multiplying the entire set by some element, and you keep chasing the "next" element, then eventually you have to end up in a loop, and therefore there is an inverse

however this fails for infinite sets. so can you figure out a way of defining multiplication, such that the "next" element "escapes to infinity", without ever being stuck in a loop?

[crystal vale]

Yes

[frigid epoch]

The counterexample is such a G

No empty words

[crystal vale]

What are empty words?

[velvet hull]

can you just think about this for a moment

[crystal vale]

Natural number N with operation multiplication

[velvet hull]

sure, that works

[crystal vale]

Thanks HChan

[frigid epoch]

Hell, even with addition, no?

[frigid epoch]

The empty word is the neutral element of free groups/monoids

[crystal vale]

With multiplication

[crystal vale]

How can I show if H and K are subgroups of G and H \subset K with H has a finite index in G then K has a finite index?

[dim widget]

do you know the isomorphism theorems?

[velvet hull]

lemma: if H <= K <= G, then [G:H] = [G:K] [K:H]

try and prove that

(note that this works even if H, K are not normal)

[crystal vale]

Yes

[velvet hull]

they don't necessarily work here because normality is not assumed
edit: see reply, that also works, longer proof but it works

[dim widget]

You can always find a normal subgroup of a subgroup of finite index which itself has finite index

that’s a nice exercise for you!

[crystal vale]

I see then [K : H ] is finite and then [G : K ] is finite

[crystal vale]

By doing some group action, right?

Let H be the subgroup which has a finite index then we make a group action G -> S, where S is the set of all left cosets of H.

It induced the group homomorphism G -> Sym(S), now G/ker is finite because S is finite.

And ker \subset H.

But how does it help me ?

[dim widget]

Well if you have an easy time proving that [G:H] = [G:K][K:H] then it doesn’t help you at all

but if you assume H is normal then K/H is a subgroup of a finite group and [G/H:K/H] = [G:K]

i think it’s better to do this HChan’s way if you see how to do it

because that formula is useful more generally

[crystal vale]

I got it thank you

[dim widget]

probably the simplest thing to show is that G/H naturally surjects onto G/K

[crystal vale]

Yes gh -> gk

And by isomorphism we have (G/H )/ (K/H) isomorphic to G/K

[crystal vale]

I proved this, is there any other way to do this question?

This is not Ai generated I used Ai for only latex

[bitter rover]

Well, a more abstract way of stating the two hypotheses is that $f(x) = x^3$ is an automorphism of $G$. You can probably use that.

[cloud walrus]

Cufflink

[crystal vale]

Automorphism maps centre to centre, is that useful here?

I mean phi(Z(G)) = Z(G) if phi is automorphism

[crystal vale]

Is there any weaker condition for this? If phi is surjective not injective then I think phi(Z(G)) \subset Z(G)

[surreal magnet]

Is the "3 not dividing the order" condition even needed?

[lusty marlin]

Yes

[rocky cloak]

There are non-abelian groups where every element has order 3, so in particular
(ab)^3 = a^3 b^3 = 1

[surreal magnet]

Ah yeah I see

[fringe marten]

Hello, i'm working on a project and i'm trying to find square roots in finite fields. it happens to be doable with fields with a prime number of elements, but is there a way to find them in fields with p^n elements ?

[south patrol]

Do you know about the structure of (F_p^n)*?

[fringe marten]

Kinda yes

You're telling me it just works with any group?

[rocky cloak]

(F_p^n)* is cyclic

Anyway, I think it mostly depends on how the finite field is given to you

[fringe marten]

Indeed

In this case i get it from extensions of Fp

Using polynomials

So i want to get roots of polynomials

[rocky cloak]

Okay, so you just have Fp[x]/(f(x)) and you want to compute square roots from that.

I'm wondering if Newtons methode would actually work well

Maybe not...

[fringe marten]

I don't know it

Nice pfp btw

Love bulbasaur

I know that for F_p, there's tonelli shanks algorithm, i wonder if there's a similar algorithm for F_p^n

[rocky cloak]

The analogues thing of just picking random elements of F_p^n and checking if they are primitive generators should work

So factor p^n - 1 = 2^e m
pick random g(x)
Compute g(x)^[2^(e-1)m] mod f(x) to check if it equals -1. Should happen 50% of the time.
Then g(x)^m should generate the 2-sylow subgroup.

[fringe marten]

I see

So it's a lot like the original

[rocky cloak]

Yeah, it should be pretty much exactly the same.

[fringe marten]

Except the last line i didn't quite understand but maybe i can figure it out

Ok ty for your help

[rocky cloak]

It's exactly this, just replace random number with random polynomial

[plucky harness]

I'm having trouble understanding a proof in Artin M's algebra. I've highlighted in yellow the line I don't fully understand.

It doesn't quite make sense to me why multiplication by g_i means that the g_iH will be a partition of the coset g_iH. I do understand that because multiplication by g_i is an invertible operation, a bijective map will exist between g_iH and H, and similarily a bijective map will exist between g_ih_1K and K, etc etc.

But I don't quite understand how that comes together to mean that g_iH is a partition of the coset g_iH.

[south patrol]

Oh lol I misunderstood

I assumed you meant like asking which fields have which square roots or smth lol

[rocky cloak]

The image of a partition under a bijection is again a partition

[plucky harness]

Thanks! I shall try to prove that statement myself, and ill come back here and ask some followup questions if I haven't managed to understand it.

[thorn jay]

Well, f(x) = x^2 is an automorphism of a finite group iff it is abelian and of odd order

So a finite field has all square roots iff it is char 2

Huge

Oh

QwQ

[crystal vale]

Commutator subgroup of D_2n is subgroup generated by { r^2k | k = 0, 1, 2,...,n}, right?

[trail cave]

For R a commutative ring and J its set of nilpotent elements, how do I show that R/J has no nonzero nilpotent elements?

I let alpha = a + J and show that a^n is in J for n s.t. alpha^n = 0, but I get stuck there on showing that a itself is 0

[storm kiln]

What does it mean for a^n to be in J?

[trail cave]

it means that there's a natural number m such that a^(nm) = 0

[crystal vale]

If a^n in J

That means (a^n)^m = 0

So a^(nm) = 0

[storm kiln]

Yeah

[crystal vale]

Implies a is nilpotent element

[trail cave]

a is nilpotent, so there exists q such that a^q = 0. If q > n, then alpha^n is not zero. So q <= n. Is this the right process?

am I missing something obvious that makes a zero from the fact that it's nilpotent?

[storm kiln]

Think about what you want to show

[trail cave]

I want to show that alpha is 0+J

alpha = a + J, so that's equivalent to showing a = 0

right?

[crystal vale]

You want to show a is zero in G/J

[crystal vale]

No

You want to show a + J = 0 + J, right?

[trail cave]

oh

so showing a is in J is enough

[crystal vale]

Yes

[trail cave]

thank you

[crystal vale]

How to show that SL_n(R) = [ GL_n(R) : GL_n(R) ]?

I know [GL_n(R) : GL_n(R) ] \subset SL_n(R)

[cloud walrus]

Bottlecap Desu~(Bottlecap Gang)

[trail cave]

deleted this but can't delete the texit, I think I figured it out

[hard hearth]

here, alternatively, i can define a bijection $G/N_{G}(P) \rightarrow \Omega$ given by $gN_{G}(P) \mapsto gPg^{-1}$?

actually this might just be unnecessary because a similar argument proves orbit stabilizer theorem which is just what they used.

[cloud walrus]

fastrack_and_backtrack

[crystal vale]

Yes if you already show that every Sylow p-subgroup is conjugated to each other

[hard hearth]

that was proved previously

thanks!

[crystal vale]

https://math.stackexchange.com/questions/4379275/g-is-a-group-and-t-in-operatornameautg-if-n-lhd-g-such-that-nt-sub

My doubt is T(N) ≠ N, so how can I induce Aut(G/N)?

I mean the element of Aut(G/N)?

[crystal vale]

| Aut (Z/3Z × Z/3Z) | = 48 ?

[rocky cloak]

You cannot induce an automorphism, but you can induce an endomorphism.

[candid patrol]

= |Gl(2,F3)| = (3^2 -1)(3^2-3) = 48

[crystal vale]

I see

[crystal vale]

Yes

[candid patrol]

??

[crystal vale]

I mean I have to show that SL_n(R) is the commutator subgroup of GL_n(R)

[candid patrol]

Oh ok mb, i forgot that [G:G] means D(G)

Did you find the solution ? Is it for n = 2 also ?

[crystal vale]

Yes, for arbitrary n

[rocky cloak]

I think the easiest way is to show that SLn is generated by elementary matrices and explicitly write those as commutators

[crystal vale]

Yes I found the same idea on MSE

[crystal vale]

3, by 1 we just need to find the image of (123...n) and (12), so let the image of (123..n) be a and the image of (12) is b, then we have a^n = 1 and b^2 = 1.

Since a and b in C* so a has n choice and b has 2 choice so total number of homomorphism are 2n, is it correct?

[rocky cloak]

Not quite.

There are at most 2n homomorphisms, but you have to verify that these choices of a and b actually give rise to homomorphisms.

Part (4) can work as a hint

[crystal vale]

For homomorphism we need f(xy) = f(x)f(y), but since a and b are in C* so they commute to each other, doesn't work?

[candid patrol]

a^n=1 and b^2 = 1 are not the only condition, you have to check if a and b generate Im phi. And it's false because phi is not necessary an isomorphism, we don't necessary get n choices for a.

It's weird to consider (1 2 3 ... n) and (1 2) as the generators of Sn, considering all the transpositions is easier imo

[rocky cloak]

Well, whether it works or not is something that needs to be checked

[rocky cloak]

The ideal generated by some elements will consist of linear combinations of the generators

What is "the mi"?

Well being generated by monomials is the definition of being a monomial ideal

I guess not

[cloud walrus]

meinebow

meinebow

[rocky cloak]

This seems kinda unrelated to what you asked before.

But if you look at the quotient ring of a polynomial ring over a field it will have a basis given by monomials

[barren sierra]

your original question was about elements of J

and now you're talking about R[x] / J

so no it's not the same question

[rocky cloak]

Every element in R[x] is a linear combination of monomials. That doesn't really change when you go to R[x]/J

[barren sierra]

For a ring with coefficients in a field (and perhaps in some more general rings but definitely not every ring) this is easier to answer

For an ideal $I \subseteq k[\overline{x}]$, the elements of $k[\overline{x}] / I$ (known as \emph{standard monomials}) are $k$-linear combinations of the monomials $\overline{x}^\alpha$ which are not in the ideal generated by the leadings terms of elements of $I$, i.e. $\overline{x}^\alpha \notin \langle \textsc{lt}(I) \rangle$.

but if k is not a field then you probably can't make a general rule like this

no

shit lemme fix

[cloud walrus]

Spamakin🎷

[barren sierra]

my bad

this is a standard fact you can find in Cox, Little, and O'Shea Ideals, Varieties, and Algorithms Chapter 5, Section 3, Proposition 4

again this is for k a field

if k is a ring, then the description will heavily depend on the ring used and idk if we can say anything general

[glad osprey]

Pinging a random user, then insisting they are misreading your wrongly written question is kind of rude, the least you could do is thank them for the help @grand ridge

[acoustic igloo]

a binary structure and a group that act identically on a set are isomorphic and therefore the binary structure is a group?

what is this argument?

[velvet hull]

those matrices are special matrices, called permutation matrices

notice that all they do, given any vector, is just swap around the entries of the vector and that is it

[acoustic igloo]

yes

[velvet hull]

so instead of thinking about the matrices themselves directly, think about what the matrices do the vector [1;2;3]

[crystal vale]

I don't get your point

[acoustic igloo]

they permute it - but i wonder why that makes them a group

[velvet hull]

well, do you know about the symmetric group?

[acoustic igloo]

yes

[velvet hull]

tell me what your definition of the symmetric group is

[lusty marlin]

You can verify the existence of an identity element, the existence of an inverse, and associativity of your binary operation

[acoustic igloo]

S_n is the group of all permutations of {1,2,...,n}

under composition

[velvet hull]

exactly - and the set of 3x3 matrices you just gave, consists of all the possible permutations of the three entires of the vector

[acoustic igloo]

it consists of matrices, not of permutations

[velvet hull]

so instead of permuting {1,2,3}, you're permuting [1;2;3]

it really isn't that different

[acoustic igloo]

hmm

[velvet hull]

this is going to be the core lesson you learn in abstract algebra - that what the actual thing looks like, matters much less than how it behaves

I don't care if you're permuting fucking oranges and bananas, as long as it's permuting 3 distinct objects in all the ways, then it, for all intents and purposes, is S_3

[torpid knot]

in the above case those two groups are isomorphic

[velvet hull]

this is arguably more of a group action / representation

[torpid knot]

epic i still cant wrap my head around group actions and i have an exam on it tmrw 😭

[velvet hull]

group actions are just fancy group homomorphisms

really, that's it

it's just sometimes more convenient to represent the action as an action instead of the underlying homomorphism

[acoustic igloo]

the permutation matrices may permute the vector, but it's important how the matrices multiply together

they need to combine in the same way as the permutations in S_n combine

[velvet hull]

sure, another way to interpet the permuation matrices, instead of permuting the entries of a vector, is that they permute the columns of matrices

[velvet hull]

here

https://cdn.discordapp.com/attachments/844681108473118750/1342381449449308191/530674466028650516.png?ex=67b96da5&is=67b81c25&hm=76c596e227a84bd85f4c9bc74e53a8878daf2068cf558a6cb995f1e16ef5ac45&

verify that this is true, and is in fact how every single permutation matrix works

[acoustic igloo]

i don't know what that notation means

[velvet hull]

v1 is just the first column of the second matrix

v2 is the second

v3 is the third

[acoustic igloo]

ok

[velvet hull]

but here's the thing: you don't really need to do all that

when the book proved that S_n is a group, did the book go through every single possible combination of elements in S_n to check that they are all associative?

we know that permutation under composition is an associative operation

so, if we show that something permute things, and that is all that thing does, then it has to be associative under composition

because we already know permutation is associative under composition, or else S_n would not be a group

the exact same thing is happening with inverses

[acoustic igloo]

do you mean to say permutation is an associative operation, or composition is an associative operation?

[velvet hull]

permutation under composition is associative, sorry, too used to that being implicitly understood as the operation

[acoustic igloo]

i'm trying to figure out why this works

[velvet hull]

how about this: actually don't, you don't need to prove that that works to show that they are S_3

[acoustic igloo]

ok

[velvet hull]

S_3 is defined as the collection of all permutations of {1,2,3}, that is true

but that doesn't mean anything set isn't explicitly the set of all permutations of {1,2,3} is not S_3

because that's not the point of group theory

the idea of it, is that if any three things even remotely looks like {1,2,3}, and it is being permuted by something, then the combination of "that thing" + "those three things" is S_3

because the symmetric group is permutation

it is the very idea of permuting

[velvet hull]

so, here's how the proof works:

first, I see that every single given matrix permutes the three entries of the vector [1,2,3].
second, I know that there are exactly 3! = 6 total permutations of three things, and I see 6 different matrices.
So, the group formed by those matrices formed under multiplication is S_3

at this level, that's it

it's like saying a=2b+3 is not the equation of a straight line because you were taught that straight lines had the form y=mx+b, and you don't see a x or a y

[candid patrol]

a^n = 1 just doesn’t mean that the order of a is n, so as jagr said, u get at most 2n morphisms

[crystal vale]

But we don't need order of a is n

Btw I see the problem

[candid patrol]

That’s why we get at most 2n morphisms

[crystal vale]

But then how can I verify they make homomorphism or not

[cloud walrus]

UGOBEL

[candid patrol]

I've technically shown that transpositions generate S_n, that's why I have some doubts about my solution.

[rocky cloak]

One possible solution is to notice that
(12)(123...)(12)(123...)
has order relatively prime to n, so we must have abab = a^2 = 1

That still doesn't narrow it all the way down though

[serene dune]

what does it mean by "fixes"

sorry "fixed"

[frigid epoch]

No k with σ(k) = k

[serene dune]

ahhh, shit

alr

[frigid epoch]

So I guess in S9, (abc)(def)(ghi) is regular, but (abcde)(fghi) isn't

In which case (a)(b)(c)(d)(e)(f)(g)(h)(i) is the only option for fixed points anyway so I'm not sure why they included the second part lol

[serene dune]

oh he asked to prove this

[serene dune]

yeahyeah

this is the statement

"nice cut"

[serene dune]

yeah, i mean always they would be writing the most pointless shit ever

[frigid epoch]

Ah interesting

[serene dune]

solving rotman's

[frigid epoch]

Cool exercise

[serene dune]

let's see

feel free to jump on https://discord.com/channels/268882317391429632/1334457895563169792 and leave comments

[frigid epoch]

I'll do what I can on my 1% remaining

[serene dune]

hahah, sure thing

i meant it for not just now tho

[frigid epoch]

Just gotta show that powers of the maximal cycle are regular

Ah

[serene dune]

oh yeah i got this one

[frigid epoch]

Sorry but I'm not bookable 😛

[serene dune]

np, i am writing down all in latex

every cool line and ofc the problems

[serene dune]

how crazy that the convo was about "same" thing

[coral spindle]

A function f fixes x if f(x) = x.

Woof, scrolling is hard

[serene dune]

nw

i appreciate

[candid patrol]

not really the same haha

[serene dune]

"same"

[candid patrol]

"""""same"""""

[delicate orchid]

f fixes x if x is in the limit over f

[coral spindle]

10 000 years of jail <3

[delicate orchid]

the worst part is I'm not even wrong

[crystal vale]

If I want to find the number of homomorphism Z/nZ -> C*, isn't if I send 1 to a, we need a^n = 1.

So they will be n homomorphism.

[coral spindle]

Yes

[crystal vale]

Okay thank you

[south patrol]

Lol

[candid patrol]

Correct answer but u need to explain

[delicate orchid]

what is there to explain

[cloud walrus]

UGOBEL

[crystal vale]

Because 1 generates Z/nZ

[candid patrol]

-_-

[crystal vale]

What I need for the image of 1 is to follow that the order of image must divide n

And n = 0 is the only relation in Z/nZ

[coral spindle]

They were asking for a check, not an answer review, I think :)

[frigid epoch]

,, 1 = \varphi(\bar{0}) = \varphi(\bar{1}^n) = \varphi(\bar{1})^n

[cloud walrus]

PKThoron

[frigid epoch]

So the image of 1bar is an n-th root of 1!

[coral spindle]

Ugobel's question was a prompt for someone else, not a request for help

[cloud walrus]

UGOBEL

[candid patrol]

Try to prove that Hom(Z/nZ, C*) is cyclic 🙂

[crystal vale]

They are all distinct because the image of 1 is distinct

[tardy hedge]

[tawdry venture]

How do I prove that N(r) < N(b)?

[lusty marlin]

What is there to prove? This is a definition.

[tawdry venture]

sorry this is the problem

[bitter rover]

Prove it meets conditions (a) and (b). What have you done so far?

[tawdry venture]

I already done for part a

I'm not sure how to do part b. I know division algorithm yields that remainder r norm must be less norm of b

[bitter rover]

Is there an example in your textbook proving anything like this?

[tawdry venture]

I think there is but idk what it is saying 😭

[bitter rover]

It's probably worth understanding, then.

[rocky cloak]

You already have a guess for what q should be, so just plug that in, solve for r and simplify until you get a bound for N(r)

[bitter rover]

Also, if $u = a + b\sqrt{-2}$ and $v = c + d\sqrt{-2}$, I'd work out what $\frac{u}{v}$ looks like in terms of $a,b,c,d$. If you express it as a single fraction of (complex) numbers, you'll find $N(c + d\sqrt{-2})$ in the denominator.

[cloud walrus]

Cufflink

[golden turtle]

I was hoping someone could check my proof

I can show the lemmas I cite if thats needed

[golden turtle]

Please ping me if you look it over

[tawdry venture]

Thank you!

[tawdry venture]

This is the problem I'm trying to solve

Here is my work:

Is there supposed to be more cases?

[floral crane]

Hello. I hope this is the right place to post this. I am trying to learn and implement algorithms for finding irreducible modular polynomials. My implementation seems to be working, but I would like to understand the mathematics better. Is there anyone around knowledgeable about that kind of thing and willing to talk?

[tawdry venture]

Here is my updated work!

[subtle crypt]

The two square theorem tells us that any prime congruent to 1 mod 4 can be written that way.

For example, 7^2 can't be represented as x^2 + y^2.

[tawdry venture]

Sorry not fermat's theorem, I meant a theory from number theory

that any prime p can be written as x^2 + y^2

[wraith cargo]

That isn't true lol

It's only true for primes congruent to 1 mod 4

[subtle crypt]

Yeah, see https://en.wikipedia.org/wiki/Fermat's_theorem_on_sums_of_two_squares.

[tawdry venture]

ohh it's an if and only if

[subtle crypt]

For the rest, you should say why x or y has to be even (if they were both odd, ...).

Also, if they were both even, ....

So, one is even and one is odd.

And then you can't say y = 2m quite yet.

You have to prove that an even perfect square has a square root with the factor of 2.

Like 2 | y^2 -> 2 | y.

You've only got 2 | y^2 so far when you say that y^2 is even.

Then, the rest is fine.

If you want, you can also shorten it to one case by saying "WLOG [without loss of generality] let us make x the odd square and y the even square".

[tawdry venture]

that's impossible. That means we don't have a prime p in mod 4

if x^2 and y^2 are both even, then p isnt even prime

[subtle crypt]

Yes, and you can say that to justify it.

[tawdry venture]

oh ok

[subtle crypt]

The three cases are both are odd, both are even, or one of each.

Prove that the "both" cases are each wrong, and you're left with one of each.

[tawdry venture]

oh yea if they both odd, then we get even, which can't be prime

[subtle crypt]

It's easier to prove that it can't be congruent to 1 mod 4.

To go the not prime route is a bit more involved because there's an even prime that's the sum of two odd squares.

Like 1^2 + 1^2.

[spice whale]

i mean

usually you just talk about odd primes

[tawdry venture]

My updated proof after your feedback! @subtle crypt

[subtle crypt]

You have to be careful, because there's an even prime.

For example, you say that if x, y are odd, then odd + odd = even, which isn't prime. However, x = y = 1 is odd + odd = even, which is prime since 1 + 1 = 2.

[tawdry venture]

ohh you're right

[subtle crypt]

Since your problem only cares about p = 1 mod 4, you can say that if x, y are odd, then odd + odd = even, so it doesn't count as a number = 1 mod 4.

[tawdry venture]

ohh true just mention that 2 doesn't count since we only care about p = 1 mod 4

[subtle crypt]

Right.

It can be either 0 or 2 mod 4, but either way, it's not 1.

Hmm, it might be better to say that x^2, y^2 odd.

Let me think.

[tawdry venture]

Ok I simply added a line that says we dont count 2

Thank you so much!

[subtle crypt]

You're welcome.

[night tartan]

are you sure ur can just quote fermat's 2 squares theorem?

if this is given as a problem in ring theory then it feels like ur meant to prove that any prime =1mod4 can be written as a sum of 2 squares

cus like after that it's just really easy

[night tartan]

also this is really wordy

if ur taking like an introductory class then i think maybe it's fine, but certainly for a ring theory class you shouldn't need that much writing

something like

From Fermat's christmas theorem, we know $\exists x,y \in \mathbb{Z}$ s.t. $p = x^2 + y^2$. Checking mod 4, we see that one of $x$ or $y$ is even. So $p = n^2 + 4m^2$ for some $n,m \in \mathbb{Z}$

[cloud walrus]

LY

[night tartan]

would suffice

(also this is why are you sure the problem isn't asking you to prove the 2 squares theorem?)

[serene dune]

need hint

[velvet hull]

every permutation in S_n can be decomposed into a product of disjoint cycles

any two disjoint cycles commute with each other

what is the order of a cycle? what is the order of the product of disjoint cycles?

[serene dune]

i can follow up to this point

but is that valuable ?

for our case

i was thinking first of all

there are 2 cases

odd, even then

[velvet hull]

well, the question you asked essentially boils down to finding all order 2 elements in Sn

[crystal vale]

What is I here?

[lusty marlin]

Yes, because the disjoint cycle decomposition directly gives you the order of the permutation

[crystal vale]

Yes

[serene dune]

identity

yeah and transpositions and identities fit for it

and upto factorization they are unique

[crystal vale]

I think you can write every element as disjoint cycles, use that

[serene dune]

alr let me try to count

[velvet hull]

but can you prove that those are the only possible ones?

it's not hard to find a example

the point is to find them all

[serene dune]

i have proven it

it was previous exercise

lol

[velvet hull]

what about (1 2) (3 4)

I claim that this is a order 2 permutation

[serene dune]

lcm of all the orders

[velvet hull]

sure, can you prove why that's true?

[serene dune]

chat feels trippy, will you provide me the statement again ?

[velvet hull]

.

[serene dune]

order is length(not doing the explicit calculation here)

disjoint cycle means only one cycle will move any arbitrary element

so each cycle acts "independently"

so we have to find a number which satifies all the cycles

and lcm would the least number to do it

idk if that is what you have asked

[velvet hull]

good enough

[serene dune]

alr

[velvet hull]

so, how would you go about counting all the possible permutations of order 2

[serene dune]

first of all i was corncered how to consttruct them

out of n elements

coz its like

(12)(3)(4).......

or

(12)(34).......

so on and so forth

[velvet hull]

the lesson to learn from the proof you just gave, is that if you want to count them in a good way, first start by figuring out all the possible cycle types of order 2 permutations

[serene dune]

right

[velvet hull]

and then do the combinatorics

[serene dune]

and 1,2 would fit here

only

[velvet hull]

so 2 would work

2,2 would also work

so does 2,2,2 etc.

[serene dune]

yeah ywah which i tried to convey but chat is too trippy

ok i would try to argue for the count

thanks for the time

[candid patrol]

If [a,b] = e, then the ordre of ab is lcm(a,b), you need to know this before studying Sn

Disjoint cycles means that [cycle a, cycle b] = e for every a and b in your décomposition

You get it I think

[serene dune]

thanks for input, i hope i got it, yeah

[coral spindle]

This isn't true

E.g. let a = (1 2 3), b = (3 2 1)

[candid patrol]

b different to a^{-1} of course, mb 👍

[coral spindle]

This isn't true either

[candid patrol]

Oh really ?

[coral spindle]

E.g. let a = (1 2 3 4 5 6 7 8) and b = a^3

[candid patrol]

[a,b] = e but o(ab) = 2

you right thanks

Where am I wrong then ?

[coral spindle]

If you additionally assume that <a> n <b> = {1} then it works.

[velvet hull]

commutativity does not imply disjointness, which is what you're looking for

[coral spindle]

Also, this has nothing to do with Sn. This is a general fact.

[candid patrol]

Yeah ok ok I fogot

We need [a,b] = e and o(a) coprime with o(b)

[coral spindle]

Haha so why refer to the lcm at all? You are saying o(ab) = o(a)o(b)

[coral spindle]

You may want to see the additional hypotheses necessary here, btw

[candid patrol]

True, so <a> cap <b> is better thanks

[serene dune]

so

if it's a single transposition ways are
nC2 (1/2.1)

for two

nC2 (n-2)C2 1/(2^2 \times 2.1)

[south patrol]

Thoguh S_n is sufficient lol

[coral spindle]

Um not for infinite groups 🤓

[south patrol]

Sure but if a, b have finite order and commute then they generate a finite subgroup

:chad:

[serene dune]

not explaining the degeneracy here

but then if i add that i would get the answer right ?

[coral spindle]

Hmmm I wonder what facts we could use to prove that......

[south patrol]

Well you definitely don't need what has been done here

since the bound o(ab) <= o(a)o(b) is trivial

[coral spindle]

That's actually a cool observation

[serene dune]

all good?

[coral spindle]

Nastasya I have no idea what you're trying to do 😭

[coral spindle]

Oh

So you just need to count things that are a single transposition, and things that are a product of 2 disjoint transpositions, a product of three etc. OK

[serene dune]

your ping made me leave the album in midways to solve it

[coral spindle]

[serene dune]

it's fine, my lazy ass would have ignored it for few another hours

[coral spindle]

Something like this

Does this look about right to you?

Shit I'm just giving away the answer. Mb. I'll spoiler

[serene dune]

i dont care for this number

[coral spindle]

You don't?

[serene dune]

did my argument sound ?

i mean im happy if my argument sounds

[coral spindle]

You haven't shown an argument...

[serene dune]

add them all up

[coral spindle]

But fwiw no, I don't see where you're getting this 2^3 thing from for example

[serene dune]

3 2-cycles

interchanging elements in them

[coral spindle]

Ah but that's not what you want to do

The 2-cycles' elements are already unordered (because you wrote nC2, (n-2)C2)

What you need to account for isntead is swapping them, because you chose the cycles in a certain order

so you need to divide by 2! for example

[serene dune]

right

[coral spindle]

Then once you choose 3 two-cycles, you need to divide by 3! etc

So the number of 2-cycles in S_n is just nC2, for example

We don't need to divide by 2

[serene dune]

got it

"swapping inside the cycles" has been taken care of

[coral spindle]

Right, but more importantly, (1 2) = (2 1) anyway so you shouldn't worry about it

Or wait, no I see what you're saying

You have got the point, yes

[serene dune]

yeah i went for the general formula haha

[serene dune]

that is why i was dividing, lol

yeah very trippiy indeed

any other thing i need to take care of ?

[coral spindle]

P sure that's all

[serene dune]

alr

[candid patrol]

What's I ?

[serene dune]

identity, lol

i shouldn't have done it that way

[candid patrol]

sorry for the ping mb

[serene dune]

fine fine

[candid patrol]

U got the answer ?

[serene dune]

i hope so, yeah

is there any argument/trick here ?

i hope he doesn't want me to blindly check for suitable candidates

[candid patrol]

$\sigma = (1~2~3), \beta = (4~5)$ and $\gamma = (1~2)$ ?

[cloud walrus]

UGOBEL

[serene dune]

ahhh alr

common with one

grr, that was easy

[rocky cloak]

Could also just pick beta the identity I guess

[candid patrol]

fact

[serene dune]

oh yeah

reminds me of matrices

[rocky cloak]

I guess the more interesting question would be 3 elements such that each commutes with exactly one of the other two.

Not sure if that's possible...

[candid patrol]

Or to find all the (alpha,beta,gamma) possibles

[crystal vale]

in S_5?

[rocky cloak]

Sure, or in whatever group

[crystal vale]

i think it can be possible in large group

[serene dune]

Sn somehow feels like too smoky, need to get a better grasp

[frigid epoch]

If a and b commute, what would c commute with of the two

[velvet hull]

The cycle type decomposition is really useful in understanding Sn

[frigid epoch]

It's just coloring a 3 node graph innit

[serene dune]

yeah, JJR goes through it but idk if it is enough

[rocky cloak]

Yeah you're, it doesn't even make sense.

Could do 4 though I guess

[crystal vale]

yes

[frigid epoch]

With four it becomes trivial, cause it's just two pairs of commutating objects

Like a, a^-1, b, b^-1

[serene dune]

right!

[rocky cloak]

Well with 4 I would want it to not come with exactly one other

[frigid epoch]

Yeah but since commuting is a symmetric relation, you're not gonna get interesting stuff here

Now let's say 4 elements so that each commutes with exactly TWO of the others...

[rocky cloak]

Yeah, I guess there's nothing interesting going on

Kinda puts into question what was the point of the original problem

[serene dune]

would you mind unpacking the first comment here?

[frigid epoch]

If you want n elements, each one commuting with only one other

Then let's pair up two of them, say a and b

And then they're gone. Both of them are now spent

[serene dune]

ahh, ok

[frigid epoch]

In S5, {(1 2), (2 3), (3 4), (4 5), (5 1)} is a set of 5 elements, each of which commutes with exactly two of the others

So that's cool

[serene dune]

yeah, i mean also if it commutes or not is very "deterministic" (possessing common element or not!)

amirite

[frigid epoch]

Yeah I think so

[serene dune]

fairs

[frigid epoch]

Would now be interesting to see if such a 4-tuple exists

[serene dune]

"such" ?

haha, this is being mentioned now!

[proud hedge]

Hey guys i have a problem.
If G is group and for any a and b of G we have the equation (ab)^2=(ba)^2. (and for every x of G if x^2=e then x=e).
prove that G is abelian.

[candid patrol]

Try to show that [a,b]^2 = e, then [a,b] = e because of x^2 = e => x = e

[a,b] = ab(ba)^{-1}

[velvet hull]

You mean aba^-1b^-1

[candid patrol]

yeah mb, thanks

[cloud walrus]

UGOBEL

[somber sleet]

Do you have some algebraically closed fields in mind which is not C over R/Q?

[coral spindle]

The field A of algebraic numbers in C is a countable, algebraically closed field

[south patrol]

Algebraic closure of Fp lol

[night tartan]

if you have 2 algebraically closed fields F_1 and F_2 of the same characteristic, then either F_1 embeds into F_2 or the other way round

so that means all your algebraically closed fields over R or Q can be thought of as some subfield of C, or some extension of C

[night tartan]

(you can prove this using zorn)

[coral spindle]

When I saw F_1 for a second I was like