groups-rings-fields

[velvet hull]

And composition with a bijective perseveres the “ive-ness” of any function

Well a bijective matrix is invertible, but that’s a LA result that’s not too hard to show

Here’s one way to prove this: let B represent the inverse function (which must exist due to bijectivity). Prove that B is linear, and thus B can be represented by a matrix and you get that AB = I

[chilly ocean]

can I use some injectivity argument perhaps? like saying T is bijective --> T is injective --> Ax = 0 has a trivial nullspace --> A is invertible

[velvet hull]

Sure I don’t see why not

You’d have to assume that A is a square matrix which is something that ive been trying to sidestep

[chilly ocean]

i think that's vacuously true since otherwise it wouldnt have an inverse

[tough raven]

Incomparability can't be reflexive, and is often not transitive...

A Galois connection between intermediate subfields and subgroups of the automorphism group?

[rotund aurora]

yeah I wonder if it's just stating what a Galois connection is in general. If it's something else, then I'm curious

[tardy hedge]

Wow everyone wants the notes, i would love to send it to ppl hopefully my prof posts them

Otherwise i have my handwritten notes buttt not the prettiest lol

[serene dune]

well the individual objects are functions right ?

a) so the natural way to counter this would be to find a function ?
idk can't think of anything

[velvet hull]

The kernel is just the set of all continuous functions that is 0 on [0,1]

[serene dune]

for b) i think i have a intuition, which is to make the ideal consisting of those functions which is zero everywhere but a subset wisely chosen to contain I

[velvet hull]

So suppose that f*g is 0 on [0,1]. Can you figure out a way for neither f nor g to be 0 on the entire interval?
Certainly ||at least one of f and g has to be 0 for every point in the interval||

[velvet hull]

For b, here’s a hint. An ideal I in R is maximal iff R/I is a field

[serene dune]

i thought about that but it seemed that to create a ideal bigger than I be easier !

[serene dune]

hmm lemme see

oh yeah i got it, inside [0,1] split it,
then make f to be 0 on one, g to to be 0 on another

too bad that it didnt click

[potent condor]

stupid question: If $k$ is an algebraically closed field and I have an algorithm that can find a root for any polynomial in one variable over $k$ (besides a nonzero constant), does it follow that there's an algorithm for finding a root for any polynomial in $n$ variables?

If $f\in k[X_1, \dots, X_{n-1},X_n]$ has a root $\vec x = (x_1, \dots, x_n)$, then $f(x_1, \dots, x_{n-1}, X_n)\in k[X_n]$ has a root $x_n$ which we can find with our algorithm. But do we have a way of finding $x_1, \dots, x_{n-1}$?

[cloud walrus]

EdgarAlnGrow

[lapis thorn]

Hi! I'm just learning about quotient groups

Am I right in this computation:

[velvet hull]

That doesn’t look right. At the very least any matrix in N can’t be in U/N, but there isn’t anything stopping from x=c=v=1 and u=v=0 in your U/N

[lapis thorn]

Oh so there's a \mathbb{R} in the (1,3) position of that definition of U/N

this was my scratch work

[thorn jay]

Oh right i was confused with a weak order smh

This is the second time I've confused them 😭

Also incomparability is reflexive in any strict order

[glad osprey]

It's true that each coset of N is of the form [x, y, w, etc...]N, but just multiplying them together doesn't really give the quotient group. For comparison, think about the quotient Z/2Z; any element is of the form x + 2n, but the quotient group is not literally the set { x + 2n, | x, n \in Z }. Maybe someone else can explain better why this doesn't work, but I think the point is that each coset is a set, but in your scratch work you're looking at each element inside N, so you just end up with U/N being equal to U

To see what U/N looks like, it's usually better to find a surjective homomorphism from U with kernel N, then use the first isomorphism theorem

[lapis thorn]

You're right and I agree that {2n+z} is wrong. It is supposed to be {z+2\mathbb{Z}} which gives the quotient group as z varies over integers, as per the definition of quotient group

In the same fashion I continued this problem-

As for using a homomorphism to understand U/N, I had it pinned down the other way when I learned the theorem, in the sense that, if I'm asked to find any homomorphism \phi G->G' with a given kernel N, I can reduce it to the easier problem of finding any homomorphism \psi on G/N->G' (no restrictions, any kernel) and just report \psi composition \pi (projection map G->G/N)

So what I'm trying to say is that I'm finding the quotient group U/N for the sake of ease in finding a homomorphism U->U with kernel N, and thus trying to find a homomorphism to understand the quotient group is a catch 22!

[regal zodiac]

If i have finite fields extensions L/K and F/K such that L = K(a_1, ... a_n) and F = K(b_1, ... b_m) then how is defined the field LF? Is is the smallest field containing both L and F ? If so, we can see it as K(a_1... a_n, b_1 ... b_m) no ?

[glad osprey]

Yeah, you can use the first isomorphism theorem in both directions; the projection G -> G/N has kernel N. Note that \psi comp \pi may have kernel bigger than N if \psi has non-trivial kernel. But I'm not sure how often you need to find a homomorphism with a specific kernel, usually you're given G/N and you want to find out if it's isomorphic to some group you're more familiar with

[lapis thorn]

Ah my bad - nontrivial kernel indeed. Okay, thanks!

[proper jolt]

for the second sentence does the image of $\phi$ have to be contained in R''?
If this is true then there are subrings of R' that cannot be subalgebras right?

are we thinking of the subalgebra R'' as a R-submodule where scalar multiplication $rx = \phi(r)x$?

[cloud walrus]

thebirdsandbees

thebirdsandbees

[torpid knot]

the subgroup part is trivial but im stuck on which subgroup it would be a centralizer of

(herstein 2.5)

would it not just be G?

[velvet hull]

not the centralizer, the center. the centralizer is a different (but related) thing

[velvet hull]

but I'm not sure what exactly you mean by your question

[torpid knot]

oh here's the previous question that introduced the C(H) notation

it's asking for what subgroup T is Z the centralizer of T

[proper jolt]

yeh then i think it just G

[velvet hull]

not necessarily (?) let me check

[torpid knot]

oh lmao aight

they want the centralizer of a subgroup here rather than the centralizer of an element

so ye

[velvet hull]

mmmh, I'm not sure. inclusion is certainly true, but how are you sure that there's nothing else in the centralizer? oh lmao you said G I'm dumb

[velvet hull]

one note I want to make is that the centraliser of any subgroup / element / subset always contains the center, but it doesn't have to be just the center

for example, if x is not in Z(G), C(x) still contains at least Z(G) AND all the powers of x, because x certainly commutes with itself. but the powers of x may or may not actually commute with everything else in the group

[torpid knot]

oh icic

[tough raven]

Assuming that L and F are both subfields of a larger field extension (say) E/K, yes.

Without that, LF is not well-defined.

[proper jolt]

i don't even know what a R-subalgebra is

[tough raven]

Just saying but subalgebra is defined here.

[vagrant zinc]

definition (xhy^{-1})*h

[proper jolt]

i dont understand the definition, that's why i asked

[tough raven]

Do you understand the definition of an R-algebra?

[proper jolt]

yeh ring R' equipped with ring map R-> R'

[tough raven]

So a subalgebra is a subset which can be made into an algebra with the same operations.

That means it (call it R'') has to contain identities and be closed under addition, negation and multiplication - so a subring - and that the ring map phi: R → R' can also be treated as a ring map from R to R''. This is possible if and only if phi(R) is a subset of R''.

So a subalgebra is a subring containing the image of the ring map, phi(R).

[proper jolt]

this is from the sc above, so then does this sub algebra use a ring map different from $\phi$ because $x_{\lambda}$ may not even be in the image of $\phi$

[cloud walrus]

thebirdsandbees

[tough raven]

There's nothing requiring it to be in the image of phi.

[proper jolt]

so then here we can define a different ring map?

[tough raven]

Nope, it's the same map.

The codomain R'' is bigger because of containing the x_\lambda's, but the domain and the map are the same.

If it helps, imagine R is a subset of R' and phi is the inclusion. Then a subalgebra is a subring containing R as a subset. This clearly does not prevent it from containing more elements.

[proper jolt]

but then your extending the map because your extending the codomain. Isn't what they're doing something like defining the sub algebra as the resulting image subring from the map R-> R[X_1, X_2, ..., X_n] -> R'' Where the first map is inclusion, second map sends X_\lambda to x_\lambda

[proper jolt]

then the resulting sub algebra wouldn't be a subset of the image of the inclusion map

[tough raven]

The image needs to be a subset of the subalgebra, but not the other way round.

[proper jolt]

that makes sense

tahnks

thanks

[serene dune]

idempotents in Z_n

what should be the approach

to count or find them

holy shit, aint reading all that

i also got this, but i'm really confused and don't have the patience to go through it now

[sudden dust]

Does the proof of the forward of this statement seem fine?

Let n ∈ Z where n > 1, and let G = {[a] : a ∈ Z, 1 ≤ a ≤ n-1}. Then G is a group under multiplication mod n ([a][b] = [ab]) if and only if n is prime.

Proof of forward:

Suppose G is a group under multiplication mod n, and assume for a contradiction n is not prime. Then there exist a,b ∈ Z such that 1 < a < n, 1 < b < n, and n = ab. Thus [a],[b] ∈ G. Since G is a group, we have that [a][b] ∈ G. Observe that [a][b] = [ab] = [n] = [0], so [0] ∈ G. Thus [0] = [k] for some k ∈ Z, 1 ≤ k ≤ n-1. Then k ≡ 0 (mod n), so n | k. Since k is positive integer, it follows that n ≤ k, which contradicts 1 ≤ k ≤ n-1. Therefore n is prime.

[lusty marlin]

Yes, this is fine

[sharp ice]

any small hints to prove 4 implies 1 ? i have 1 -> 2 -> 3 -> 4. just need 4 -> 1

[wispy stag]

Try contrapositive, if you have a submodule which is not finitely generated you should be able to pick elements that then create a sequence which does not satisfy the rule in 4

[rocky cloak]

Chinese remainder theorem should take you most of the way

[serene dune]

alr, i shall try

[sharp ice]

thank you ! i have it now, thanks

[regal zodiac]

We can always see L and F as subfields of the algebraic closure of K no?

[rocky cloak]

They can always be embedded in an algebraic closure, but not necessarily in a unique way.

You can for example think of something like L = Q(cuberoot(2)) and F = Q[x]/(x^3 - 2)

What should LF be in this case?

[candid patrol]

Hello, Is it possible to determine all the representations of the dihedral group $D_{2n}$ of order $2n$ as a semidirect prpduct for any $n$? I think so, since such a construction depends on the choice of a non-trivial normal subgroup. If I start with even $n$, I know that the normal subgroups of $D_{2n}$ are the subgroups of $\langle r \rangle$, $H = \langle r^2, s \rangle$, and $H = \langle r^2, sr \rangle$. Should I try, for each of these subgroups, to find if there exists another subgroup of order equal to the index of $N$ (the normal subgroup) in $D_{2n}$, and verify that its intersection with $N$ is trivial? Or is there a less linear method?

[cloud walrus]

UGOBEL

[regal zodiac]

Oh OK i see the problem ! Thanks 🙂
Suppose this is well defined, then if L is generated by a_1, a_n and F by b_1 , b_m, then LF is just K(a_1, a_n b_1, b_m)?

[serene dune]

Every countable group has only countably many subgroups.

can I get a counter with Z?

if not what are some other ways to not only porduce this sort of thing but to generalise the pheonomenon if that makes sense

i can kinda smell free group, but no idea how to really go ahead

[rocky cloak]

Well you still need them to live in some common extension.

Like in the example I mentioned above you wouldn't say LF = K(cuberoot(2), x) probably.

[regal zodiac]

Yeah assuming they live in a common extension 🙂

[rocky cloak]

So I guess the way to think about it is that a countable generated group is countable, but the amount of subsets of the generators will be uncountable.

So in situations where you have few relations between generators this should give you an uncountable amount of subgroups.

For example the free group would work, or the free abelian group, but also things with more relations like Q would work.

[proper jolt]

for this proof can someone explain to me why $\Lambda_1$ is finite?

[cloud walrus]

thebirdsandbees

[rocky cloak]

By definition of free module, every element is a finite linear combination of basis vectors

Like finite linear combinations are the only things that make sense

[proper jolt]

yeh i guess that makes sense, but i was thinking something like converging sequence in the rational numbers and such

[rocky cloak]

Well then you're imposing a lot of structure that is not stated in the theorem

[proper jolt]

then doesn't that mean that every module should be a finite linear combination of its generators (if it has one)

[rocky cloak]

I don't what you mean by a module being a finite linear combination.

But a set generates a module if every element is a finite linear combination of those generators yes

[proper jolt]

cuz i thought that free module just means that it has a free basis. These are the definitions im refering to

[rocky cloak]

Indeed, and there is no notion of infinite sum here.

The notation isn't super clear, but the sum means finite linear combination, because infinite linear combinations is not a thing that is defined for modules

[proper jolt]

ok gotcha. i guess, also, cause there's nothing that tells us if infinite linear combinations are even closed under modules right?

[south patrol]

3 is always false by considering the empty set (jk)

[rocky cloak]

Exactly

[sharp ice]

(tell that to eisenbud)

[south patrol]

Well there's nothing telling us it even makes sense

there needn't be any topology/metric etc

[proper jolt]

no but even if there was the axioms of a module doesn't even guarantee that a module is closed under infinite linear combinations, i think

[south patrol]

Well that is hypothetical, like to he closed under an operation the operation should make sense first

[thorn jay]

Infinite linear combinations do not make sense, they are not defined

The only reason we can define them in the real numbers, is because of the fact that the real numbers are a complete metric space, and we consider the limit of finite partial sums

Dealing with that fact is basically the core idea of analysis, among infinitesimal things

[serene dune]

ok i can comprehend the Q coz of its Z \oplus Z like structure, but i will try to look at other two too

[rocky cloak]

So the other two should be the more obvious ones.

If you take the subgroup generated by any subset of generators in the free group, it won't contain any of the other generators.

For Q it's a little more subtle, but has to do with unique prime factorization. If you look at the subgroups generated by things of the form 1/p for p a prime, then every subset of primes gives you a different subgroup.

[proper jolt]

oh ok thanks

[thorn jay]

Yw!

[serene dune]

oh right!

got it now

its so trivial after i see it, does a demon open up some door in other dimension or what

[tough raven]

Yes. In functional analysis it is common to consider vector subspaces V of a vector space W where there are infinite series which are well-defined in W but whose sums don't lie in V even though all the terms do (e.g. V could be any dense subspace of a normed space W other than W itself).

[velvet hull]

nooooo clutches my Hilbert space

[tough raven]

Dense subspaces of Hilbert spaces matter too.

[candid patrol]

Can someone confirm this ?

I tried to find all the semidirect product représentation of D12

[tardy hedge]

Holy Guacamole!

[candid patrol]

🤣

[sly crescent]

Looks right to me

[candid patrol]

Hope so haha

It took me 1 hour 😵

[thorn jay]

Damnn thats a long time

[candid patrol]

I made to much mistackes

But it is not that hard u right

[thorn jay]

I mean, i suppose you just gotta go along every normal subgroup right?

[candid patrol]

Yeah 👍

[thorn jay]

(Assuming every quotient has a natural embedding)

(Aka assuming every induced exact sequence splits)

[tardy hedge]

I spend one hour on two sentences sometimes -_-

[candid patrol]

Let’s try D44 now !!

[thorn jay]

Hell

Also, you forgot one

[candid patrol]

What really?

[thorn jay]

1 \rtimes D_12

:3

[candid patrol]

…

[thorn jay]

Oh and
D_12 \rtimes 1
Too!!

[candid patrol]

You right haha

[thorn jay]

Felt

[tardy hedge]

🙏

[tardy hedge]

The splitting field of x^3-2 over Q is not just Q(sqrt3(2)), because there are also complex roots, but, if you adjoin a complex root w instead and consider Q(w), Q(w) should be isomorphic as fields to Q(sqrt3(2)), right?

[wraith cargo]

are you sure?

Okay wait lol I have to think about this

yeah that's true

I had a moment there for a sec lol

[glad osprey]

I'm reading about the exact same thing right now

[tardy hedge]

Its also true about the field isomorphism i said?

[tardy hedge]

Oh wow nice description. What text?

Dummit and foote didnt make it as clear as that

[glad osprey]

Aluffi's Notes from the Underground. It's so fun to read

[wraith cargo]

yeah that's what I meant
The idea is that if you adjoin any roots it's just Q[t]/(t^3-2)

[south patrol]

Indeed for any two roots of an irreducible polynomial there's a homomorphism taking the extensions by adding one element to the other since you can do this ting

[thorn jay]

yeah, what I'm guessing happens is that x^3 - 2 isn't irreducible anymore in Q(sqrt3(2)), so that's where these two new roots "pop up" out of nowhere

[south patrol]

you don't get any more roots, just one

Writing omega like this had me very confused tho lol

Cause often omega denotes a third root of unity and sqrt3(2)w is a cubic root of 2 lol

[thorn jay]

yeah, well, I've seen zeta used

[south patrol]

Both are often used

[tardy hedge]

Ohhh yeah cool good point

I was trying to think about what makes them different if the field adjoin “”root”” are all isomorphic

[tardy hedge]

Wdym by this?

I guess im still trying to figure out what this all means

[south patrol]

Well this makes it sound like x^3 -2 splits after you add one root

And I don't really understand what is meant by it if that isn't what is meant

[tardy hedge]

Idk i was gonna say smth

On the tip of my tongue

[tough raven]

One way to think about it is that (call the roots x, y, z) the roots x and y are indistinguishable, but the pairs of roots (x, x) and (x, y) are not.

Actually that doesn't really capture it.

The splitting field is larger because the pairs of roots (x, y) and (x, z) are also not distinguishable.

[tardy hedge]

I think i got a bit lost

[tough raven]

(Thinking too hard about this probably leads to reinventing Galois theory.)

[tardy hedge]

Lolll

[glad osprey]

@tardy hedge have you learned about splitting fields yet?

[tardy hedge]

Thats the section im on in dummit and foote

13.4

Just started reading about it

I was reading thru example of splitting field for x^3-2 over Q

I guess the “over Q” part is not necessary to say

[glad osprey]

I see. Just learning this myself, but I think the idea is that the "nicest" extensions, in particular the Galois extensions, are algebraic extensions where if you add one root of the minimal polynomial you simultaneously get all roots

[tardy hedge]

It sounds like if i learn more things will make more sense in general

I havent gotten to anything “galois” yet

[tardy hedge]

Oh cool

[glad osprey]

Me neither, but Aluffi keeps teasing about "coming attractions" (ie Galois extensions)

[tardy hedge]

I see. Aluffi sounds nice

Algebra chapter 0?

[glad osprey]

Notes from the Underground

[tardy hedge]

Oh yea

I heard algebra chapter 0 is also good

Presents algebra with category theory from the start

[glad osprey]

Is it okay if I post another screenshot? He writes about the difference between Q(\sqrt(2)) and Q(\cbrt(2)) and why the latter isn't Galois

[tardy hedge]

Why wouldnt it be okay?

[glad osprey]

Just in case it felt spammy

[tardy hedge]

We are indeed discussing groups-rings-fields in the channel groups-rings-fields

Lol

Fair

[grizzled spindle]

for a general hermitian inner product and the norm induced, is it true that norm(v+w) = norm(v) + norm(w) iff v is a scalar multiple of w?

[tardy hedge]

I would like to see this

[tardy hedge]

This exposition sounds so nice

I feel like there really is a benefit from swapping and learning from different texts

[tough raven]

A non-negative real scalar multiple, yes.

[glad osprey]

Yeah, I'm swapping between Aluffi, Fraleigh and a little bit of Field Theory by Steven Roman

[tardy hedge]

Fraleigh like the intro book? “A first course in abstract algebra”?

[glad osprey]

Yeah, there's a little bit of Galois theory at the end

[serene dune]

aluffi is chapter 0 right?

[glad osprey]

In this case it's Notes from the Underground

[serene dune]

by Fyodor dostoevsky

[grizzled spindle]

thx

[south patrol]

Lol yeah this is what I think every time

[serene dune]

hey potato would you look at my recent modmail?
for some thread related concern

[glad osprey]

I have a quick question relating to the above discussion: so a1 = \cbrt(2) is a root of x^3 - 2, and let a2 and a3 be the other complex roots. The difference between Q(a1) and Q(a2) only comes from the way it is embedded in C, right? Because the construction of the extension field Q[x]/<x^3 - 2> is unique, but it can be embedded in C three different ways, correct?

[thorn jay]

it means that, while two roots may be indistinguishable (same minimal polynomial), the relationship between these roots may not be

[thorn jay]

they differ by an automorphism of their algebraic completion, so to speak, if you take the conjugate of every element in Q(a1) you get Q(a2)

oh wait I misread

well, yeah you're correct

it's just that the embeddings of Q(a2) and Q(a3) "differ" by an automorphism of C

while the embedding of Q(a1) is wholly different

[glad osprey]

I see, thanks crazy that it's so asymmetrical I'm so used to thinking of isomorphic objects as "the same", I feel like this is the first time I'm forced to think about how things are isomorphic (or embedded in this case)

[thorn jay]

Ah well it's the same with groups/modules

I mean thats what the Ext functor is all about

You quotient a group/module by another group/module, but if N and M are isomorphic, then G/N and G/M need not be

[glad osprey]

Yeah, I just realized, I have seen that before Z/2Z vs Z/3Z for example

[thorn jay]

Ongg

For modules it's already bad, but for groups it's hopeless

Considering how many different products exist

I mean, of course you can work with isomorphic objects, but then you've gotta consider the different embeddings

Which is arguably more of a pain

[tardy hedge]

Bro said he wasnt well versed in field theory

[tough raven]

No, it also differs from the other two by an automorphism.

[tough raven]

It's not asymmetrical at all.

[tardy hedge]

Bro came to shake shit up^

[thorn jay]

I meant of the field Q[i]

[crystal vale]

how to find the number of 5-sylow subgroup in S_6 ?

[candid patrol]

I would find the 5-Sylow of S_5 at first

[crystal vale]

why?

[candid patrol]

I don’t know if it’ sa french notation, but v_5(5!) = v_5(6!)

[crystal vale]

what is v_5(5!) ?

[candid patrol]

The 5-Sylow of S_5 embed into the 5-Sylow of S_6

[candid patrol]

5! = 2^3 • 3 • 5
6! = 2^4 • 3^2 • 5

So v_2(5!) = 3, and v_2(6!) = 4 for exemple

It is called the p-addique valuation i think

[crystal vale]

i don't know about it

[cloud walrus]

UGOBEL

[candid patrol]

For S_6, the anser is n5 = 36

[glad osprey]

It's a little asymmetrical: Q(a1) is a subset of R, but Q(a2) and Q(a3) are both dense in C

[cloud walrus]

UGOBEL

[candid patrol]

In fact, it's possible to find the p-Sylow of any Sn by using wreath product if I remember

[cloud walrus]

UGOBEL

[crystal vale]

where i can learn these all ?

[serene dune]

only seen some wreath in 2^n order

[serene dune]

try rotman's book(the group theory one)
idk if he covers this

but a good text i believe

[candid patrol]

i'm still trying to understand the definition lmao

$\wr$

[cloud walrus]

UGOBEL

[candid patrol]

That is magic !!

[serene dune]

hahaha

i will try to recall and further dive into wreath tonight

good remainder

[candid patrol]

☠️

[crystal vale]

there is a many book by Rotman, which one? and i did one of that but i didn't found this type of question

[candid patrol]

Try to find the p-Sylow of S4 and S5, i got the correction if you need (in french but I can translate with GPT)

[candid patrol]

I was wrong mb

[serene dune]

i would dm you, let me reach home

i would give you my group theory directory

[rocky cloak]

25 doesnt divide 6!, so you're just counting the number of subgroups of order 5.

So that's the number of 5-cycles divide by 4 (each subgroup consider of the identity and four 5-cycles)

[crystal vale]

why i have to divide it by 4?

[candid patrol]

4 = phi(5)

The number of distincts 5-cycles in a 5-Sylow subgroup is the generators of the 5-Sylow

[candid patrol]

I wrote it

[pearl fog]

is p-2 always a generator of zpx?

[crystal vale]

i see

thank you

[hidden cairn]

i know the ideals of the quotient correspond to an ideal in F[x] containing (p(x)) by the isomorphism theorems for rings. but can i also describe the ideals as follows:

Let $\deg F = q, \deg p = n$ Then the quotient space is a vector space with basis $\overline{1}, \overline{x}, \dots, \overline{x^{n-1}}$ Any ideal of the quotient space is spanned by a set of the form ${ \overline{x^k}, \overline{x^{k+1}}, \dots, \overline{x^{n-1}} }$ for some $0 \leq k \leq n-1$.

[cloud walrus]

pink_panther

[potent condor]

I have a ring $R$ and ideals $I,J\leq R$. Intuitively, I want to compute $R/(I+J)$ by ``first modding out by $I$ and then modding out by $J$". Formally, I imagine this is done roughly as follows:

$$\frac{R}{I+J}\cong \frac{R/I}{(I+J)/I}\cong \frac{R/I}{J/I\cap J}$$

where the first relation comes from the third isomorphism theorem and the second relation comes from applying the second isomorphism theorem to $(I+J)/I$. Of course, this doesn't make sense because $J/I\cap J$ is not even an ideal of $R/I$. But is there a simple tweak I can make to the numerator $R/I$ to get this to work?

[cloud walrus]

EdgarAlnGrow

[thorn jay]

You can take the quotient of two ideals where one is contained in the other, just like how you'd quotient modules

So, if I is contained in J, then
J/I := { j + I | j in J }
Which is an ideal of R/I

[potent condor]

but im not assuming I is contained in J

[thorn jay]

Yeye

Lemme finish

[potent condor]

no

u had ur chance

u are now discredited as a mathematician and banned from the server

[thorn jay]

So then you have the third isomorphism theorem, saying
R/(I + J) = (R/I) / ((I + J) / I)

Then you can compute that
(I + J) / I = { j + I | j in J }
Which is still an ideal, but not called J/I, this has to with some preimage stuff

[thorn jay]

NOOO

I mean what you're essentially doing is working with the ideals and R as R-modules

[wintry sluice]

are there rings where one of the operations is function composition? or just groups

[glad osprey]

Yes, for example the set of endomorphisms of an abelian group is a ring under composition and pointwise addition

[thorn jay]

even better, any ring has a canonical embedding h : R -> End_Z(R), as it is a left-module over itself, making every ring isomorphic to some ring of functions under pointwise addition and composition

[tough raven]

I don't think you can get what you want by changing the numerator R/I. Personally, I think of the second form (R/I)/((I + J)/I) as expressing the intuition you stated; note that (I + J)/I = {j + I : j in J} is the image of J under the quotient map R → R/I, so quotienting by it can be reasonably considered "further quotienting by J". (Applying this image under a canonical homomorphism is something that also comes up when you try to show that quotients and localisations commute and it can be justified as the correct interpretation based on universal properties if necessary.)

[white oxide]

Are there no non-injective homomorphisms from Z to Z (apart from the 0 map)

[coral spindle]

What do you think? Remember that a homomorphism out of Z is just a choice of where to send 1

[white oxide]

No there aren't

[coral spindle]

Why?

[south patrol]

There is only one homomorphism from Z -> Z, namely the identity
||if you mean rings, since you didn't state the category. Jk trolling over||

[thorn jay]

||because Z has no torsion elements which aren't 0? h(n) = 0 => n h(1) = 0 => h(1) = 0 => h = 0||

[coral spindle]

Yup

||In general there is a bijection {non-injective group homs Z -> G} <-> {torsion elements of G including the identity}, which you have figured out here||

[chilly ocean]

Woah

[coral spindle]

In fact ||the order of an element g in G is n, where nZ is the kernel of the group hom from Z to G sending 1 to g||

[thorn jay]

hmhmm that makes sense!

[coral spindle]

Hence the order of 1 in Z should be 0

[thorn jay]

right

[coral spindle]

C.f. the characteristic of a ring and the unique map Z -> R for any ring R

[thorn jay]

mfw the characteristic of a field creates a division between which fields homomorphisms can exist hence is racist characteristicist

[chilly ocean]

For discussing groupoids, would this or #advanced-algebra be the correct channel?

[south patrol]

Probably #category-theory lol

[serene dune]

however your category my honour

[chilly ocean]

Ic

[dull tiger]

Calculate $\frac{2X^2}{2X^2 +1}$ in $\mathbb{F}_7/(X^3+X+1)$
So I know how to do polynomial division in Rings like $\mathbb{Z}/n\mathbb{Z}$, but for this one i have no idea how to even start.

[cloud walrus]

mathrie

[violet spade]

this is a hard question i think

Well firstly it's easy to see that $\frac{2X^2}{2X^2+1}=\frac{2X^2+1-1}{2X^2+1}=1-\frac{1}{2X^2+1}$.

[cloud walrus]

joel

[violet spade]

let's try adding $2X^3+2X+2$ to the top part of the fraction (this is allowed, because it's equal to zero in this ring)

[cloud walrus]

joel

[violet spade]

So $\frac{1}{2X^2+1}=\frac{2X^3+2X+3}{2X^2+1}$

[cloud walrus]

joel

[violet spade]

Notice that $X=1$ is a root of $2X^3+2X+3$, so we may factor as follows: $2X^3+2X+3=(X-1)(aX^2+bX+c)$. Now we just need to find the coefficients of a,b,c and hope it cancels nicely

[cloud walrus]

joel

[violet spade]

$2X^3+2X+3=(X-1)(aX^2+bX+c)=aX^3+bX^2+cX-aX^2-bX-c=aX^3+(b-a)X^2+(c-b)X-c$. Comparing coefficients yields the following: $\begin{cases} a=2 \ b-a=0 \ c-b = 2 \ -c =3 \end{cases}$

[cloud walrus]

joel

[violet spade]

Therefore $a=b=2$ and $c=4$. Thus $2X^3+2X+3=(X-1)(2X^2+2X+4)$.

[cloud walrus]

joel

[violet spade]

$=(2X-2)(X^2+X+2)$. Notice that $X=3$ is a root of $X^2+X+2$. It holds that $X^2+X+2=(X-3)(X+4)$. In the end, we have:
$\frac{(2X-2)(X-3)(X+4)}{2X^2+1}$. Now try to work this out for yourself.

[cloud walrus]

joel

[violet spade]

@dull tiger

[dull tiger]

So essentially now we want $(2X-2)(X-3)(X+4)=1$

[cloud walrus]

mathrie

[violet spade]

Well this is always true in this ring

[dull tiger]

yes

[violet spade]

Maybe doing a smart trick where you have a common factor of 2X^2+1 inside of the numerator would help...

Like, $2(X-3)(X+4) = 2X^2+2X+4=2X^2+1 + 2X+3$. So
$\frac{(2X-2)(X-3)(X+4)}{2X^2+1}=X-1+\frac{(2X+3)(X-1)}{2X^2+1}...$

[cloud walrus]

joel

[violet spade]

it's kind of just trying stuff and hoping it works...

[dull tiger]

i'm so not used to this stuff haha

ah wait i think i see now where its going

okay so i got now $X-1 + \frac{(2x+3)(X-1)}{2X^2+1} = X-1 + \frac{2X^2+1+X-4}{2X^2+1} = X-1 +1 + \frac{X-4}{2X^2+1}$

[cloud walrus]

mathrie

[dull tiger]

what do i do with the last part?

[violet spade]

Okay, perhaps i have taken the wrong path, i apologise.
Let's calculate the inverse $(2X^2+1)^{-1}=\frac{1}{2X^2+1}$. So we want to find a polynomial $aX^2+bX+c$ such that:
$(aX^2+bX+c)(2X^2+1)=1$. \ Re-write:
$2aX^4+aX^2+2bX^3+bX+2cX^2+c=1$. \
Note that $X^3=-X-1$, so $X^4=-X^2-X$. \
Thus
$2a(-X^2-X)+aX^2+2b(-X-1)+bX+2cX^2+c=1$. \
Now:
$(-2a+a+2c)X^2+(-2a-2b+b)X+(-2b+c)=1$. \
By comparing coefficients, we have:
$\begin{cases} 2c-a = 0 \ -b-2a = 0 \ c-2b=1 \end{cases}$ \
So $a=2c$, and $b=-2a=-4c$. \
We have $c-2b=c+8c=9c=2c=1 \Longrightarrow c = 4$. \
Now directly we have $a = 2(4)=1$ and $b=-4(4)=-16=5$. \
So $\frac{1}{2X^2+1}=(2X^2+1)^{-1}=X^2+5X+4$. \
Now, $\frac{2X^2}{2X^2+1}=1-(X^2+5X+4)=-X^2-5X-3=-X^2+2X+4$.

@dull tiger

[cloud walrus]

joel

[violet spade]

I leave the verification for you to do yourself 😄

[rocky cloak]

So computing the inverse of polynomials in a quotient ring works the same as finding inverses in Z/n with Euclid's algorithm. So here it would be

x^3 + x + 1 = 4x(2x^2 + 1) - (3x - 1)
2x^2 + 1 = (3x + 1)(3x - 1) - 5

So 4x(3x + 1) multiplied by (2x^2 + 1) is 5 modulo x^3 + x + 1. So you're fraction simplifies to
2x^2(4x)(3x+1)/5
and 5*3 = 1, so it's
6x^2(4x)(3x+1)

Which can then be simplified further if you expand and so on

[violet spade]

crazy!

didnt know that trick

nice

[dull tiger]

ah okay, thanks. so in the end i am supposed to get something that is not a fraction?

[dull tiger]

ouh thats cool! Thank you 🙂

[violet spade]

Yes that's what "work it out" means. And since X^3+X+1 is irreducible in F_7[X], we have that F_7[X]/(X^3+X+1) is a field so it is definitely possible.

[dull tiger]

great, thanks for all the help!

[empty perch]

Hey folks, just a quick reminder that the Abstract Algebra Lecture series is meeting regularly again ^_^
We will meet to discuss Cycles and Presentations in about 45 minutes over in the ⁠⁠#1055201711679082516. More information can be found in our thread: ⁠⁠#1317307081535000606

[tardy hedge]

if a polynomial is not symmetric, it cannot be written as a polynomial in terms of elementary symmetric polynomials right?

because a polynomial f(s1, s2, .. sn) where the s's are the elementary symmetric polynomials over n variables is always symmetric

This was introduced in my number theory class but i have no idea what its useful for

[rocky cloak]

The fact that if a polynomial is not symmetric it's also not a polynomial of elementary symmetric polynomials is not particularly significant. And should be fairly immediate, just following from the argument you gave.

What is much more interesting is that if a polynomial is symmetric, then it will be a polynomial of elementary symmetric polynomials in a unique way!

[tardy hedge]

yea i see, my prof wrote that theorem but skipped the proof

[static mauve]

I was trying to figure out if Z_2 x Z_12 is isomorphic to Z_4 x Z_6. They both have the same number of elements, and the max order of an element from each seems to be 12. I painstakingly wrote out the the order of all the elements of each group to see if there orders match up between these 2. It ALMOST seemed like they matched up, but I got that the order of elements with 12 and 6 didnt quite match up (I want to believe I made a mistake, but couldnt actually find one.)

Is there an easier way to see if said groups are isomorphic? I have a similar set of problems where the number of elements is WAY more than 24 so this process I tried wouldnt really be feasible.

[coral spindle]

Yes, there is a much easier way

You need the Chinese remainder theorem

Z_a x Z_b is isomorphic to Z_ab iff gcd(a, b) = 1.

Now factorise 12 and 6 and see what you get out of the isomorphisms you obtain

I should add that if you're being asked these questions, then chances are you've either seen the Chinese remainder theorem or the classification of finite Abelian groups and you're being asked to use them here. You should probably go back and check your notes.

[static mauve]

Yeah, I am familiar with: Z_a x Z_b is isomorphic to Z_ab iff gcd(a, b) = 1

But since 2 and 12 have a gcd greater than 1, then this tells me that Z_2 x Z_12 is not isomorphic to Z_24

Maybe Im just not being clever here 🙃

[thorn jay]

It tells you that Z_2 x Z_4 x Z_3 is the unique maximal decomposition of that group

For Z_4 x Z_6 you can do something similar

[coral spindle]

Isomorphisms work both ways. So Z_ab is isomorphic to Z_a x Z_b. Maybe you can notice that 12 = 3 * 4

And if it's not clear, if A is isomorphic to A' then A x B is isomorphic to A' x B.

[static mauve]

Gotcha. Ill stew on this a bit. Thanks for giving me some direction!

[static mauve]

Ok so just to be clear, I worked through the example I mentioned above, and it does seem that these are isomorphic, correct?

Additionally, another problem asks if: Z8 x Z10 x Z24 is isomorphic to Z4 x Z12 x Z40... and these DONT seem isomorphic, right?

[coral spindle]

Yes, the groups you listed before are isomorphic.

Why do you think these new groups are not isomorphic?

[static mauve]

Z8 x Z10 x Z24 is isomorphic to Z8 x Z2 x Z5 x Z8 x Z3

and

Z4 x Z12 x Z40 is isomorphic to Z4 x Z3 x Z4 x Z8 x Z5

[rocky cloak]

Yes this is correct. In general finite abelian groups can be factored as a product of cyclic groups whose order is a prime power in a unique way.

Alternatively it's also possible to calculate that the former has 96 elements of order 8 while the latter has 64.

Or perhaps easier to calculate, the former has 32 elements whose order divides 4, while the latter has 64

[velvet hull]

any update?

[minor fulcrum]

if a finite group's elements are all order 2, it's isomorphic to (Z/2Z)^n for some n (I think this is the case). Can we say anything about an infinite group whose elements are all order 2, aside from the fact that it must be abelian?

[coral spindle]

Sure. It's an F_2-vector space. So in some sense we know the structure entirely.

[rocky cloak]

As Boytjie is saying it's the same as the finite case.

It's isomorphic to some direct sum of copies of Z/2

[sly crescent]

What can we say about a finite group whose elements (besides the identity) are all order 3?

[minor fulcrum]

Neat! That's very clean

[coral spindle]

Hm ok so I don't know

But

I know about the Heisenberg group

Now

[rocky cloak]

I'm not sure you can say much more than you can about p-groups in general

[coral spindle]

Question: are there indecomposable, non-Abelian groups of arbitrary size whose elements are all of order 3?

Surely yes

[minor fulcrum]

Interesting, the group with three elements and the groups with 9 are all abelian

So this is kinda like the smallest possible non abelian such group

[coral spindle]

Indeed

[sly crescent]

Does every Tarski monster group have exactly 2 conjugacy classes of elements?

[rocky cloak]

You can generalize the Heisenberg group to get larger groups. I would assume these are still indecomposable.

[coral spindle]

Oh duh doy

I feel silly

[sly crescent]

(These are the extraspecial groups of plus type)

[sly crescent]

Okay I guess not

[rocky cloak]

It appears none of them do

https://math.stackexchange.com/a/2795604/306319

[sly crescent]

Yeah

Do any Tarski monster groups have exactly 3 conjugacy classes of subgroups?

[rocky cloak]

Seems plausibel, but I don't know much about them

[finite turtle]

how to show that to get a presentation on factor group G/N from G you just need to add the relations for N?

[lusty marlin]

Since conjugation preserves the order of a subgroup, the subgroups of a group with exactly 3 conjugacy classes of subgroups can have at most 3 distinct cardinalities, and since we need to include the trivial subgroup and the whole group, there can only be one more subgroup cardinality. This implies that the cardinality of the group can have at most one prime factor, and must in fact be the square of a prime.

[dull tiger]

How do i know if there is a non-abelian group of order n?
Let's say i have a group of order 15. This can be C_15 which is isomorphic to C_3 x C_5. But how can i make sure that there are no more non-abelian groups of order 15?
Is it because there is only one sylow-subgroup of order 3 and order 5 respectively?

[thorn jay]

When you've got any non-prime greater than 4 you can take a nontrivial semidirect product C_n \rtimes C_m where n ≠ 2, n >= m and n * m is your order i think

So that guarantees at least one non-abelian group

[dull tiger]

but groups of order 15 are always abelian

[thorn jay]

Ah, nevermind then

Groups are weird

[rocky cloak]

The cardinality if Tarskis monster is countably infinite

[rocky cloak]

In general it can be hard, but for 15 or more generally for order pq a product of two primes p>q, then
The number of p-sylow subgroups must divide q, but also be 1 modulo p. Then it has to be 1.

So Cp is a normal subgroup. That means your group is a semidirect product of Cp and Cq.

The semidirect products are classified by maps Cq -> Aut(Cp) = C[p-1].

So a non-abelian group exists if and only if q divides p-1

Similar arguments can work in general, if you can argue enough sylow subgroups are normal, then the group must be a product of its sylow subgroups.

And groups of order p or p^2 are always abelian

[boreal inlet]

Let's say I have the canonical projection of a commutative unital ring R to R/I for an ideal I of R.

Let's assume we have an arbitrary collection of ideals in R/I. If we take the sum, is the preimage of the sum wrt the canonical map is equal to the sum of the ideals of R we get via preimage?

[thorn jay]

Yes, you get a complete lattice embedding of the lattice of ideals of R/I into the lattice of ideals of R

[cloud walrus]

Riku

[boreal inlet]

So this is fine right?

[thorn jay]

Should be

[boreal inlet]

I was trying to show the topological properties of zero sets of these ideals in the affine coordinate rings

Thanks, this makes showing the results easier

[worn kiln]

this is just axiom of distribution isnt it?

[thorn jay]

:3

Lattice theory ftw, take that Bourbaki

[boreal inlet]

Oh I'm not familiar with the term, what are you referring to?

[worn kiln]

i dont know the proper english term for it, but i was referring to this. is it called law of distribution? let me know @boreal inlet
(a+b)c=ac+bc

[lusty marlin]

Do all of its elements have infinite order?

[coral spindle]

People would usually call this the distributivity of multiplication

[worn kiln]

oh alright thanks

[coral spindle]

But the screenshot you are replying to is not the same thing, it should be added

[boreal inlet]

Nono, it's not the case here

[worn kiln]

is pi^(-1) an inverse function?

[boreal inlet]

Here the map pi is surjective, and it's required

[coral spindle]

pi^-1(X) is notation for the preimage of X under pi

[boreal inlet]

It is denoting the preimage of some set. What elements of R when mapped through via pi gives the set

[worn kiln]

oooh i see, i thought this was just 1/pi lol

i should stfu when i have no clue xd

thanks tho

[boreal inlet]

Yeah you won't be able to define that when you don't have the map to be bijective

[rocky cloak]

No all elements have order p.

A Tarski monster is a group where every proper subgroup has order p.

[lusty marlin]

Ah ok

[rocky cloak]

Tarskis monster only exist for sufficiently big primes, so by pigeonhole x would have to be conjugate to x^a for some a.

If you take y such that yxy^- = x^a then x and y generated a non-abelian group of order p^2, which can't exist.

In other words Tarskis monster must have at least p conjugacy classes

[graceful dome]

Is this just bad notation by my textbook?

[rocky cloak]

What part of the notation do you find bad? It's all pretty standard

[graceful dome]

From my understanding nx here refers to x+x+x+x... (n summands of x)

but couldnt you easily misinterpret this as the multiplicative operation of the ring

[rocky cloak]

Yes, it's common to write
2x = x + x
3x = x + x + x
etc

If your ring contains the positive integers and the multiplication doesn't satisfies 2x = x+x, then there's something weird with your ring not your notation

Notice a ring satisfies
(1+1)x = 1*x + 1*x = x+x

So this will only be a problem if 1+1 doesn't equal 2

[graceful dome]

I suppose that makes sense

then why is it earlier in the book they say

in what situation would there then be potential for confusion

[rocky cloak]

I guess if you don't specify that n is an integer.

Or if you're in some funky situation where the additive operation of your ring is multiplication or something

[graceful dome]

ah okay then lol thanks for clarifying

[regal zodiac]

Hello, i need to compute the length of A = Z[x] / ( 6, x^2 ).
I found a short exact sequence :
0 -> (x) -> A -> (x) -> 0.
Now, (x) is isomorphic to Z/6Z and thus has length 2 , so the length should be 4, is that correct ?

[near gazelle]

Let X be a set, $Y \subset X$, G a group, and $G^X$ the group of G-valued functions on X.

[cloud walrus]

Ante0417

[near gazelle]

Then how can we say that $G^X$ forms a group? How is composition defined when domain and range are different sets?

[cloud walrus]

Ante0417

[rocky cloak]

You just use the group structure of G, and do operations pointwise

[near gazelle]

Should we understand it as X is G but without its added group structure?

[rocky cloak]

So for two functions f,g:X -> G you define
(f*g)(x) = f(x) * g(x)

[near gazelle]

Oh i see

Of course

Thanks 🙏

[proper tide]

Can I form/get a ring with addition and multiplication operation being same??

[potent condor]

The zero ring

But that’s the only one

Bc you need 0 = 1

[proper tide]

Which axiom will be invalid in this condition??

[proper tide]

How??I didnt get it..Could you please explain it.

[potent condor]

1 = 1 + 0 = 1x0 = 0

[south patrol]

I guess one funny thing here is that the zero ring has characteristic 1

cute

[rocky cloak]

Z/1 is indeed the 0-ring

[proper tide]

We can say like this :Char R is 1 iff R={0}..is it true ??

[rocky cloak]

It's true.

Proof:
For all x, x = 1*x = 0*x = 0

[proper tide]

Yeahh

[tardy hedge]

Is any finite extension of Q a subfield of C?

[rocky cloak]

I guess it depends how literal you interpret the word "is", but they're definitely isomorphic to a subfield of C

[tardy hedge]

Yeah i guess thats what i meant. Why is that exactly? I feel like i “almost” know the answer but not quite lol

[vapid vale]

what do you know is true for any finite extension of a field?

(C is a hint)

[tardy hedge]

Every element of the extension is algebraic over the base field

[vapid vale]

right

so where can you say any finite extension of Q lies in

put another way, whats the "largest" algebraic extension of Q? (in a way which makes sense for this question)

[tardy hedge]

I mean, im guessing its C (its the algebraic closure of Q?) but i dont really know why

Largest alg extension in what sense?

[vapid vale]

well C isn't the algebraic closure of Q (how can you get, say, pi + 4i)

[vapid vale]

how do you expect any arbitrary algebraic extension of Q to compare to the algebraic closure of Q?

[tardy hedge]

Always a subfield

[vapid vale]

right

so you have that any finite extension of Q is a subfield of the algebraic closure of Q

you are one step away

[thorn jay]

Every finite extension is algebraic hence
Q < E < Q[i] < C

[tardy hedge]

The alg closure of Q is subfield of alg closure of R which is C?

And that last statement is fundamental thm of algebra that can be proven with other methods

[thorn jay]

Yee

[thorn jay]

It is assumed as a general factoid :3333

[tardy hedge]

what da flip

[mint seal]

doesn't the algebraic closure of Q consist of {a + bi | a, b are real and algebraic} ?

[coral spindle]

No

The algebraic closure of Q contains only algebraic numbers

There are many transcendental numbers amongst the reals.

Oh. fuck. I'm sorry

I misread that so badly

Hm. I am not sure it is sufficient, what you've written

[mint seal]

haha, I once asked this to a professor during class, and she said something like "Hm, interesting idea. Why don't you try to prove it?" and refused to confirm or deny it on the spot

I thought maybe it was false

[toxic zephyr]

"Let I be an ideal which is maximal with respect to being non-principal"
this means that I it not a proper subset of any non-principal ideal?

[coral spindle]

It's not immediately clear if it's true or false to me

[mint seal]

but I think it's actually true

[minor fulcrum]

semidirect product knowledge check: if I have a group N and a group H and action ϕ: H -> Aut(N) of H on N; let G = H⋉_ϕN, will it always be the case that

1. H is a subgroup of G
2. N is a normal subgroup of G
3. H∩N = {e} and HN = G
4. the action of H on N is given by hnh^-1 for h \in H and n \in N

I am pretty sure of 1, 2, and 3; is 4 also true?

[coral spindle]

Yes

[toxic zephyr]

thank you!

[coral spindle]

All are true.

It is important to note that it is required that hnh^-1 = phi(h)(n)

If indeed you write phi

[toxic zephyr]

so then if we have $I_a=\gen{I,a}$ (for $a\notin I$) then that means $I_a$ has to be principal right? since $I$ is a proper subset?

[cloud walrus]

eigentaylor (STfFGMOaPID)

[minor fulcrum]

when you say "required" you mean that is part of the definition of the semidirect product?

[coral spindle]

If you write phi there, you are saying that hnh^-1 = phi(h)(n)

That's what the phi in the definition means

OK, let me actually say this

OK, you're definitely aware of the difference between an internal and external direct product, right?

If you like, 1), 2), 3) are sufficient to describe the internal semidirect product. If you want to make an external one, you need the map phi so you can figure out what the conjugation action is.

So if you're writing phi, you're prescribing what the action is -- phi of course

[bitter quiver]

Could anyone explain where the 1/p in A assumption might be used here?

[mint seal]

If z = a+bi is algebraic over Q, then so is \bar{z} = a-bi. And we have a = (z + \bar{z})/2, b = (z - \bar{z})/(2i). So then a and b are also algebraic over Q

And if given a,b are algebraic over Q, so is a+bi

[coral spindle]

Nice

[soft tiger]

Is the group of symmetries of the letter “Z” trivial or isomorphic to C_2

[barren sierra]

what do you think?

and why?

[soft tiger]

If I have the identity and a 180 degree clockwise rotation that should form a group

[soft tiger]

But I was trying to make sure that made sense, because it might just be trivial

[barren sierra]

Ok so I agree that a 180 degree clockwise rotation of "Z" gives back "Z"

and what is the inverse of that rotation?

[soft tiger]

It should be involutory

[barren sierra]

cause groups need inverses

yea

so do you agree that this group, whatever it is, isn't trivial?

because you found a non-identity element

[soft tiger]

Right. So it’s just two elements I see, each involutions

The identity and that single rotation

[barren sierra]

so is it isomorphic to C2?

[soft tiger]

Yea

[chilly ocean]

Is C2 the cylic group of order 2

[soft tiger]

Yea

[chilly ocean]

Ic

Z/2Z and Z_2 are both also notation for that, right?

[barren sierra]

yes

[chilly ocean]

Thanks

[soft tiger]

Thanks for the help guys

[barren sierra]

errr sometimes Z_p, for a prime p, is the multiplicative group of units modulo p

so Z_2 would be the trivial group

whereas Z/pZ would be the group with addition modulo p

and so Z/2Z would be the 2 element group

[chilly ocean]

Ic

[barren sierra]

but this is only notation of course

if this ever comes up in context, doesn't hurt to clarify with others

[wary rover]

just the 16cos 2pi/7 or all the roots of unity from z1 to z16?

[proper jolt]

The sum with a circle symbol is direct sum

what im confused about is that when they say

$M=M' \oplus M''$

[cloud walrus]

thebirdsandbees

[proper jolt]

it means that M is isomorphic to it

since i thought $M' \oplus M'' = {(m' , m'')}$

[cloud walrus]

thebirdsandbees

[proper jolt]

so how can $M', M''$ be a submodules of $M$ while $M=M' \oplus M''$ at the same time

[cloud walrus]

thebirdsandbees

[prisma ibex]

Z_2 is bad notation for this

Z/2 or Z/2Z are standard, C_2 is also fine

Some people write Z_2 but this is bad and conflicts with standard notation for something entirely different

[chilly ocean]

Ic

[worthy solar]

I don't know anything about axiom of choice but it was asked in lecture about if every ring has a maximum ideal. How do I know if I am using axiom of choice.

[rocky cloak]

The axiom of choice is always assumed, unless you're specifically investigating nonstandard axiomatic systems.

[worthy solar]

Good thing I'm not investigating any axiomatic systems

[rocky cloak]

Well all math is rooted in some fundamental assumptions.

[proper jolt]

is the dual of this sequence the same sequence?

[worthy solar]

Yeah but I'm not going that far down 😔

[proper jolt]

like when it's saying dually isn't it talking about the same sequence

[rocky cloak]

You don't have to go down to the fundations, but if you find some scaffolding that says "depends on AC" you can still safely step on it

[next obsidian]

Yes

[serene dune]

holy shit it's everywhere

[tight brook]

Yes when they say dually they just mean a dual definition. A different way to phrase the same thing. But both of these are not the best formalism for exact sequences. A sequence of algebraic structures is called exact if and only if the image of a map is exactly the kernel of the next. A short exact sequence is an exact sequence of 5 structures with the 0 group at both ends.

This formalism allows us to generalize the idea of a short exact sequence to more interesting things that have major implications in homology theory.

Ill leave you as an exercise to prove that a short exact sequence is exactly the same as these two dual definitions already given.

[proper jolt]

ok ill keep this in my mind when i get there, but i just started learning about exact sequences

[tardy hedge]

Kian’s lemma lol

Prof gave me too much credit 😂

[serene dune]

algebra with that handwriting is a crime

Ici est ne evariste galois illustre mathematicien francais

[rocky cloak]

This is honestly one of the better handwritings I've seen.

https://youtu.be/naj0kjZgE0c?si=JJwIoKg5mdvT_zCt

[serene dune]

https://media.discordapp.net/attachments/1054780485408133170/1284071255137456128/attachment.png?ex=6799e98c&is=6798980c&hm=7a62ff2a931f7329b6a9cd47d1d290e8b10edaf1e4586a7da0adf3e007e194d5&

[thorn jay]

Fym \hookrightarrow \cong

[rocky cloak]

People writing \mathbb{Z}_4 : 🤡
People writing \mathbb Z_4 :

[candid patrol]

$\mbb Z_4$

[cloud walrus]

UGOBEL

[candid patrol]

🥱

[white oxide]

Potentially silly question, but is an ideal radical if and only if its equal to its radical

[elfin wraith]

Yeah

[white oxide]

thx

[elfin wraith]

I’ve usually seen it presented the other way around, defining the radical of an ideal then saying an ideal is radical if it is equal to its radical, but you should be able to check quite easily that your definitions agree with that

[proven swift]

hey can anyone explain me what exactly exp G means? the author havent used this notation yet so im kinda lost

[wintry sluice]

what kind of background is needed for the Nielsen–Schreier theorem?

[white oxide]

In the definition of the radical presented there, do you know what n is? Is it an integer, natural number or positive integer?

Oh looks like positive integer

Why is $\sqrt{(x^2)} \subseteq (x)$? If $f^n = x^2 p(x)$ for some $p$, I'm not sure how to see why $f \in (x)$

[cloud walrus]

okeyokay

[coral spindle]

It's probably the exponent of a group. That is, the lowest integer $n$ such that $g^n = e_G$ for every $g \in G$. The exponent is the lowest common multiple of the orders of the elements of the group.

[cloud walrus]

$\mathbf{Boytjie}$

[rocky cloak]

If x^2 divides f^n, then x divides f^n which means x divides f since x is prime

[rocky cloak]

I think the usual proof uses algebraic topology.

So you would need to know about the fundamental group, and the relationship between covering spaces and subgroups of the fundamental group.

[proven swift]

oh that makes sense, thank you!

[potent condor]

Can someone give me a hint on showing that the maximal ideal $\mathfrak m = (X^2-Y,Y^2-2)\subseteq \mathbb Q[X,Y]$ cannot be expressed in the form $(f(X), g(Y))$?

[cloud walrus]

EdgarAlnGrow

[rocky cloak]

Q[x, y]/m is a field. Which field is it? What does x and y correspond to in the field? What are their minimal polynomials? What does that say about what f and g could be? Can that give a maximal ideal?

[potent condor]

• Q[x, y]/m is Q[2^{1/4}]
• x corresponds to +/- 2^{1/4} and y corresponds to +/- 2^{1/2}
• the minimal polynomials over Q are x^4-2 and y^2-2 respectively

But I am dumb so I dont see how that says anything about what f and g can be. presumably they have to be those minimal polynomials, but i dont see why

[rocky cloak]

If f(x) is in m, that means f(x) is 0 in Q[x, y]/m.

But what is f(x) in Q(2^1/4)?

Maybe it helps to write
f(x) = a0 + a1x + a2x^2 + ...
and think of this as a homomorphism Q[x, y] -> Q(2^1/4)

[rotund aurora]

why do you think this requires background? idk the proof but you can find it in Robinson's book, for example

[tardy hedge]

Every hom from that quotient to some field K is in bijection with the solutions to q(x) in K. Can someone help me with this oke

One

[potent condor]

How is this a homomorphism Q[x, y] -> Q(2^1/4)? Such a homomorphism consists of an image for X in Q(2^1/4), an image for Y in Q(2^1/4), and a homomorphism from Q to Q(2^1/4). I don't see how f contains any of that information

[rocky cloak]

What do you mean f containing the information?

Let's just step back for a second. Say f(x) = x + 1. Do you know what element f represents in Q(2^1/4)?

[potent condor]

2^(1/4) + 1?

[rocky cloak]

Indeed, and you know that's not 0.

What about f(x) = x^2

[potent condor]

2^(1/2)

[rocky cloak]

What if
f(x) = ax^2 + bx + c?

[potent condor]

a2^(1/2) + b2^(1/4) + c

[rocky cloak]

Indeed, so in general. What would be a reasonable way to describe the element f(x) represents?

[potent condor]

evaluating at 2^(1/4)

[rocky cloak]

That's right. It's just
f(2^1/4)

So then, for which f is f(2^1/4) = 0?

[potent condor]

x^4-2

[rocky cloak]

That's one example sure

Could it be any other polynomial?

[potent condor]

it could be a constant multiple of this

but nothing else

oh wait

ignore that

[rocky cloak]

Yes, so any polynomial with f(2^1/4) is a multiple of the minimal polynomial

[potent condor]

oh sure polynomial multiple

[rocky cloak]

In particular
(f(x), g(y)) < (x^4 - 2, y^2 - 2)
Whenever f and g are taken from m

[tardy hedge]

Rip

Im looking at the field homs from F[x]/(q(x)) to any field K is in bijection with solutions to q(x)=0 in K

I think also i was unsure about the greater context of why we are looking at this

[potent condor]

ok i got it, thanks for the patient explanation!

[rocky cloak]

Maybe start with what are the homomorphisms F[x] -> K? When does such a homomorphism map q(x) to 0

As for greater context, we are kinda interested in roots of polynomials. So it's nice to have something that corresponds to that

[tardy hedge]

You can do this sort of thing for any map out of a quotient right? I think i start getting nervous any time someone says something like “so this must be in the kernel so the map “”””factors””” through” sort of things

I usually get a bit lost at that point

[rocky cloak]

Well, that is the main idea behind quotients. So if it makes you nervous you may want to befriend quotients a little bit

[tardy hedge]

Yeah

[proper jolt]

in this prove im confused why $\alpha '$ factors through a unique map $\phi '$

[cloud walrus]

thebirdsandbees

[proper jolt]

i thought factor through means $\alpha '=\phi ' * f$

[cloud walrus]

thebirdsandbees

[proper jolt]

but it seems like you can get $\phi '$ by the map

[cloud walrus]

thebirdsandbees

[proper jolt]

idk if im correct though

[toxic zephyr]

say $I\subseteq I_a=(\alpha)$ and $J$ is an ideal such that $I\subseteq J$. if i want to prove

$x\in I\implies x=s\alpha$ where $s\in J$

is the contrapositive that if $x=s\alpha$ where $s\notin J$ then $x\notin I$ (i.e. $x\in I_a\setminus I$)?

[cloud walrus]

eigentaylor (STfFGMOaPID)

[toxic zephyr]

not sure if im logicing properly

[velvet hull]

so are you just trying to prove that all elements in <alpha> are of the form s*alpha?

[toxic zephyr]

no, that elements in I are of the form s alpha with s in J. i.e. I subet Ia J

still working on that so i don't want hints but i just wanna know if my contrapositive is right or wrong/flawed

[proper jolt]

i think it's wrong

because there could exist multiple $s$ where $x=sa$

[cloud walrus]

thebirdsandbees

[proper jolt]

and one of them could lie outside of J

[toxic zephyr]

what if it's an integral domain? then wouldn't that be unique?

[proper jolt]

yeh it would be unique then

cuz you can cancel out $\alpha$

[cloud walrus]

thebirdsandbees

[proper jolt]

and your contrapositive would make sense

[serene dune]

snake lemma is neat

[proper jolt]

yeh it is but i don't even understand the application

i mean i don't know its application

[serene dune]

honestly me neither in its full generality

https://math.stackexchange.com/questions/2508678/proving-the-strong-four-lemma-using-the-snake-lemma

i find it through this, tho it was introduced in the book in the later chapter

maybe you would find something interesting here

the moreover part is presented in this book as a separate lemma

[south patrol]

I would say the main use of the snake lemma I've seen is for creating other various "long exact sequences" such as the standard ones on (co)homology

[serene dune]

one tangent question, i kinda got the 'motivation' for homological algebra, is there any specific prerequisite i need to be aware about ?

lmao

[thorn jay]

Mfw functors dont preserve exactness

[proper jolt]

this is not something i can read with only 3 hrs of sleep lmao

[serene dune]

alr sleep more and take a look thereafter

[serene dune]

sanity check while trying to understand this

${ x \in \mathbb{R}^n \mid p(x) = 0 }$

[cloud walrus]

nastasya

$\{ x \in \mathbb{R}^n \mid p(x) = 0 \}$

[serene dune]

is this the zero set of polynomial ?

and p is fixed right ?

i mean for a given p the say V_p = { x \in \mathbb{R}^n \mid p(x) = 0 }

[bitter rover]

Yep

[serene dune]

alr

[rocky cloak]

Like before you learn homological algebra you mean?

Not really. You just need the basics of rings and modules, and then you should pick up some category theory on the way.

And I guess it may or may not be useful to have some reason to care about it. Like knowing some algtop/alggeo/reptheory or something

[thorn jay]

Rawdogging homological algebra

Going in without any reason to care abt it

[rocky cloak]

It is interesting in it's own, so you could totally rawdog it

[thorn jay]

Definitely!

Another valid reason is because you get cool diagrams

Mathematicians hate this one little trick

[serene dune]

gatdamn, alr coach o7

demoted in priority list

[glad osprey]

Is the motivation for commutative algebra also algebraic geometry/topology? Do commalg and homological algebra overlap in some way?

[arctic trail]

given a subgroup H of G, we can talk about the 'kth-root" of H as being the subgroup K of H generated by the set {h: h^k in H}.
Does this have a common name?

[rotund aurora]

why should it be motivated by something else? And yes, see Hilbert's syzygy theorem for example

[glad osprey]

Are you saying people study algebra just for its own sake? jk, but it seems like all the subjects come together at higher levels, like algebraic geometry for example uses topology, commalg, homological algebra, etc.

[rotund aurora]

subjects intersecting doesn't mean one is motivated by another

[thorn jay]

Case and point: logic and geometry in topos theory

Wait thats not how the saying goes

[glad osprey]

Obviously not, but often when you study something, it's because you want to use it for something else. I don't care about point-set topology by itself for example, I want to use it for other things

I guess it's a bit inaccurate to ask for "the" motivation for commutative algebra though, a more appropriate question would be "where is commutative algebra used"

[rotund aurora]

that's fine. But that's a very personal experience. Some topologist will tell you that they don't care about algebra, but they have to learn some to study the homotopy of spheres or whatever. jagr2808 will tell you that homological algebra is cool.

Perhaps you could have mentioned what you are interested in and asked what possible connections are there with commutative algebra

[tardy hedge]

Croquet do u think hom algebra is cool?

[rotund aurora]

Yes I do, specially group cohomology as used for class field theory/number theory

[rocky cloak]

I'd say the motivations for commutative algebra are alggeo and algebraic number theory. And maybe to a lesser extend algebraic topology.

And yes it overlaps a lot with homological algebra. Or perhaps it's better to say that homological algebra is a critical tool to doing commutative algebra

[delicate orchid]

go learn about obstruction theory for non-commutative extenstions and tell me that shit ain't cool

[glad osprey]

I feel like it's still a bit early to say what I'm truly interested in, but algebra by itself is fun and I think I've fallen victim to Aluffi's algebraic geometry-propaganda, so maybe I'll try studying that eventually

[rotund aurora]

I can suggest chapter 1 of Eisenbud's book on commutative algebra to learn a little bit about the context of commutative algebra. Although I guess his view might be slighly biased towards AG, not sure. But it's still a nice read

[thorn jay]

hey if you think algebra is cool, try universal algebra, it's totally worth it i swear!!!

[delicate orchid]

this user is attempting to lead you astray do NOT fall for their tricks

[thorn jay]

Bbut

Finitary monads

[glad osprey]

Cool, I happen to have that book, so I'll check it out

[glad osprey]

this sounds like a trick, I'm not falling for it

[bitter rover]

haha

[thorn jay]

Honestly being genuine here it's definitely not for everyone, and sometimes rather hard to work cuz sometimes proofs really just come out of left field, but i still think that malcev conditions are some of the coolest things

[delicate orchid]

isn't model theory just universal algebra but good

[thorn jay]

False, it's universal algebra but diluted with logic taking away from the goodness of working with solely equational identities and equational quantifiers

[delicate orchid]

ok so it's unviersal algebra but more general

[rocky cloak]

Why use existential quantifier when universal quantifier does trick?

[thorn jay]

In a way, but most things or constructions one uses in UA aren't possible in model theory

It's also what gives UA it's algebra flavour, instead of literally being logic

[thorn jay]

(Namely, the congruence/subalgebra lattices, free algebras, quotients, the like)

[delicate orchid]

I see

[thorn jay]

Free algebras / term algebras are especially important because the free algebra generated by n variables basically tells you exactly which equational identities hold:
every K-free algebra is a quotient of algebra of terms over a type (or signature, depending on literature), and
p/~ = q/~ <=> every algebra in K satisfies the identity p ≈ q

Anyhow, i should stop talking, UA triggers some neuron activation in me or smt

https://tenor.com/view/neuron-activation-neuron-gif-1455356329209493514

[worthy solar]

if i am talking about say the quotient ring z_2[x] / <x^2 + 1>
is it okay to say x^2 = -1 mod (x^2+1)
or should i say x^2 + <x^2+1> = -1 + <x^2+1>

[worthy solar]

the problem i see with the first one is im also working with integers from z_2 ie z/2z

[south patrol]

Saying the first is better honestly

More readable lol

[arctic trail]

I'd write like F_2[eps], eps^2 = 0

[south patrol]

You can even just say "x^2 = -1 in [ring name]"

[arctic trail]

your quotient is just adding a commutative nilpotent element

x^2 +1 = (x+1)^2

[thorn jay]

The most based option, and the [ring name] can often be dropped in the next couple lines as it is implicit you're working in that quotient ring

[worthy solar]

[prisma ibex]

they have a lot of overlap not least because the best behaved categories to do homological algebra in are categories of modules over rings

and loads of applications in geometry/topology make use of techniques from commutative/non-commutative/homological algebra

algebraic geometry especially, the entire foundations of that field is commutative algebra

[worthy solar]

Is there an analog for the classification of finite abelian groups to the classification of say finite rings or fields. Have not gotten far into my studies but surely it must build off finite abelian groups given finite rings and fields have an additive abelian group in it.

[coral spindle]

There is indeed a classification of finite fields. There is a unique field (up to isomorphism) of order p^n for a given prime p and natural n>0.

[velvet hull]

there's also the classification of finitely generated modules over a PID

can't think of anything else

[coral spindle]

I think there is a classification of finite rings. I cannot quite remember how it works, I apologise.

[velvet hull]

(its finitely generated abelian groups, not finite abelian groups but close enough)

[coral spindle]

Ah, yes, here it is: https://mathoverflow.net/questions/7133/classification-of-finite-commutative-rings

only commutative rings unfortunately

[thorn jay]

Well, any finite abelian group is trivially finitely generated

It's not completely related, as it's not necessarily a classification (general algebraic structures are far too complicated for that), but I have a book by Ralph McKenzie and David Hobby about the structure of finite algebras, most of which is going over my head sadly, but i dont think there's in general a classification for finite algebras

[rocky cloak]

Fields are sufficiently restrictive, that I wouldn't say the classification of finite abelian groups really comes up. There are just very few abelian groups it would even seem likely could be fields.

As for rings, every abelian group can become a ring (possibly in several different ways), so there it might be useful.

[candid patrol]

A hard one (to me) : classify the groups of order p^2q with p and q primes

[rocky cloak]

Hmmm, if you have square free characteristic you can split it into a product of finite dimensional algebras over Fp. And all of them appear as endomorphism rings of projective generators over a species with admissible relations.

Not sure if that counts as a classification...

[coral spindle]

It’s a start at least!

[rocky cloak]

It's almost unique. In that the species is uniquely determined, but the relations are not, and I think it's tricky to determine when different relations give isomorphic algebras

[chilly ocean]

Relevant:Non-commutative(higher dimensional) algebra is also useful for AT. If you want more info about that, Ronald Brown wrote quite a bit about it

[glad osprey]

I see Nonabelian Algebraic Topology? Or Topology and Groupoids?

[chilly ocean]

The former, for the most part

I wouldn’t recommend starting out with that though tbf, unless you already know some alg top

[glad osprey]

Yeah, probably a year or two in the future

[thorn jay]

Where people found joy and happiness in commutative algebra, mathematicians sought to uncommutatify the nation, bringing chaos upon everyone.
Now my question is when are we getting noncommutative algebraic topology for arbitrary algebraic structures

[chilly ocean]

Lmao

[chilly ocean]

If you really wanna maximize your Groupoid knowledge, I recommend reading T&G then NAT(as opposed to reading some other alg top textbook then NAT)

[glad osprey]

Yep, T&G looks nice, I think I'll read it next year (probably alongside Rotman)

[chilly ocean]

W

[elfin wraith]

Commalg and Homological algebra overlap a lot, combinatorial commalg uses a fun mix of the two

Miller and Strumfels book is great, as is Hertzogs and Hibis

[proper jolt]

can someone explain to me why isomorphisms between exact sequences are defined this way?

like why take the direct sums

[mint seal]

I think they're defining "essentially isomorphic" rather than isomorphic

but I couldn't tell you why it's done that way

[proper jolt]

or is there any intuition as to why you would do that?

that you know of

[graceful dome]

Can someone help me with 4?

Since F(x) is a field of quotients of F[x]
Then an element of F(x) lets call it a can be written as a quotient of elements of F[x] lets say g(x) and h(x).

So a = g(x)/h(x)
Wts we cannot have a = x s.t a = (g(x)/h(x))^2

[crystal vale]

I selected module theory as my Bachelor's thesis, it can be a reading project but can anyone give some directions what can I do with module theory?

[velvet hull]

so let n = deg f and m = deg g

what can you say about the degree of f^2 and g^2?

[thorn jay]

Module theory is mainly used to study the ring it's a module over, as properties of a ring translate to properties of it's category of modules.

Furthermore it's the nicest example of smt called an abelian category, which captures the "abelianness", for example the fact that the set of homomorphisms between two modules naturally forms an abelian group, or the fact that every submodule is a "normal subgroup", so to speak.

Because of that they form the natural environment for commutative and homological algebra. They're also used in rep theory because linear representations of a group G in a field K correspond to modules over the group ring K[G]

[proper jolt]

anyone know why snake lemma implies that $coker \lambda=0$ as $coker \mu =0$

[cloud walrus]

thebirdsandbees

[thorn jay]

You get an exact sequence
... -> ker 1_N -> coker \lambda -> coker \mu -> ...
ker 1_N = 0 and coker \mu = 0, so by exactness the desired thing follows

[crystal vale]

Okay

[rocky cloak]

The idea is that a short exact sequence like
0 -> P -> P -> 0 -> 0
is kinda "trivial". So adding this gives "essentially the same" sequence.

The bigger picture here comes when you look at longer sequences. Two projective resolutions of M are always homotopy equivalent. And two complexes of projectives are homotopy equivalent if and only if they differ by these trivial complexes
0 -> P = P -> 0

[graceful dome]

thank you i ended up figuring out exactly what you explained

can someone clarify prime ideals for me though real quick

[velvet hull]

this only works because F is a field (or integral domain, mainly)

[graceful dome]

because of cancellation?

Is the "or" for prime ideals a and not b?

or b and not a

or can it be a and b also

[velvet hull]

no reason why not

just at least one of a or b is in P

that's it

[graceful dome]

okk cause im trying to prove 2Z[x] is a prime ideal of Z[x]
and my argument is that 2Z[x] is a strict ideal since any product of an element of Z[x] with that of 2Z[x] is an element of 2Z[x] (proving its an ideal)

Then for a,b in Z[x] if ab is in 2Z[x],
Then it is clear ab has even coefficients, if both a and b are not in 2Z[x]
then both a and b have at least one odd coefficient then so must its product, i.e ab cant be in 2Z[x] unless either a or b is in 2Z[x]

@velvet hull

[velvet hull]

that looks like it works

[graceful dome]

i think so too 😂

but please tell me if it doesnt or if its not rigorous enough

[thorn jay]

Yeah thats a valid argument, 2Z[x] is the set of polynomials with even coefficients

[velvet hull]

but the main point of a prime ideal, is that the quotient ring is an integral domain

so an alternative argument would be that Z[x]/2Z[x] is the ring {0,1} which is indeed an integral domain

and you're done

[graceful dome]

Hmm that may be a better way to go about it

[thorn jay]

Hell, it's a field, so it's even maximal

[graceful dome]

can you explain the part that Z[x]/2Z[x] is the ring {0,1}

[velvet hull]

actually proving it is kind of a pain in the ass

[rocky cloak]

It's actually (Z/2)[x], not a field

[graceful dome]

so should I just stick to my original proof?

[rocky cloak]

Seeing why is not too hard. It's just that two polynomials become equal iff all their coefficients are equal modulo 2

[rocky cloak]

Your proof is correct

[graceful dome]

Okay thank you

[rocky cloak]

It's prime but not maximal

[serene dune]

Rings and modules

[lime junco]

Let a be a nilpotent element

is there some prime ideal P such that for the localization R_(P), a/1 is not equivalent to 0?

[proper jolt]

how did you get the sequence ker 1_N -> coker \lambda? I thought snake lemma only gives you coker \lambda -> coker \mu

[proper jolt]

shoot this is too advanced for me i don't even know what homotopy equivalence means yet

but ill look into it when i learn thanks

ill just save it in my read later list lmao

[rocky cloak]

Alright, well just think about it like the fact that L(+)P' = P(+)L' says that the sequences are pretty similar right. You kinda have "all the same pieces" just distributed a little differently.

[proper jolt]

yeh i guess in a way because i think it's kinda saying that if you combine L and P' you get that they are equivalent if you combine P and L'

so kinda if you ignore order than they can be similar to each other idk

[proper jolt]

wait does the snake lemma also have a loop

actually nvm

[rocky cloak]

Another thing this means, you have what is called the stable category. Which is just where you "ignore" projective modules.

Then this lemma is saying that taking syzygy (M |-> L) is a well defined functor in the stable category.

Or said another way, if we treat all projective as 0, then L and L' are equal.

[proper jolt]

im going to learn about functor soon

like tomorrow probs

so maybe ill be more well equiped tomorrow lmao

[rocky cloak]

This is very useful when working with things like Ext(-, N) which doesn't see projectives. So behaves the same on things that are isomorphic in the stable category

[proper jolt]

ok thanks ill look into it tomorrow, i'll prbably know more by then

how did you get the sequence ker 1_N ->

[crystal vale]

can i conclude that in direct sum we have x_i = 0 for all but finitely many whereas in direct product it is not necessay to be?

[mint seal]

yup

[valid night]

I'm doing an exercise that wants me to prove that a field automorphism $\sigma$ on $\mathbb{R}$ satisfies $\sigma(y) = y$ for all $y \in \mathbb{Q}$.
This is the first proof I was able to come up with, but this holds for all real $y$ and not just rational, right?

[cloud walrus]

Kroros

[valid night]

Oh nvm I see the mistake

[rocky cloak]

So a few things going on here

Number 1: you have a mistake by your second equal sign you've just gone
sigma(x) z = sigma(x z)
which usually isn't true.

Number 2: it actually is true that the only automorphism of R is the identity, but the proof would have to use properties of R.

Number 3: the statement that any automorphism fixes Q is true for any field that contains Q. But you will have to use some properties of Q to prove this.

[valid night]

Yeah the next part of the exercise wants me to prove that the only automorphism on the reals is the identity. I assumed it wants me to use that it fixes Q and that it preserves positiveness.

My plan was to use that a real number is a limit of a Cauchy sequence in Q, along with the fact that the identity is continuous, is that the correct approach?

[crystal vale]

i proved that if ring R is Noetherian ring then every ideal is finitely generated, but any hint how can i prove converse part?

[rocky cloak]

You're onto something at least. But you want to prove that any automorphism is the identity, so the fact that the identity is continuous doesn't really help, since you don't yet know your automorphism is the identity.

[valid night]

I do know that it is the identity on Q, no?

[thorn jay]

Ideals correspond exactly to submodules of R over itself

[rocky cloak]

Sure, but you can cook up discontinuous functions that are the identity on Q. Like f(x) = x for x rational and 0 for x irrational is not continuous.

So you have to use something more

[valid night]

Ah I see, thanks

I think I have an idea

I think I found it

[molten viper]

Use contrapositive maybe

No modules required

ie. Using a non-stabilizing chain of ideals you can get a module which isnt fg

[storm kiln]

You could take the union of an infinite chain I think, since it will also be a finitely generated ideal

[molten viper]

Uh yeah for the converse that may be the way to go

One direction is messy to prove directly and I forget whoch

[crystal vale]

yes but i don't know for modules too

[crystal vale]

got it, thank you

[serene dune]

how is A \oplus B defined for groups
i was always under the impression its same as \times which is a isomorphism for abelian groups

but how is \oplus defined

this is the drawback of never opening a book, need to follow a book for algebra now

[thorn jay]

Its called the free product of groups

[serene dune]

alr

[thorn jay]

Basically you take the every word consisting of elements in G and H, but you reduce them using the structure of G and H

So in
Z_4 \oplus C_2 = { 0, 1, 2, 3 } \oplus { e, g }
the words
10g0g and 1 would be the same

For rings the coproduct is the Z-tensor product when the rings are seen as Z-algebras

[serene dune]

yeah the ring makes much more sense to me for whatever reason

also the free group is defined at the 11th chaper in rotman's so everything is blurry to me

[thorn jay]

Right

[serene dune]

should i try to follow that book from from front to end ?

[thorn jay]

You can try sure, but i havent done that with many books tbh

[tardy hedge]

For ppl who were curious about my galois theory class… my notes are very messy so im not sure if its useful for me to show them

[serene dune]

which book is it following if it is following anything

[tardy hedge]

Dummit and foote but not really because my prof is speaking mainly in category theory

Or at least he’s teaching it from that framework / pov

Like today he was talking about how we want to study the category of field extensions over F by studying functors to Set , i guess the functor being Hom(-,E)

But then apparently if u consider the category of Ring over F then there some adjunction you can use. Idrk honestly

The pic here was something like the functor factors through a functor to category of G-sets (thats where automorphism group of field extension comes up)

Galois group

[thorn jay]

Based

[tardy hedge]

It kind of is tbh

Im locked in sooo hard during lectures to try and follow but its been fun

[thorn jay]

Thats so cool though tff

[tardy hedge]

Dummit and foote’s chapter on commutative algebra actually looks kinda nice

I feel like literally no one learns commutative algebra from dummit and foote lol

They do cover a lot of the essentials at least

[rotund aurora]

I think from this POV the relationship between usual Galois theory and Galois coverings becomes clearer? Then there's Grothendieck's Galois theory, which iirc is like coverings but for schemes, etc. Maybe if you ask your prof he tells you something about it

[crystal vale]

Which book is good for learning all the stuff which is necessary to understand The Quillen Suslin theorem? If this message is not to belong here I will delete it?

[velvet hull]

that's tragic. any book that the prof is going off of, or any notes I can look at on his personal website?

[viral halo]

$r$

[cloud walrus]

uginp

[viral halo]

Given a representation $\rho: G\rightarrow GL(V)$ we get the dual representation $\rho^:G\rightarrow GL(V)$ by $\rho(g)^=\rho(g^{-1})^T$. This must satisify $<\rho^(g)\phi,\rho(g)v>=<\phi,v>

[cloud walrus]

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

[viral halo]

on wikipedia they define $<\phi,v>=\phi^Tv=\phi(v)$. Where $\phi$ is an element of the dual space, now how does $\phi^T v$ make sense? If we consider the elements of V as columns vectors, aren't the elements of $V^*$ row vectors?

[thorn jay]

You forgot to close the first mathmode

[cloud walrus]

uginp

[thorn jay]

Also, $\langle$ and $\rangle$ look nicer than $<$ and $>$ imo

[cloud walrus]

.enpeace_music

[thorn jay]

But idk how angled you usually write em

[viral halo]

sometimes as (,)

🙂

This makes sense if we just write down the definitions, but idk what to make of $\phi^T$

[cloud walrus]

uginp

[viral halo]

I guess it makes sense to consider the elements of the dual space as column vectors since we want to act on them on the left, but I'm used to elements $V^*$ being row vectors if the elements of $V$ are column vector.

[cloud walrus]

uginp

[thorn jay]

Honestly i think < \phi, v > just interprets them both as column vectors

As it's the standard inner product notation

A little notational annoyance

[viral halo]

that might be it. On the wiki page they have written this, I'm not sure what to make of the equality with an extra row in this context.

[thorn jay]

Is defined as? Taken as? Idk

[soft tiger]

I’ve been working through this interesting problem. I’ve been trying to solve part 3, where they ask us to prove that G has n elements.

I know that the identity mapping on K is in G

[wintry sluice]

is there anything like calay’s theorem for other algebraic structures?

[rocky cloak]

Think about the possible values alpha(k1) can take. Use assumption (vii) and (viii).

[coral spindle]

There’s already a good answer, but for example, every ring is a subring of a ring of endomorphisms of an Abelian group (namely its own underlying Additive group).

[coral spindle]

Sharing here an interesting, reportedly yet unproven, but very elementary conjecture in finite groups

[lusty marlin]

Very nice

[radiant remnant]

nice

[candid patrol]

Let n < 2000 odd, then if G isn't cyclic, he is not simple 🙂

[tough raven]

Has this actually been checked for a reasonable number of groups, say for |A|, |G| < 10 (or <100 with a computer)?

It seems kind of unlikely to me, because it's like the formula was derived by just assuming all homomorphisms factor through the abelianisation.

[south patrol]

Lol yeah I wonder if it can just be bashed

[lusty marlin]

Time to bash it and speedrun a publication lol

Although I get the feeling this has probably been verified by computer for small orders

[rocky cloak]

There seem to be a lot of partial results that maybe indicate it should be true

https://repository.kulib.kyoto-u.ac.jp/dspace/bitstream/2433/84161/1/0877-01.pdf

[lusty marlin]

Ah

[glad osprey]

Are any two fields of the same characteristic contained in some common field extension? I'm 90% sure it's true, but I don't see how to prove it

[south patrol]

Yes (but the proof I have is mind is silly)

[delicate orchid]

r u just gonna ||take the algebraic closures lol||

[south patrol]

Well that needn't work

[delicate orchid]

does in positive characteristic

[south patrol]

But yeah I had smth sillier

[delicate orchid]

none of this Q bar bullshit

[south patrol]

Fp and some massive extension of Fp bar

[delicate orchid]

how tf are you extending Fp bar! It's extended all the way!

jingle fields jingle fields extended all the wayyy

ok but fr ||the actual move is to just extend one field by the other right?||