groups-rings-fields

[dull ginkgo]

@crystal vale if you want some good exercises along these lines, here are some ones I enjoyed a lot

[crystal vale]

I did this one, from Jacobson right?

[dull ginkgo]

yep

[crystal vale]

7th one is interesting

[dull ginkgo]

Took me a while

[crystal vale]

And 6th is a tricky one

[dull ginkgo]

6 and 7 are like a package deal lol

6 makes 7 way easier,

[crystal vale]

I love 8 one direction is trivial but other one is very good excercise

[crystal vale]

Yes

I am talking about 7's counterexample

[dull ginkgo]

oh the counterexample the best route is to literally take the most simple (FREE) route possible

Let M be the monoid in 3 letters: {x, a, b}: Quotient by the relation e ~ xa ~ xb.

[crystal vale]

[dull ginkgo]

every string reduces to some string of a's and b's with an x power on the end. you can't have any other right units of x besides a and b since x is not torsion

[tardy hedge]

U are dedicated af bro

U are constantly here working problems

[hidden wind]

i love left and right inverses of functions

[tardy hedge]

Yuhhhh

[dull ginkgo]

Sections and pullbacks right

[south patrol]

This also works more generally. You can consider multiplication by a on R. In this case you've assumed it's injective so it must be bijective since we have a finite set. But also this works more generally e.g. if you have a finite dim k-algebra A then multiplication by a will be an injective linear map on a fin dim space and hence an isomorphism

[graceful dome]

Why is that a group under self conjugation, if we have class number of 2, we have that the group is isomorphic to C_2?

From what I understand the conjugacy classes are {e} and the remaining class cl(g) contains all the other elements of G and they are all conjugate to each other

but im kinda stuck there

[rocky cloak]

So note that this is only true if G is assumed to be finite, so it will use a counting argument of sort.

Maybe think about what the orbit stabilizer theorem says about the action of G on itself.

[graceful dome]

Oh yea I forgot to mention that G was finite

[graceful dome]

So then for some element g in G, o(G) = |cl(g)| * #stabilizers of g

[rocky cloak]

That's right, but what is |cl(g)|?

[graceful dome]

1? Because we have cl(e)

[rocky cloak]

Well let's say g is not equal to e,

What would it be then

[graceful dome]

however many numbver of elements in G/e

[rocky cloak]

Remember your assumption. What you assumed was that G had only two conjugacy classes. You've already established that one of them has size 1.

[graceful dome]

yea to be honest im a bit stumped about what that tells me about the other class

other than all non identity elements in g are conjugate to each other

[rocky cloak]

Yeah, but you know how many non-identity elements there are.

Say o(G) was 5 for example, what is |cl(g)|?

[graceful dome]

That would depend on the stabilizers right?

oh wait

are you trying to get at the fact that groups of prime order are cyclic or am I completely gone

[graceful dome]

I'm just gonna say this is either a gap in my knowledge or im just blanking

[rocky cloak]

So you know that every element is part of some conjugacy class right?

[graceful dome]

yes

oh nevermind yea so |cl(g)| = 4

since cl(e) has order 1 right

[rocky cloak]

That's right, so in general it would be o(G)-1

So now you're in the position where o(G)-1 divides o(G)

[graceful dome]

ohhhhh

took me a second

but that can only be the case if cl(g) was 1 and o(G) was 2

[old hollow]

Oh, I see. The idea is that Fbar contains all elements which are algebraic over F. (Consider why the polynomial root definition captures this.) Therefore it is equal to the union of all algebraic extensions E_i — intuitively, not actually; see note below.

Once you see it in this way, it is clear that for a field F, any algebraic extension E of F must be contained in the algebraic closure of F

There is a big subtle issue with what I just said, which you should be very careful of

it is possible to have two algebraic closures which are isomorphic to each other

Then when you have algebraic extensions Ei, it is possible that Ei is a subset of one algebraic closure, but not a subset of the other algebraic closure

the key is that Ei will be isomorphic to a subset of the other algebraic closure

so you have to be very careful about saying "subset" and "union"; instead work with isomorphisms, or otherwise use about some fixed global extensions and talk about subsets of those

[alpine island]

This should be trivial but I can't find a lemma or something to imply it

[cobalt vessel]

are you taking 113 or 250 rn?

bc my 113 hasnt covered this yet

[lone niche]

For a polynomial of degree 3, do you know the possible Galois groups?

For this polynomial, what are the roots?

Is the polynomial irreducible?

[alpine island]

It's order 6 or lower. If I can rule out S_3 I'm fine, everything else is abelian

[crystal vale]

If the ring is an integral domain of order 2^n, can I say if char R exists then it must be 2?

Because char R is the LCM of all additive orders of elements of R so it must be of the form 2^k, and because it is an integral domain it must be prime.

[crystal vale]

Interesting but I am not sure about dim k-algebra A

[lone niche]

Do you know about the cube roots of unity?

Well in general about nth roots of unity

[alpine island]

Well yeah I can find the roots here in C easily

but they aren't in Q

and every lemma about this that I find requires that there is at least one primitive root of unity in your base field

[lone niche]

That's good

What are the types of roots in C? Do you have any real root?

Do you have complex roots?

[alpine island]

yeah $\sqrt[3]{10}, \zeta_3\sqrt[3]{10}, \zeta_3^2\sqrt[3]{10}$

[lone niche]

And what happens if you adjoin that first root to Q?

[cloud walrus]

Resaturated nHail

[lone niche]

Is the splitting field equal to that field?

[alpine island]

No, it'll be contained in R

which the splitting field isn't

[lone niche]

Correct

And what is the degree of this field ?

[alpine island]

6, then

[lone niche]

I assume you mean of the splitting field yes

[alpine island]

yeah, this would be 3

[lone niche]

Yup, so the answer is...

[alpine island]

Well this tells you you have a subextension given by just the first root

but you would have that no matter what?

Like that gives you a subgroup in the galois group

I'm clearly missing something, I haven't been able to keep up with this class given everything else this term

[lone niche]

If I understood your question correctly, then $F(a) \subseteq F(a,b)$ yes.

[cloud walrus]

きよし

[alpine island]

Ah, I found the relevant part.

We should have that [Q(zeta, cbrt10) : Q(cbrt10)] = [Q(zeta):Q], but these are 2 and 3

so it's not galois

[lone niche]

There's something funky going on here

[alpine island]

and therefore not abelian

[lone niche]

No, the splitting field of a polynomial is always Galois

Over Q at least.

[lone niche]

I think you are trying to use tower lemma, but it's missing some parts.

[alpine island]

I can share the note I'm referencing from the class notes

[lone niche]

sure

[alpine island]

but it's not uh, the most readable

(these are the notes the professor uploads btw )

[old hollow]

It’s giving S3 vibes tbh

Cuz x^3 - 2 has Galois group S3

And sqrt2 vs sqrt10 are not so different

[alpine island]

I could just compute the roots, look for automorphisms, and do this by brute force

[old hollow]

These are the roots

[alpine island]

well yes

[old hollow]

Wait how much Galois theory do you know

The main theorem?

[alpine island]

Yes

subgroups of Gal(K/F) <-> subextensions of K/F

[old hollow]

The order of the Galois group is equal to the degree of the field extension

(Since it’s a normal extension)

^ Are you familiar with this fact

[lone niche]

Ah I see what you did. This gives you 2=2

[alpine island]

Yes.

[lone niche]

This here, both sides are 2 by the theorem in the notes

The relevant theorem you want is: https://en.wikipedia.org/wiki/Degree_of_a_field_extension#The_multiplicativity_formula_for_degrees

[old hollow]

Ok so you’re trying to show the Galois group has order 6 (the current guess is S3)

You can do this by showing the field extension has degree 6

Right

[alpine island]

Yeah I have this pretty easily

I just need to see S_3 vs Z/6

[old hollow]

Ohh

So true

[alpine island]

to show it's abelian

or not

[old hollow]

What’s the fundamental diff between S3 and Z6

I mean honestly it’s only 6 automorphisms, you could hammer them out and show it’s not cyclic right

[alpine island]

I'm debating this

[old hollow]

Do it

It’s good practice

See how it works

[lone niche]

Oh, the relevant theorem is that for a polynomial of degree n, the galois group must be a subgroup of Sn

[old hollow]

Oh shit that’s it

Z6 is not a subgroup of S3

[lone niche]

I thought we were stuck in showing degree 6 lol

[alpine island]

Oh.

Yeah, that does it then

I should have seen that honestly

thank you

[crystal vale]

I proved that if the commutative ring R has no zero-divisor then every non-zero element has the same additive order.

Can I conclude that if R is finite then the ring must be of order p^k, p is a prime.

Because since additive structure is abelian so if there are two distinct prime divisors it then there exists elements which has order p_1 and p_2.

I think this result is still true when I remove Commutative and add unity

[old hollow]

This is super easy to prove as well. The automorphisms are completely decided by where the roots go. But the roots can only be sent to other roots. So each automorphism is a permutation of the roots. Therefore it’s a subgroup of Sn

[alpine island]

True.

[crystal vale]

https://math.stackexchange.com/questions/4181182/let-r-be-a-ring-without-zero-divisors-then-show-that-all-nonzero-elements-of

I found that we don't need R to be commutative

[warm ember]

yes

the characteristic is p

then you get order is power of p by vector spaces

[alpine island]

is this really as easy as "it must be a subgroup of (Z/10)^X = Z/4, which is cyclic"

[lone niche]

I think so yes

Perhaps mention subgroup structure of cyclic groups

[rocky cloak]

Seems you figured this out, but a useful piece of information is that the Galois group of a degree n polynomial is a subgroup of Sn.

So here you have a subgroup of S3.

[crystal vale]

Okay thank you

[alpine island]

frankly I should know things like this

[lone niche]

:v)

[alpine island]

but this class is a bit hard to follow sadly

even though it's an interesting topic to me

The prof has a thick accent and mumbles, and the lecture notes are... you saw them above.

handwritten by the prof with 1s that look like 2s.

[lone niche]

https://kconrad.math.uconn.edu/blurbs/

I like his field & galois theory notes

[old hollow]

Let A, B be free abelian groups of rank n contained in a field K. Assume that B ⊂ A. Prove that A/B is a finite group.

I have no idea where to start with this

I know that free abelian group really means free Z-module, but I'm not sure where to go from here lol

[rocky cloak]

I don't know what machinery you're working with but you could take the short exact sequence
0 -> B -> A -> A/B -> 0
and tensor with Q

[old hollow]

This is what I saw on SO

Seems like an elegant solution

But I wish I knew what tensoring was

[rocky cloak]

Well, there is a simpler way using the same idea. So:

You can think of A/B as generated by the generators for A, so if you can show that each of them have finite order you're good.

Now think about the rational vector space spanned by A and B.

Then sp(B) is a subset of sp(A) and these are both n-dimensional vector spaces.

You can think about elements in sp(B) as b/m for b in B and m an integer. What more can you say about the relationship between sp(B) and sp(A)? What does this mean for the order of elements in A/B?

[rich granite]

ahh this makes more sense thanks

[topaz rivet]

im new to group theory and having some trouble with understanding if a lie algebra is simple or not. why are there no semi simple algebra in dimension 2 ?

[pearl fog]

in PID R can i say $a, b \in R$, if $aR \subseteq bR$ then $b | a$?

[cloud walrus]

Cro

[pearl fog]

nvm $br_1=a*1_R$ for some $r_1 \in R$, dumb me

[hot pebble]

whats the logic here? The first iso is clear but i just dont see how to manipulate the ideal to get to that

I know both 5 and x+3 is in the ideal (1+2x,x^2+1) but how to show the other way?

[cloud walrus]

Cro

[hot pebble]

and also, if i dont have this answer how should I go about "simplifying" an ideal like this? usually when both the polynomials are monic i just use euclidean algorithm but now idk how to do

[pearl fog]

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

essentially divide original expression by the polynomial with highest degree first, then work your way down

[hot pebble]

right, so how was I supposed get 5 and x+3 in the first place?

i got x+3 from pure luck so idk whats a systematic way to do it

[languid trellis]

Can someone explain why we're quotienting out by the diagonal matrices pls?

[pearl fog]

try writing euclid's algorithm in symbols, organize in a way so that linear combinations of x^2+1 and 1+2x stands out

[hot pebble]

but what dividing what

[pearl fog]

if you find out for example 1+2x cant really divide x^2+1 up to highest order term notice 2(x^2+1) is in the ideal too

[hot pebble]

oh 2(x^2+1)

ok sure let me try

[pearl fog]

use the remainder of x^2+1 divided by 1+2x to divide 1+2x

and youll notice everything is arranged in linear combinations of 1+2x and x^2+1 hence in ideal

[night tartan]

at/a is the same as t

we'd like to send at + b / ct + d to [a b c d] (i'm too lazy to matrixify)

but that's not quite well defined cus i.e. (2t+0)/(0t+2) is the same as (t+0)/(0t+1)

so we need to quotient out by diagonal matrices

[languid trellis]

I see now, thanks

[night tartan]

nw

[crystal vale]

If R is a principal ideal domain and I is an ideal of R, prove that every ideal of R/I is principal.

We know that every ideal of R/I is of the form J/I, where J is an ideal of R containing I. Since R is the principal ideal domain so J = (a), we can write J/I = (a+I), right?

[crystal vale]

If R is a commutative ring, is it true that (a)(b) = (ab) these are ideals?

[rich granite]

We know that the splitting field of $x^3 - 2$ is $\mathbb{Q}(\sqrt[3]{2}, i\sqrt{3})$, and is of degree 6. How do I find its basis of six elements? I can't seem to grasp the idea of roots of unity (those with $\omega$).

[elfin wraith]

If you have the definition of the product of two ideals its very straight forward to verify that it is also an ideal, you should try to do this

[cloud walrus]

Inxi25

[crystal vale]

No I mean (a)(b) = (ab) ?

[bitter rover]

What's your definition of (a)(b)?

[crystal vale]

Product of an ideal, finite sum of elements of the form xy, where x in (a) and y in (b)

[elfin wraith]

Wondering if anyone can help me with a step in the algorithm for computing Groebner bases, my notes arent that clear. To give an example,

Let $f = xy+x$ and $g =y^2+1$ and $I = \langle f,g\rangle \triangleleft \mathbb{C}[x,y]$. Fix the monomial ordering to be deglex.

So computing the $S$ polynomial we have that $S(f,g) = xy-x$ and then the algorithm we're given says to reduce this, and im not quite sure what this means. The example in the notes gives that $S(f,g) = -2x$ but im really not sure where this is coming from

[cloud walrus]

Nope

[elfin wraith]

Like im not sure what they actually mean by reduction, it doesnt seem to be defined in my notes as far as I can see

[warm dove]

my teacher explained euclidean domains horribly

😭

why hsow that they're a PID to get UFD

when you could do it all without ideals

and shwo the euclidean algorithm structure

💀

thank god keith conrad exists

I knew it was true that we could do it without ideals

[elfin wraith]

are we just setting f=0 and then saying that xy = -x?

[south rain]

Let $C_i(G)$ be the central series of $G$ such that ${1} \subseteq C_i(G) \subseteq C_{i+1}$ and let $G$ be nilpotent, i.e. $C_n(G) = G$ for some $n \in \mathbb{N}$. Further, let $D_k(G)$ be the lower central series such that $D_{k+1}(G) \subseteq D_{k}(G)$ defined through $D_{k} := [G, D_{k-1}(G)]$. I want to show that $G$ is nilpotent if and only if $D_k(G) = {1}$ for some $k$. For this I try to show that $D_k \subseteq C_{n-k}$ inductively. But that's where I'm stuck already. I mean I know that $D_{k+1}(G) \subseteq D_{k}(G)$ and $ C_{n-(k+1)}(G) \subseteq C_{n-k}$ but that does not imply $D_{k+1} \subseteq C_{n-(k+1)}$ necessarily, so I'm not sure what to do.

[cloud walrus]

dellinger

[dull ginkgo]

For commutative rings yes

Each element in the principal ideals takes the form xa and yb respectively,
So the products would be xy(ab) and finite sums of those would just factor out the ab and the sums would be in the R in Rab = (ab) lol

[dull ginkgo]

I’ve had a horrible realization

What if you could do similar shit on the lexicographic ordering on the free ring R[F_2]

[tough raven]

IIRC you do polynomial long division of S(f, g) by the existing generating set {f, g} and if the remainder isn't 0, you add it to the generating set.

[dull ginkgo]

Just pondered something.

The N-indeterminate polynomial ring is literally the monoid ring over N^n. But isn’t this ordered monoid isomorphic to the generated ordered monoid in N generated by the first n primes?

Since the “prime” elements would be the indeterminates

[elfin wraith]

Yeah I’ve kinda worked out that the reduction thing is just setting each of the functions to 0 and reducing the leading monomial to the lower degree term

I wish I hadn’t worked it out though it’s incredibly tedious to do this, but also kinda easy so guess I’ll do a couple for the exam then never worry about it again

[tough raven]

If you mean partially ordered by componentwise ordering and divisibility respectively, then sure, they're isomorphic (partially) ordered monoids. But I don't see the point (or how it's related to what I said).

[old hollow]

Neither. I took 113 last semester and I studied field extensions and Galois theory through the DRP this semester

[crystal vale]

Okay thank you

[old hollow]

Well first let me get some stuff out of the way. It seems like we’re thinking of B and A not as abstract modules, but instead as the Z-span of some basis vectors b1,…,bn and a1,…,an in K. Therefore when we say sp(B) and sp(A), we are really saying the Q-span of those basis vectors. Is that right?

[rocky cloak]

Yup

[old hollow]

Ok I believe that sp(B) is contained in sp(A), and they are both dimension n

Since they are vector spaces, doesn’t this mean sp(B) = sp(A)?

I’m trying to convince myself that the elements in sp(B) can be written b/m

My guess is that you’re collecting the denominator:
(1/2)b1 + (2/3)b2 + (5/4)b3
= (6b1 + 8b2 + 15b3) / 12

Ok cool makes sense

Same applies to sp(A), the elements can be written a/m

[old hollow]

Man A/B is still a mystery to me

I know it consists of cosets of the form a + B, but I’d like to understand it in terms of generators

And I have no idea where the sp(A) and sp(B) come in here

My intuition for all this is lattices in R^2

[rocky cloak]

So if a is in A, then it's also in sp(A).

And what was the relationship between sp(A) and sp(B) again?

[old hollow]

oh…

Dude I’m off my game today

That was right there

Ok let me see

a = b/m for some integer m……..

So A/B consists of cosets of the form b/m + B

We’re trying to get that there are finitely many cosets

But you can pick any integer for m, so I don’t see why there aren’t infinitely many choices of cosets

Why is this so difficult to wrap my head around

[rocky cloak]

And what happens if you multiply by m on both sides

[old hollow]

am = b

dude I feel like an idiot

Okay so

[rocky cloak]

So what's the order of a+B in A/B?

[old hollow]

Oh my god

I forgot order is additive

m

Damn

wait so every element in A/B has finite order

Thus the group is finite

[rocky cloak]

Ding ding ding

[old hollow]

Hahaha

Thank you

Does tensoring help here or is that really abstract machinery

[old hollow]

Wait why can’t there be an infinite group where every element has finite order, like sequences in Z2

[rocky cloak]

So tensoring by Q kills elements of finite order and embeds the rest in a Q-vector space.

So the tensor argument is really the same argument, but with fancy words

[rocky cloak]

It's also finitely generated, because A is

[old hollow]

Oh

Finitely generated + every element has finite order = finite set

Well I suppose if the generators have finite order, then there’s only finitely many unique choices for the coefficients

[rocky cloak]

Indeed

[old hollow]

If you have infinitely many generators, can they each have finite order, but there is still an element in the group with infinite order?

[rocky cloak]

Assuming we're still talking about abelian groups, then no.

[old hollow]

Hm

[old hollow]

Oh so the sequence would look like
0 —> BxQ —> AxQ —> 0 —> 0

Tensoring a sequence by Q means tensoring each object with Q right

I really shouldn’t be talking since I have no idea what a tensor product does lol

[rocky cloak]

That's right (think you swapped A and B though)

[old hollow]

Whoops yeah

And that’s it? Then every element has finite order, and combined with A/B being finitely generated, you’re done?

What is there to prove when you tensor with Q

That seems so quick

Surely you have to prove A/B x Q is 0

[rocky cloak]

So it's assuming you know quite a bit about tensor products, but the proof goes like this:

Q is flat, so when you tensor an exact sequence you get another exact sequence

0 -> B(x)Q -> A(x)Q -> A/B(x)Q -> 0

Since B(x)Q = A(x)Q = Q^n, then by linear algebra the first map must be an isomorphism, hence A/B(x)Q = 0.

It's true that X(x)Q = 0 if and only if every element in X has finite order.

But as you can see the outline of the proof is pretty much the same, just using some established tools to prove each step.

[old hollow]

Oh wow

That’s amazing

That’s so nice

I have subtle questions about what’s going on when you make those steps

The sequence
0 -> B -> A -> A/B -> 0
has its objects as Z-modules and its arrows as module homomorphisms

when you tensor with Q, the object B(x)Q for instance, aren’t the objects now Q-vector spaces?

Why is this jump from Z-modules to Q-vector spaces allowed

Seems like they are fundamentally different objects

[rocky cloak]

That's right. In general when you have a ring homomorphism R -> S, then tensoring by S takes R-modules to S-modules. This is called extension of scalars, and it's a "universal" way of producing an S-module from an R-module (c.f. adjunction in category theory)

[rocky cloak]

You'll notice that all Q-vector spaces are in particular abelian groups, so their not fundamentally different.

[old hollow]

Oh interesting, this reminds me of a homomorphism R->S giving rise to a homomorphism R[x]->S[x]

Extension of scalars is a good name lol

Anyway thank you

[dull ginkgo]

Here's a nugget of information, a lot of that can equivalently be stated using ideals, typically using principal ideals:

Greatest Common Denominators: Intersections of principal ideals are principal
Unique Factorization: GCD and "Principally-Noetherian", i.e every strictly increasing chain of principal ideals is finite, or equivalently each set of principal ideals has a maximal principal ideal.
Bezout Domain: Sums of principal ideals are principal (compare this to GCD domains)
Principal Ideal: Every ideal is principal lol

the whole "principal ideal" thing here really is just another way of saying the same thing lol

[warm dove]

Yeah but sometimes it’s nice to not invoke abstraction

In particular

We can get UFD without invoking thr ascending chain condition in Euclidean domains

Which is nice

Because that’s a kind of abstract pain in the ass

[dull ginkgo]

I prefer to work in the ideal sense or the element sense one at a time and not to hop between them like crazy

[elfin wraith]

Equally though, a lot of the power of ring theory is actually to do with ideals (well modules really, but same thing) so it’s not a bad thing to get used to

[warm dove]

Hopping is terrible yeah 💀

[warm dove]

Yeah, I’ve definitely gotten used to ideals after the ascending chain homework. Super useful stuff. Unfortunately, my class doesn’t have time to cover modules so that’ll be saved for an algebra class in college

[elfin wraith]

Yeah there’s a lot to cover, you can’t do everything, but having covered Noetherian rings in highschool I’d say your pretty set for uni haha

Ideals are just submodules of a ring viewed as a module over its self though, so really you have been working with modules!

[south rain]

I should prove that if $G$ is generated as $G = H_1 \cdot H_2 \dots H_n = K_1 \cdot K_2 \dots K_m$ and all $H_i, K_i$ are simple normal subgroups of $G$, then $n = m$ and $H_i \cong K_{\pi(i)}$ for some permutation. My proof:

Since all subgroups are normal and simple we can construct a composition series by
$$ F_i = \prod_{j = 1}^{n-i} H_j$$, since now $F_i/F_{i+1} \cong H_{n-i}$ is simple. The same construction can be done with the $K_i$, call the chain elements $T_i$. Now, by Jordan-Hölder both must be equivalent , i.e. the lengths are the same (n = m) and there is a bijective mapping between all factors, ie $H_{n-i} \cong F_i/F_{i+1} \cong T_{\pi(i)}/T_{\pi(i)+1} \cong K_{n-\pi(i)}$.

Now does this seem correct?

[cloud walrus]

dellinger

[tardy hedge]

You are taking abstract algebra class in highschool? Your high school offers that?

[warm dove]

💀 this is terrible I miss groups (I don't) 💀

[warm dove]

yes - magnet high school

dw the class is as hard as a regular university class (judging by the C- average on the Group Theory test before the curve 💀 )

[elfin wraith]

No don’t be silly, groups are boring and tedious, rings are cool fun and useful

Grouprings toe the line dangerously

[warm dove]

💀what’s really scary is field theory

If we end up doing Galois it’s cooked

[tardy hedge]

That’s ridiculous! Never heard of a hs offering that

[warm dove]

Google proof school (not my school but it’s ludicrous)

[fallen geyser]

Why does a linear transformation of a polynomial ring have to preserve degree?

[south patrol]

E.g. the zero map is linear and doesn't preserve degree

And there's the question of "linear over what"

E.g. consider a general k-linear map on k[x] where you send x^k to p_k(x) for any choice of p_k

Definitely needn't preserve degree

[dull ginkgo]

Galois theory to me is exceptionally boring

[warm dove]

Yeah but it’s probably hard

[tardy hedge]

“Needn’t”. Only a mathematician would say something like “needn’t”

Lol

[elfin wraith]

I think maths is probably the only place whence has been used in about 200 years

[south patrol]

Idk I use needn't a lot tbf

I feel like many people misuse whence where they could say hence

I wonder if it comes from French influence since d'où is common

[hidden wind]

i don’t recall seing whence much (or at all) in math

hence very common though

[grizzled spindle]

does the quotient of a ring or module distribute over direct sum?

[coral steeple]

I am told that Aut S3=S3. Since for each element of S3 I get a unique automorphism (conjugation by that element), this means that all the automorphisms of S3 are inner automorphisms. How can I show that none of the other bijections of S3 respect the group structure without going through them all and finding pairs of elements making that condition fail?

[wraith nexus]

consider generating elements of S3

||where can (123) and (12) get sent to||

[coral steeple]

Shoot I completely forgot that those generate S3; that's still weird to me

[coral steeple]

Well they get sent to a 3- and 2-cycle respectively. Does only one pair of such cycles generate Sn for each n or something?

No that wouldn't help

[bitter rover]

Orders have to match. How many elements can each be sent to? That puts an upper bound on the number isomorphisms, which is certainly no greater than the number of bijections satisfying that condition.

[coral steeple]

Hmm yeah that works. Why does it actually matter that (123) and (12) generate S3 then? The orders have to match, so (123) can be sent to two elements [itself or (132)] and (12) can be sent to any of the three transpositions. Hence there are at most 6 automorphisms. I could have used (132) and (12) or what have you.

[wraith nexus]

ywah

[coral steeple]

It seems kind of mysterious that I get a worse bound than by noticing that the set of 3-cycles is fixed by an automorphism and likewise for the transpositions, so there are at most 2!3!=12 automorphisms, since that packages the argument you suggested

[bitter rover]

Because homomorphisms are determined by choice of where generators go

Well, those two also generate S_3

[coral steeple]

Right, but I only use the fact that the orders have to be preserved in the argument

Maybe I actually misunderstood your idea somehow

[bitter rover]

It works out here to be the same thing because of the orders involved and number of generators involved.

[coral steeple]

But maybe in a different group if I tried to generalise your argument like I did here it would break?

[bitter rover]

Yes, I think so, once the details are spelled out.

Anyhow what you said is correct. That's the idea.

Then you can exhibit all 6 inner automorphisms directly if you want.

Well, 6 distinct ones. Together that means those 6 must be all of them.

[coral steeple]

Well I try to skirt around that by saying that the conjugations are relabelings of the permutations, meaning different conjugations correspond to distinct relabelings, so the map sending alpha to (x |-> alpha x alpha^-1) is injective. No?

[bitter rover]

Sure

[coral steeple]

Nice

Thanks for the help!

[hidden wind]

why would anyone use the notation $M_{m\times n}(\mathbb R)$ when the notation $\mathbb R^{m\times n}$ exists ?

[cloud walrus]

rødbet-jens

[hidden wind]

[south patrol]

Eh well M_n(R) is already common tbf so makes sense to keep it similar

but also like R^{m x n} is slightly abusive notation since like lol

R^{2 x 3} is not R^6

so the x is sort of silly lol

[hidden wind]

hmm

[glad osprey]

I don't think so; I can't think of a counter example, but if it were true it would make the proof that the quotient of a semisimple module is semisimple too easy

[south patrol]

What do you mean by this

For the question to make sense you need extra conditions which make the answer simple

[grizzled spindle]

(M+N)/S = M/S + N/S

[south patrol]

Yeah so like

What does that really mean

There you are viewing S as a submodule of M, N and of M (+) N

[grizzled spindle]

yeah

[south patrol]

like how does that set up work

[grizzled spindle]

well you take the direct sum of S with itself id assume

at least i believe thats the standard notation

so in reality it would be (M+N)/(S+S)

[glad osprey]

Wouldn't you consider M and N submodules of M (+) N, and S a subm of M and N?

Taking S (+) S would be different, it seems more natural to just take (M (+) N)/S

[grizzled spindle]

oh like identifying N with N x {0}?

[glad osprey]

Yeah

[grizzled spindle]

well, either coordinate works

hmm im not sure, it wasnt precised in the notes given by my prof

can i send the example directly maybe it would help

im a bit confused by it because we keep changing which structure we use to analyze it

[grizzled spindle]

nope its this one we want here

[elfin wraith]

I would absolutely read the latter as an mn tuple not an n by m matrix

[hidden wind]

interesting

[hidden wind]

this is not a point of confusion for me, as i never use x to denote scalar multiplication, only for like set products or the cross product

[glad osprey]

Okay, I just realized it's easy to prove that (M + N)/(S + S) is iso to M/S + N/S, maybe that is the most natural construction. I'll have to think about it more tomorrow

[old hollow]

M_ emphasizes matrices which is nice

[somber sleet]

Does there always exist a complement for vector spaces? Also for infinite dimenisonal vector spaces?

[ashen heron]

do you mean for a subspace?

[somber sleet]

yes exactly

[ashen heron]

i think so

[somber sleet]

do you have a proof?

[ashen heron]

its a Zorns lemma proof for sure

[somber sleet]

Or somewhere where I could read one

[lone niche]

https://math.stackexchange.com/questions/609245/extension-of-linearly-independent-set-to-a-basis-in-an-infinite-dimensional-vect

[ashen heron]

https://math.stackexchange.com/questions/1515512/does-every-vector-sub-space-has-an-algebraic-complement

[somber sleet]

does it also work if a set is not countable?

like R/Q

[ashen heron]

yep

[south patrol]

To summarise: given a subspace U of V, you can (using choice/ zorns) pick a basis of U, extend to one of V, and then all the "extra" basis vectors you had to add in span a space W which is easily seen to be a complement of U in V

[ashen heron]

zorns lemma is equivalent to an axiom of set theory so it works for all sets (edit: poorly worded but you get the point)

[glad osprey]

If M is a direct sum of A and B, and N is a submodule of M, is it true that N is of the form N1 (+) N2 where N1 and N2 are submodules of A and B respectively?

[lone niche]

I don't think so, take Z x Z and the subgroup generated by (1,1) I don't think that decomposes into a direct sum of subgroups of Z.

[tardy hedge]

The grouping together here is using chinese remainder theorem right?

If u have R/(p^2) x R/(p^3) thats isomorphic to R/(p^3)?

I havent learned crt properly so i may be so wrong

[glad osprey]

I see, thanks

[tardy hedge]

Not every torsion R-module has a nonzero annihiltor right?

[tough raven]

Consider (+)_{p prime} ℤ/pℤ.

[tardy hedge]

Always need to think of some infinite sum …

[tardy hedge]

For the elementary divisors of a f.g module over a pid, do we count the repeat primes?

[dull ginkgo]

terror

[tardy hedge]

Yea i saw somewhere earlier that ur not a fan of this stuff lol

I guess its not that interesting idk

[dull ginkgo]

Jacobson uses proofs that involve doing matrix calculations

[tardy hedge]

Ohhh uhhh 😬

[dull ginkgo]

Which tbf if you have the time to algorithmically write it out is… admittedly kinda neat but that’s about it

[tardy hedge]

Yea

[dull ginkgo]

Just it’s annoying to manually calculate

[tardy hedge]

What IS interesting is that today we are starting some homological algebra stuff in class

[dull ginkgo]

Fun to implement if you code it yourself :3

[tardy hedge]

I c

[dull ginkgo]

The only homological stuff I’ve done is some cursed form of diagram chasing I concocted

[tardy hedge]

I should practice some diagram chasing questions before the exam on friday

Im looking forward to learning about derived functors

Im not sure how much we are gonna be able to cover

[dull ginkgo]

Actually in retrospect it’s piss easy to describe, easier than Jordan normal form

Basically you keep reducing the matrix to a diagonal of values on the top left and a lower (n - 1) x (m - 1) block on the bottom right

Sort a column by number of prime divisors, moving 0’s to the bottom. Then apply Bezout to reduce the top one to a gcd of all of the ones below it. Then wipe out the lower ones. Repeat for the row. Then you have turned the row and column into a diagonal

[tardy hedge]

For (2), the second iso thm is being used right?

Im not so sure on how its being used

[dull ginkgo]

I think it’s the second

[tardy hedge]

How is it being used?

We can just say that image of (p) in R/(a) is (p)/(p) cap (a) and then use second iso?

[dull ginkgo]

Sounds about right

[dull ginkgo]

I wouldn’t call it an abstraction.

The bread and butter of commutative ring theory is modules and ideals

We can use principal ideals to pull back structure defined on modules or ideals to the elements themselves

Principal ideals obviously allow you to map elements into the set of ideals

Which naturally have the inclusion preorder and different kinds of operations like ideal intersections, products, sums which you can sorta pull back to the elements

You can define equivalence relations and preorders from this, like x ~ y iff (x) = (y)

Or x <= y iff (x) is a subset of (y)

Which of you might work it out, are quite familiar

And frankly the “noetherian condition on principal ideals” only really deals with the principal ideals so it’s almost immediate to pull that back to the elements

[dull ginkgo]

A lot of the domain conditions are basically just closure of these different ops

As mentioned: Bezout domain when the family of principal ideals is closed under sums, GCD when closed under intersections

If you pull back the prime condition on prime principal ideals you get the classical prime element defn

Even being a unit can be rephrased in terms of ideals, i.e not lying in any proper ideal

[tardy hedge]

A prime element in a ring has no divisors? Or an irreducible element has no divisors? (Havent studied this material properly before)

Ik in pid they are equivalent concept

[dull ginkgo]

Irreducible element’s principal ideal contains no other proper principal ideals

In a domain, which is the usual context

I.e they are “atomic”, can’t get smaller principal ideals that aren’t the trivial one

[tardy hedge]

Do u mean irreducible elements principal ideal is not contained in any other principal ideal?

[dull ginkgo]

Oop yeah got it flipped

[tardy hedge]

Thanks for the mistake cuz it forced me to think it thru betyer

Lol

[dull ginkgo]

Lmao ye

[tardy hedge]

Better*

[dull ginkgo]

But yeah it’s just pulling back that “subset” relation back to the elements

[tardy hedge]

Ok. Yeah so irreducible elements dont have nontrivial divisors

[dull ginkgo]

Two elements are equivalent if their principal ideals are equal, in which for the case of a domain, you can show that it implies that they differ by a unit multiplier

If (x) is a subset of (y) strictly iff y divides x

Let’s say you have a set A, and a set B admits an ordering <, then if you have a function from A to B, you can define the pullback ordering on A where x < y iff f(x) < f(y)

For the map of the ring to the set of ideals, technically the pullback of the inclusion relation is “is divided by”

[dull ginkgo]

Yeah, maximal amongst principal ideals

[glad osprey]

In our lecture notes on modules over PIDs, it says that we can have rank F = rank G without F = G, where F, G are modules over a PID, and it gives the example (2) subset Z over Z. However, (2) is isomorphic to Z, right? F and G would always be isomorphic if they have the same rank, wouldn't they?

[lone niche]

Depends on how you are defining the rank, or more precisely for which modules you are defining it.

[dull ginkgo]

I thought PID rings satisfy IBP

[lone niche]

Are your modules free modules?

[dull ginkgo]

I don’t get why we don’t just say dim 3 or something instead of rank

Assuming freeness

[next obsidian]

Equality not isomorphism

This is just saying unlike vector spaces where if you have L < V with dim L = dim V, then L = V

You can have a proper inclusion of free modules L < V where rank V = rank L but L is not all of V

As the example 2Z < Z illustrates

[dull ginkgo]

The existence of inverses forces the equality

[next obsidian]

This is true when you have finite length modules N < M, if l(M) = l(N) then N = M. The difference is that for vector spaces, the length is equal to the rank / dimension. But for modules over non-fields, a free module is usually not even finite length

So for modules you kind of split in two two camps.

Finite length modules behave like vector spaces in terms of rank-nullity where you use length instead of dimension.

Free modules behave like vector spaces in that they have a basis and you can define maps by just saying where those go

Unfortunately over general rings you can’t have your cake and eat it too

[glad osprey]

Ah, that makes sense, thanks the remark as stated in the lecture notes seems uselessly weak, but if F subset G then I see how that is interesting to note

[next obsidian]

Yeah

[tardy hedge]

When proving uniqueness for the structure theorem for modules over pids, we know off the bat if M1 and M2 are isomorphic then Tor(M1) = Tor(M2). What is the difficulty then of just saying, for some elementary divisor of M1, say p^2, that the submodule R/(p^2) gets sent to R/(p^2) in M2?

If its isomorphic doesnt submodules map to submodules isomorphically

Im trying to understand what the problem is here with the elementary divisors

[next obsidian]

Sorry what’s an elementary divisor again

[void cosmos]

i think the prime ideals in the structure theorem for pids (the primes u get after crt)

[next obsidian]

I recall like two ways to write down the thing though

One is with like asserting that successive things divide the other

And the other one is something else

[void cosmos]

apply crt

[next obsidian]

And they’re related via CRT

[void cosmos]

those primes are the prime divisors

elementary divsiors*

[next obsidian]

Prime powers right?

[void cosmos]

yes

[next obsidian]

can you post / state what the statement you’re looking at is, and what uniqueness its asserting

[void cosmos]

think of the standard problems that involvle classifying an abelian group of order something

[next obsidian]

I suspect the problem might be that when you use elementary divisors you can permute the order of the things

[void cosmos]

u do prime decompistion of the order into powers and keep playing with them using CRT

thats the "elementary divisors" form

[next obsidian]

Whereas when you make it so each successive thing divides the next it fixes the order, up to the only ambiguity being writing something up to a unit

[tardy hedge]

Theorem 9 is asserting uniqueness

[next obsidian]

So what’s your question?

[tardy hedge]

If M1 and M2 are isomorphic, their submodules are isomorphic right? So if M1 has an elementary divisor of like p^2, then R/(p^2) is a submodule so then why cant we just say by the isomorphism R/(p^2) is a submodule of M2

In the proof though, apprently we can only say off the bat that their p-primary submodules are isomorphic

Which are when u group together the repeat primes into their highest power, like R/(p^3) iso to R/(p^2) x R/(p^3)

[next obsidian]

Can you show the proof and where you’re asking why you can’t do this instead

[tardy hedge]

“If M1 iso M2 then the p-primary submodule of M1 is iso to p-primary submodule of M2”

[next obsidian]

You have to essentially prove that you can only write a module in this way in only a single way

Even if M1 has some R/p^n summand, it isn’t clear that when you write M2 in this way that R/p^n has to show up in that

Like what if somehow you can write down
Z/4Z x Z/2Z x Z/9Z ≈ Z/4Z x Z/2Z x Z/3Z x Z/3Z

Then there’d be a Z/9Z submodule on the right, but it hasn’t shown up in the decomposition on the right

And you’d probably go “no that can’t be true because Z/9Z isn’t isomorphic to Z/3Z x Z/3Z” and that’s essentially the crux of the issue, but you don’t know that this specific isomorphism would have to take Z/9Z to Z/3Z x Z/3Z a priori

But when you look at the p-primary part of both sides, on the left you get Z/9Z and on the right you get Z/3Z x Z/3Z, so actually wait this isomorphism would imply their isomorphic. So now you can do your like, kinda CRT proof or whatever to get your contradiction

[tardy hedge]

Ok i think i kind of see what u mean

[next obsidian]

Basically, I’m saying that the notion of “invariant factors” or “elementary divisors” as written a priori depends on how you decompose M

And the theorems at first only assert there’s some decomposition

But you don’t know that this is instrinsic to just M, that you can only write it in one such way. In which case you’d have two different lists of these things

Think about composition series and composition factors or whatever

You need to prove Jordan-Hölder to say that the list of composition factors only depends on M itself, not on the specific composition series you had

[void cosmos]

u mean that they are unique?

yeah

ig to be more annoying proving jordan-holder is easy if you know the schrier refinment theorem

haha

[still flare]

Can anyone explain or hint to why alpha is a root of f

[dim widget]

Well what happens when you evaluate f on alpha

[still flare]

I would say we get zero if f was a homomorphism. But idk what's supposed to happen if f is just some polynome

[dim widget]

why does it matter if it is a homomorphism?

suppose f(x) = x^2 + 1

can you see why it works in this case?>

[still flare]

Since alpha is 0 in that constructed field, f(alpha) = 0 would follow

[dim widget]

well alpha is not 0

alpha is the coset x + (x^2 + 1) which is certainly not the same as 0 + (x^2 + 1)

[still flare]

right mb

[still flare]

This is confusing. What even is the 1 in "x^2+1" in K[x]/(x^2+1) ?

[dim widget]

1 + (x^2 + 1)

[still flare]

ah thats good

[still flare]

I claim f(x +f(x)) = f(x) + (f(x))

Ok but why didn't my lecturer at least say that

How does life justify the hardships it imposes on me

[dim widget]

that's true

in fact for any polynomial g(x) we have that g(x + (f(x))) = g(x) + (f(x))

[still flare]

It's as clear as day now for me

thanks

[minor fulcrum]

why is the trivial ring excluded from being a field
what about field theory gets more annoying if we just let it

[bitter rover]

https://en.m.wikipedia.org/wiki/Field_with_one_element

[coral spindle]

Uh I'm not sure it is a good choice to reply to that question with that article

[coral spindle]

Anyway, the trivial ring is excluded from being a field bc it is annoying to say "let F be a nontrivial field..." every time.

N.b. the "field with one element" referenced above is not in fact a field and does not actually exist.

[minor fulcrum]

what is "every time"

how many theorems would actually require that

[coral spindle]

Instead of saying "let F be a field" I would instead have to say "let F be a nontrivial field" to say anything interesting.

Ok here is a simple one

There is a vector space of dimension n for all fields and integers n.

Or idk even more straightforwardly

dim F^n = n

[minor fulcrum]

mm, maybe it would break a lot of linear algebra yeah

[elfin wraith]

This does actually raise an interesting question to me, why do we allow the trivial ring and trivial group? In my experience they only really appear as a counter example, are excluded from a theorem or a nice base case for an induction

[coral spindle]

Why do we allow the empty set?

I think there are things to say here but I will say this: the trivial group turns up infinitely more often in my experience than the trivial ring ever has

I want to actively ignore the trivial ring, whereas I occasionally argue that a group is trivial in various ways (e.g. by saying that a quotient group G/N is trivial)

[elfin wraith]

Yeah no that is fair the trivial group definitely has its uses as does the empty set, but I’ve literally never encountered the trivial ring beyond intro ring theory books saying this is a thing and we ignore it

Because yeah the trivial group is nice for like composition series and the like, and the empty set is useful all the time, say the closure is a set together with its set of limit points, but I’m still yet to come across an actual instance of the trivial ring, so what is lost by requiring more generally that rings have different additive and multiplicative identities?

And in that case you don’t need to make the distinction that a field can’t be the trivial ring which I have always felt is an odd distinction to make

[minor fulcrum]

the category of rings and groups would be much, much worse without them

the category of fields already sucks because 0 isn't invertible

[elfin wraith]

I wondered if that would be the case

And yeah now that I’m home and can actually think of course that’s the case, it’s the zero object in Group

[coral spindle]

"Yes, but" in practice this still isn't really an issue.

In actual study of these objects, we still want to ignore the zero ring.

[elfin wraith]

But then I suppose we allow the category of fields to be annoying by not having the trivial field, just to avoid having to say non trivial field everywhere, much like we do with rings, so I guess it’s just convention more than anything?

[coral spindle]

The category of fields is weird for unrelated reasons

And in a non-categorical perspective, fields are weird as algebraic structures for unrelated reasons.

[tardy hedge]

Im not sure how we know the Ni’s are cyclic modules

[coral spindle]

This is not a shocking fact. There is no isomorphism theorem for fields, for example. In universal algebra we would say that fields are not a variety.

[minor fulcrum]

yeah, and that's because 0 isn't invertible

[tardy hedge]

Oh yeah.

Somehow i didnt realize that tho

Everything is a shocking fact for ppl less knowledgeable in math

Funny how that works

I remember being really impressed by the idea of quotient objects in general

When i first saw them

[elfin wraith]

Yeah I suppose I’ve never really considered that, I being a simple ring is a slightly weird property

Could you elaborate more on fields being weird as algebraic objects though? I’ve not done much (any really) field theory so I guess maybe I’m missing some stuff but idk to me they’re just rings with so much structure that they become kinda simple again

[minor fulcrum]

every single field map is an injection

[bitter rover]

There's an answer which boils down to "definitional efficiency", i.e., having to say "non-trivial field" everywhere we said "field".

But one can also wonder if there's more going on.

For example, unlike groups and rings, the class of fields isn't definable over an algebraic/equational theory in the sense of universal algebra. If that were the case they'd be closed under direct products, among other things.

From that perspective, it's maybe less surprising that fields are "something different."

[minor fulcrum]

I was mostly asking how often we would actually have to specify non-trivial

if the answer is "all over the place in linear algebra" I guess that makes sense

[elfin wraith]

I guess it does just come back to them being simple rings again then, maybe I’ve just never appreciated that being quite so weird of a property

[minor fulcrum]

it is weird in that it makes them unlike most other algebraic objects

[elfin wraith]

Yeah I guess, I suppose when I proved Schurs lemma I just never really appreciated the result before

[rotund aurora]

What does this mean?

[minor fulcrum]

do you know what a category is

[rotund aurora]

yes

[minor fulcrum]

ok so in rings and groups, the trivial ring/group is the initial and final object

this makes them very nice categories

for instance, the image of a ring homomorphism is a ring, this stops being true if the trivial ring isn't a ring

many, many, many more examples

[rotund aurora]

like?

[minor fulcrum]

well that not being true means that you don't get ring or group quotients as a well defined thing

[rotund aurora]

You do get quotients, except in very specific circumstances. I think the importance of the trivial group is being overstated here, it doesn't prove anything you didn't know

[vagrant zinc]

What page is that?

[tardy hedge]

I think i just straight up dont understand this proof

[tardy hedge]

465

[tardy hedge]

I understand “in the general case Ni is a submodule of M with annihilator dividing pi^ai”

I dont understand the application of crt near the end

I dont understand why we see the annihilator of Ni is “precisely” pi^ai

[vagrant zinc]

Remove the author's comments and just leave the definition, propositions, theorems used in the demonstration and then connect the comments to what he is saying.

[tardy hedge]

Well yeah

Thats what ive been trying to do lol

[vagrant zinc]

I think it says something about a prime factorization of powers, maybe you should go back a little bit, besides it talks about comaximals, take a rest a little bit.

[dense root]

can someone tell me whats the point of PIDs

[opal coral]

aM = 0, that is M = M / aM and aM is the intersection of all (pi^alpha_i M), so the crt gives you M / aM = the sum of M / pi^alpha_i M and then you shall identify those with Ni
If you take m -> (a / pi^alpha_i) * m, this gives a mapping from M / pi^alpha_i M to N_i, it's surjective as a/pi^alpha_i is invertible in R/pi^alpha_i, so that should make this a bijection i think

[elfin wraith]

I’m not sure what you mean by what the point is, some rings just are PIDs, and PIDs have a lot of nice properties

Like they’re all Noetherian, prime and maximal ideals coincide etc, there’s just a lot of things you can say

[tardy hedge]

Tbh you’ll see the point the more algebra u study

[tardy hedge]

Thanks a lot

[dull ginkgo]

Allows you to describe the powerful ideal structure of a ring purely through the elements alone

[minor fulcrum]

no like, the quotient of two groups G/N is no longer actually well-defined.

I agree that you can technically prove anything you want without it, but the categories of rings and groups being nice is awesome and there's no reason to chuck that away

[tardy hedge]

bro im trippin. Z -> Z/(2) x Z/(4)

Has kernel (2) cap (4) = (4). So Z/(4) is isomorphic to Z/(2) x Z/(4)?

y am i trippin

Ok because this is not true for groups right

Z4 is not isomorphic to Z2 x Z4 for groups?

nah i just know im doing something profusely wrong rn

[warm dove]

2 cap 2 is 2

[dim widget]

it's just not surjective

[warm dove]

Also yeah not surjective bc gcd isn’t 1

[dim widget]

1st isomorphism theorem tells you that Z/4 maps to Z/2 \times Z/4 but the value on the first factor is determined by the value on the second factor

so the map maps to a diagonal subgroup of the product which is isomorphic to Z/4Z

[tardy hedge]

thanks'

i realize now that in the proof for crt they say IF the ideals are comaximal then ...

I should look at the proof for that

more closely

[tardy hedge]

How do we know a-priori that the p-primary submodules are the same?

Is that just because M1 and M2 are isomorphic?

[next obsidian]

Any homomorphism has to send the p-primary part into the p-primary part cuz linear, and you apply that both directions on the isomorphism to see it restricts to an isomorphism of p-primary parts

[tardy hedge]

ok yes

so lets say a p-primary part is like Ni = R/(p) x R/(p^2)

isnt R/(p) also a submodule of M? Why doesnt that correspond to the same submodule for M2 via the isomorphism?

[next obsidian]

I mean, it does (but if there’s multiple R/(p)’s you can’t say which)

But the point is that if the other guy was written as R/(p^3), there isn’t an isomorphism

If there were then you’d have an R/(p) submodule from that isomorphism, in fact a summand

But there is no R/(p) summand of R/(p^3), and as a result or “in fact”, no isomorphism from your Ni to R/(p^3)

Try writing out the proof as you want to do it and notice where you get stuck, or show the proof you wrote and have someone point out where you mess up

[tardy hedge]

what is the "other guy"?

[next obsidian]

The p-primary part of M2 or whatever

[next obsidian]

I think until you do this you aren’t gonna get what I’m talking about

Because the statements you keep asking are vague

In some sense

[tardy hedge]

Wouldnt it make more sense to say if the other guy was written as R/(p^2)? Because that has the same annihilator as R/(p) x R/(p^2) .

I am thinking that a p-primary submodule of M1 has annihilator (p^a), then the p-primary submodule of M2 is isomorphic so has same annihilator. But then the question is like, are their decompositions in M1 and M2 the same?

thx for the help btw i think im slowly understanding it a bit better

[warm dove]

does this proof look okay for showing that finitely-generated is equivalent to ACC?

[elfin wraith]

Only skimmed it so maybe some small details could be a little off but the broad strokes look good to me. Another fun exercise you could do is showing they’re both equivalent to the maximum condition if you were interested

[warm dove]

maximum condition?

[opaque finch]

Any collection of non-empty ideals has a maximal element.

Your proof is good and you used details which can be modified a bit to prove that

[warm dove]

Maximal element meaning one that contains all of these?

[dim widget]

no that wouldn’t be true, the ideals could be comaximal

[void cosmos]

a maximal ideal is an ideal that can't be contained in another non-trivial ideal

[warm dove]

Oh maximal as in that

Right

My bad I forgot that was a thing and defaulted maximal to like containment

[void cosmos]

no its fine normal thinking

[warm dove]

For a proof for this, I can get if $R$ satisfies the maximal condition, it is ACC easily.

If $R$ satisfies the maximal condition, any chain $I_0 \subset I_1 \subset ...$ must have a maximal element realized as $I_i$ for some $i$, in which case it stabilizes at that. Conversely, if $R$ satisfies the ascending chain condition, consider any set $S$ of ideals that contains no maximal element. We should be able to build some chain of ideals of S, and if we can build a chain $I_0 \subset I_1 \subset ... I_k$, and then since $I_k$ isn't maximal by construction, we should be able to find $I_{k+1}$ to continue the chain?

[cloud walrus]

marlins_karp

[warm dove]

but the issue is I don't know how to force that chain to consist of elements entirely in S

[rotund aurora]

You can't always take quotients of two groups, so idk what you are saying. And it is well defined

[minor fulcrum]

you always can if N is a normal subgroup of G

[old hollow]

is 2 irreducible in Z[i]?

[rotund aurora]

no

[delicate bloom]

hint: square 1+i

[old hollow]

ok cuz (1+i)(1-i) = 2

and I can't see any issue with this decomposition

[delicate bloom]

that shows it's reducible

[old hollow]

but I saw an exercise where "we will prove 2 is irreducible in Z[i]"

[delicate bloom]

you just reduced it yeah

[old hollow]

must be a mistake

yeah lol

ok just sanity check ty

[delicate bloom]

2 is special in Z[i] to be sure

it's the only ramified prime

[dull ginkgo]

Yeah assuming dependent choice we can turn “finite strict inclusion chains only” into “every collection of ideals has a maximum element”

[dull ginkgo]

Note to self, investigate the following:

Say a commutative, totally ordered monoid $M$ is psuedonatural if for each $x < y$ there exists a $c$ such that $x + c > y$

Does the monoid ring $R[M]$ inherit the countable chain conditions (Artinian, Noetherian) from $R$?

[cloud walrus]

shatgtp

[dull ginkgo]

I think the answer is yes. We can easily define the degree of an element based off of the largest monoid summand. We can define the equivalence relation ~ on the monoid ring where r_1 ~ r_2 if they have the same coefficient and leading degree, i.e have the same leading term.

We can then split an ideal J into these equivalence classes, and each equivalence class can be associated to a pair [a,m] where a is the leading coefficient and m is the monoid element.

For each m, the set of a such that [a, k] intersects J for each k <= m should form an ideal in R since if s in R[X] is [a,m] then we can lift it to any [a,k] for k > a by multiplying by the c given by psuedonaturality

Then we can use that map to try to squeeze an increasing chain in R[X]

Since obviously that map on the ideals would preserve the inclusion order, but would have to terminate in R, for each m

Call that map [J,m], then if we have a strict sequence of increasing ideals J_n, we should be able to show [J_n, m] terminates for each m

I’ll come back to this later I have more productive stuff to do

[rocky cloak]

Don't think so, like if you take M to be the positive really numbers under addition, then the condition is satisfied just be taking c=y, but R[M] is very infinite, even when R is a field.

[dull ginkgo]

I think I need to impose a countable condition for a diagonal argument of of the ideal maps mentioned

[tough raven]

I suspect the condition you want is that M is well-quasi-ordered (https://en.m.wikipedia.org/wiki/Well-quasi-ordering). IIRC this is equivalent to Hilbert's Basis Theorem for the special case of an ideal generated by monomials.

[tough raven]

Also you "shouldn't" expect the Artinian property to be inherited; for example, a polynomial ring over a field is not Artinian even though a field is.

[dull ginkgo]

Yeah I realized that later

[dull ginkgo]

I got something similar to that lol

[dull ginkgo]

I got something similar to that by considering a commutative ordered monoid ring over F_2 and seeing what happens to the principal ideals of the monoid elements

[misty cosmos]

can i get a hint to show there are no simple groups of order 396

[bitter rover]

Decompose 396 into prime factors and create various constraints on how many subgroups of different orders there can be using Sylow's theorems.

[regal zodiac]

If i have a module M and a submodule A, is there necessarily a submodule B such that the direct sum of A and B is M ?

[rocky cloak]

Consider 2Z/4 as a submodule of Z/4

[rocky cloak]

Some useful facts could be that if a group has n sylow subgroups, then the action gives a map to Sn.

And if P is a sylow subgroup, then the index of its normalizer equals the number of sylow subgroups.

[tardy hedge]

A projective module cannot have any torsion elements right?

[tough raven]

(assuming the base ring is an integral domain)

Yes, since it's a submodule of a free module.

[tardy hedge]

Ok thx that what i was thinking

[slender cedar]

Hi gyus hope y'all good , does anyone have or know a math french group thnx

[fallen geyser]

If a K-algebra is finitely generated, then is their a fixed size for all minimal sets of generators?

[coral spindle]

The group algebra KS_n over any field K is generated by the following minimal (by inclusion) generating sets:

1. {(1 2), (2 3), ..., (n-1 n)},
2. {(1 2), (1 2 3 ... n)}.

[worldly harness]

Let I be ideal in a ring R and J be ideal for R/I. Does J = {I + k | k is in K} for some ideal K in R?

[coral spindle]

So we write this as J = K/I

And yes, this is true. More importantly, there is a bijection {ideals of R/I} and {ideals of R containing I}.

Now go forth and prove it!

[chilly ocean]

@chilly ocean hello!! ty for offering to help u dont know how confused i am rn lmao

Pick something

Are you in a first course for abstract, algebra

yes, the first course

my test is on groups and homomorphisms

i think i understand what a group is loosely enough?

here's an example of a question i just did, that stumped me and made me go to the notes

i just dont even get what T^-1 would even look like? the markscheme was so unhelpful, it just said "(a) by inspection"

Ok

[chilly ocean]

You know these are elements of the symmetric group right

yep

You know how to write down the inverses?

no...

would the inverse be like, swapping them? so if 1 mapped to 2 would the inverse be 2 mapping to 1...?

[chilly ocean]

No like remember that g*g^-1=e

Oh yeah that would be the inverse

The example you gave

But doing that for each element

[acoustic rose]

Could I have some help with this?

It's for my abstract algebra but I'm not really seeing how abstract algebra topics really connect to this 😭

[chilly ocean]

Any progress

[acoustic rose]

I tried letting a = x^2 + y^2, b = m^2 + n^2 but I don't think that leads me anywhere

I also tried doing some examples but Im not seeing anything

[chilly ocean]

Why not lol

[acoustic rose]

I get ab = (xm)^2 + (xn)^2 + (ym)^n + (yn)^2

but how do I make this the sum of two squares

[chilly ocean]

Is there more you can do there?

[acoustic rose]

Factor maybe but doesn't that just give me back a and b?

[chilly ocean]

Yeah you dont want that

You want to show that ab is a sum of two squares

So ab=q^2+p^2 where p and q are integers

[chilly ocean]

so basically switching those 'rows' ?

[wraith cargo]

Add in 2(xymn) - 2(xymn)

[acoustic rose]

holy shit

thanks

[chilly ocean]

Yeah kinda but you need to go through each element

Have you heard of cycle notation?

[wraith cargo]

Then you have that ab = (xm+yn)^2 + (xn-ym)^2

[chilly ocean]

yep, like (1 2 3)

but idk how to find the inverse

actually, is the inverse of (1 2 3), (1 3 2)?

[chilly ocean]

Nah you read it backwards

So like (1 2 3) is one goes to two and two goes to three and three goes to one

If you want the reverse you have 3 goes to 2 goes to 1 goes to 3

yeah, so doesnt (1 3 2) do the opposite?

[chilly ocean]

ohh does it have to go in this order

(3 2 1)?

Nah it doesnt matter its the same thing

I didn’t recognize it at first

okok

if its not in cycle notation, and its in like that weird table matrix form, doesnt swapping the 'rows' work, when finding the inverse?

Id say try it

Oh wait my bad interpretting

For sure swapping the rows would work but when we write that two line notation we want the first row to be in order

So swapping the rows doesnt do that and you have to reorder which takes more work than cycle notation

do u think its better to convert it into cycle notation and then find the inverse

Yeah because sorting takes longer

okay lemme try that

[chilly ocean]

okay i put tau into a cycle of (1 2 3 8 7 6 5)

and then im confused bcs if i want to do the opposite, cant i just write it backwards?

so like, (5 6 7 8 3 2 1)

?

Ye

Also I think putting the one 1st is standard maybe?

[coral spindle]

Maybe it's worth saying this explicitly

(1 2 3) = (2 3 1) = (3 1 2).

Maybe this is why you were confused about writing it backwards. You can, but there are other ways to write it.

[chilly ocean]

Yea

[chilly ocean]

i seeeeee

[chilly ocean]

okok i think ive got inverses understood now, how did they spot the answer here was A using inspection? it feels lke magic

[coral spindle]

They just calculated it

[chilly ocean]

so like, take 1 e.g., sigma first so 1 maps to 3, then through tau^-1, 3 maps to 2?

did theycalculate it like that

[coral spindle]

Sure

[chilly ocean]

then how does that (2 3 1) cycle show that?

1 → 3, 3 → 2

but (2 3 1) is the opposite no?

[coral spindle]

(2 3 1) sends 1 to 2.

(2 3 1) = (1 2 3).

Function composition is right-to-left usually, but some authors disagree. However cycles are always left-to-right.

(1 2 3) sends 1 to 2, 2 to 3, and 3 to 1.

[rustic rapids]

is there any easy way to tell what the potential stabilizers for some group action is? For example in this problem, its not super hard to just check each: 1) stable under every element so |Gx| = 6, |O| = 1, then check proper subsets, |Gx| = 2,3, then then only identity |Gx| = 1, |orbit| = 6. But if the group has a high order I dont see an easy way to figure it out.

[coral spindle]

I'm not aware of anything that makes it easy. Perhaps the computational algebraists have some wild tricks, but they're probably not useful for humans.

This seems like a tedious exercise to me.

[runic nimbus]

is there a mistake here? shouldn't N be isomorphic to just Z mod 2?

[coral spindle]

Yes there is definitely a mistake, Z_4 x Z_2 is not cyclic

And I agree, it should be isomorphic to Z_2.

[south patrol]

I can't even find small perturbations of the question that are correct lol

e.g. <x^2, y^2> would just be Z/2 x Z/2 I guess

[runic nimbus]

yes it should be G/N

[coral spindle]

Ahhhh indeed

[south patrol]

Ah that makes more sense lol

[thorn jay]

What a world it would be if 4 was coprime to 2..

[tardy hedge]

Lol

[tardy hedge]

In a question like 12

Are we taking V (x) V as a vector space over F?

So V has a (F,F) bimodule structure in the usual way?

Do we just assume that?

[tough raven]

When taking tensor products over any commutative ring R, one typically assumes that all R-modules are (R, R)-bimodules with the left and right actions being the same. In this case, r ⋅ (m (⨯) n) = r⋅m (⨯) n = m (⨯) r⋅n = m (⨯) n⋅r = (m (⨯) n) ⋅ r, so the same is true of the tensor products. So you can view the tensor product as multiplying R-modules to give R-modules and forget about bimodules.

(You have to be careful about this when (and AFAIK only when) there are already bimodules with different left and right actions.)

[tardy hedge]

Thanks

And im a bit stuck on this question anyway. What should be the approach when starting with v x v’ = v’ x v?

[tough raven]

Personally, I'd try decomposing in terms of a basis for V; that's helpful for concluding anything from an equality in a tensor product.

[rocky cloak]

Usually the most convenient way to deal with tensor products is through the universal property.

If you can find a bilinear map such that (v, v') and (v', v) are mapped different places, then they can't be equal in the tensor product.

[tardy hedge]

Since V x V is a vector space you can use a basis argument there right?

[tough raven]

Yes, but I more specifically mean a basis of the form {e_i (⨯) e_j : i, j in I} where {e_i : i in I} is a basis for V.

[glad osprey]

Is Wedderburn-Artin really an if and only if? Specifically curious about whether a semisimple ring has no nilpotent ideals

Here's how it appears in my textbook

[elfin wraith]

Every nilpotent is in every prime ideal, if R is artinian then it’s semi simple if and only if its Jacobson radical is 0 so I think that works?

[glad osprey]

I see, so the reverse direction is also true? I'm confused by the fact that the conclusion "R is isomorphic to a finite direct sum of matrix rings over division rings" seem much stronger than "R is semisimple"

[dull ginkgo]

We need Artinian-ness

iirc

Actually no i think that's to pass from a trivial J radical

You can still decompose endomorphisms over a finite direct sum of modules

[rocky cloak]

I think your underestimating the strength of being semisimple.

One easy way to see this direction is assume R is semi-simple, then R = Sum_i Si^ni for some simple modules Si.

Consider End_R(R), in general for any ring this is equal to R, but
End( Sum Si^ni ) = Prod End(Si^ni) because Hom(Si, Sj) = 0
and End(Si^ni) = Mni(End(Si)) and End(Si) is a division ring by schurs lemma

[elfin wraith]

Is this not basically just the proof of artin wedderburn?

[rocky cloak]

It's a proof of one direction at least

[dull ginkgo]

Actually wait if we have two R-modules that are NOT isomorphic then is the endomorphism ring of their direct sum the direct sum of the endomorphism rings of them respectively?

[rocky cloak]

No, you can still have homomorphisms between modules if they're not isomorphic.

(but not if they're simple)

[dull ginkgo]

oh what does simplicity have to do with it

[tardy hedge]

I was doing 2b by using R/m is a field but i dont want to do that. What is the more basic way?

I feel like ive seen this question before idk

[rocky cloak]

Because simple modules don't have nontrivial submodules...

[dull ginkgo]

oh wait yeah doh,

brain fart

[rocky cloak]

Assume that there is no such r, what can you say about the ideal (x) + m?

I guess worth noting that commutativity is irrelevant

(assuming ideal means left ideal)

Actually it's needed for the r not in m part. So I guess not completely irrelevant

[tardy hedge]

Thanks. That makes sense

[dull ginkgo]

additional exercise, show the following are equivalent:

1. R is a division ring
2. R is left-simple, i.e no nontrivial left ideals
3. R is right-simple, i.e no nontrivial right ideals

[tardy hedge]

Somethign like that was what i was trying to get to

[dull ginkgo]

You can also bullshit a bit and say that R is a simple right-module over itself, so thus it's endomorphism ring (itself) is a division ring

[tardy hedge]

m maximal makes (x)+m = m and then u get a contradiction cause you’d get 1 is in m

[elfin wraith]

The fact that I’ve felt vaguely confidence that I could answer the last two questions asked in here is making me feel good about my upcoming exams

The noncom exam could still be hellish but we move

[glad osprey]

Interesting, I think I get the idea, but not 100% sure about the details What is Si^ni? The ni-fold product of Si by itself? And Sum and Prod is the direct sum and product respectively, right? And it is different because you could have infinitely many non-zero components of End(Si^ni)?

[dull ginkgo]

what are you guys on in noncomm rn

[rocky cloak]

Well, the sum/product is finite in each case. I just think it's weird to talk about a "sum of rings"

[elfin wraith]

Course finished last week, it’s exam time now. Last thing we covered was Berstiens ineq and defined holonomic modules

[rocky cloak]

So I guess that's technically a missing step, proving that the sum is finite.

But it just follows from R being finitely generated (generated by 1)

[elfin wraith]

I’m currently trying very hard to revise the localisation section but it just absolutely does not interest me unfortunately

[dull ginkgo]

bernstien ineq in alg?

[elfin wraith]

The man’s prolific, but yeah it gives a lower bound on the dimension of modules over the Weyl algebra

[glad osprey]

I see, thanks 🙏 and I just learned that semisimple rings are Artinian, so I guess that answers my question

[tough raven]

IG you mean that Artinian rings need not be semisimple?

[tough raven]

\newcommand*{\End}{\operatorname{End}} \newcommand*{\Hom}{\operatorname{Hom}} $\End(M \oplus N)$ is the ring of matrices of the form $\begin{psmallmatrix} A & B \ C & D \end{psmallmatrix}$, where $A \in \End(M)$, $D \in \End(N)$, $B \in \Hom(N, M)$, and $C \in \Hom(M, N)$ (where matrices are multiplied by the usual formula but composing entries instead of multiplying them). This contains $\End(M) \times \End(N)$ as the subring of ``diaogonal'' matrices, but is larger unless $\Hom(M, N) = \Hom(N, M) = 0$ (which is at least true if $M, N$ are non-isomorphic simple modules by Schur's Lemma, but in general is very false).

[cloud walrus]

Raghuram

[elfin wraith]

Maybe? Am I getting something confused?

[tough raven]

Semisimple ⇔ Artinian + (Jacobson radical = 0), I believe.

[elfin wraith]

I know if R is Artinian then semisimple iff J(R)=0 is true

[glad osprey]

I just read this on wikipedia:

[elfin wraith]

Actually yes I’m being very dumb

My reasoning for being dumb is even wrong, but yes no that is kinda trivial to show, maybe time for a study break

[glad osprey]

You can't be dumber than me, I have no idea what I'm doing

[elfin wraith]

Don’t worry I’m sure if you scroll back in this chat a few weeks you’ll find me being similarly confused by much of the same stuff

[tardy hedge]

Localization seemed cool to me

Havent learned it much yet though

Thats when u “add in” fractions or smth right

It gets rid of prime ideals or sooomething

It seems interesting

[south patrol]

Yes

[elfin wraith]

Localisation is cool in the commutative case because it a) just works and b) is useful

In the non commutative case it’s just a headache and not, to the best of my knowledge, anywhere near as useful lol

[dull ginkgo]

For commutative rings it’s chill but for noncomm you want your “fractions” to be of the form rs^{-1} for a given set S right

I.e $(r_1 s_1^{-1})(r_2 s_2^{-1})= r_3 s_3^{-1}$ so we have some sense of closure right, but that needs some form of bastardized normality on the set s

[cloud walrus]

shatgtp

[elfin wraith]

This guy uses right Ore sets, oh brother

[dull ginkgo]

Objectively better stfu

[dense raven]

is higman's embedding theorem also true of semigroups

[glad osprey]

The term "semisimple artinian ring" is weird considering semisimple rings are artinian. I'm guessing, like most terminology in this book, it's non-standard?

[crystal vale]

Determine all ring isomorphism from Z/nZ to itself.

Let 1 maps to a, a in Z/nZ then for ring isomorphism a must be generator and a^2 = a.

Since a is generator so a^-1 exists so we can get a = 1.

There is only 1 mapping, right?

[crystal vale]

I want to prove that identity is the only ring automorphism of R.

Now here I am aware of the Cauchy- functional equation so I need to prove that the mapping is continuous.

How can I show that?

I already saw this one a few months ago on MSE so the idea is order preserving here

But I don't know how to use that ?

[tardy hedge]

Is that all it is?

Is the proof for 6 like:
Assume (i)
Take D an R-module. If we show every ses with D at the start is split then D is injective:

For some ses
0 -> D -> M -> N -> 0

N is projective so it is split

Its the same thing for (ii) -> (i) right

[elfin wraith]

My class also tended to talk about semi simple Artinian rings, but I think more so because we were generally talking about Artinian rings in that chapter

So it’s not entirely non standard

[glad osprey]

I see, thanks

[topaz rivet]

why are there no semi simple algebra in dimension 2? read it in my tutor's comments on my work. tried to look it up but i dont get it

[rocky cloak]

Do you mean semi simple lie algebra perhaps?

[topaz rivet]

oops yes i meant lie algebra

[rocky cloak]

I guess, pick a basis x, y
Then show that the span of [x, y] must be an ideal.

[crystal vale]

To show Q[√ 2] is not ring isomorphic to Q[√ 5].

Idea: there is x in Q[√ 5] such that x^2 = 5 but this is not case in Q[√ 2].

Say f be the ring isomorphic mapping then f(5) = 5, but f(x)^2 = 5, but f(x) in Q[ √2], contradiction.

Correct?

[hidden wind]

i would use Q[sqrt 5] iso Q[x]/(x^2 - 5) and Q[sqrt 2] iso Q[x]/(x^2 - 2), then show a separate little lemma that for any ring A and ideals I,J in A, we have A/J iso A/I iff I = J

and of course (x^2 - 5) is not equal (x^2 - 2)

oop but i'm not actually sure if that lemma is true ehehe

[elfin wraith]

I’m fairly certain it’s not

[hidden wind]

that would be a pretty good reason for why i am failing to find a proof for it, but that is also very likely to be as i'm just bad at math

[elfin wraith]

Yeah no you can have R/I \cong R/J with I neq J in the noncom case for sure

I’m not sure about the case for commutative rings though, then it might be true

[tardy hedge]

I feel like noncommutatice algebra would be really hard to gain intuition for

[elfin wraith]

It’s too late for me to think about it much, but as stated it’s definitely not true

[elfin wraith]

There’s a reason I’ve averaged about 5 hours per homework question!

About 120 hours on hand ins for a class that’s supposed to be 100 hours of work total including teaching reading etc lol

It’s fun though, there’s just kinda more to say and different ideas than go into to commutative rings. I like those too though I’ve did a decent bit of combinatorial comalg for my UG project and it’s been really fun

[hidden wind]

ah yep not even true in the commutative case i think i have a counterexample one sec

yeah take simply A = Z x Z, and I = ((0,1)), J = ((0,1))

[tardy hedge]

Oh yea combinatorial comm alg? Did you learn stanley-reisner theory?

[elfin wraith]

Yeah looked at Stanley Reisner rings a good bit, it’s really cool, definitely something I’d be interested in doing more with

[hidden wind]

we have A/I iso A/J iso Z, but I neq J... here still I iso J but i guess there might be counterexample where the ideals are not even isomorphic

[tardy hedge]

Im gonna be doing a project in that too this year

Ive only learned the basic ideas so far but i havent gone into details

[elfin wraith]

Well I presume you’ve already come across them but Miller and Strumfels Combinatorial Comalg is fantastic as is Herzogs monomial ideals

[tardy hedge]

Hm i might not have actually

I was just looking at intro papers

Not really textbooks yet

[elfin wraith]

Miller and Strumfels starts out quite gently, there isn’t really too much prerq knowledge needed until like section 1.5 where you might need to dip out and learn some homological algebra

[tardy hedge]

I see

[elfin wraith]

But if you’re fine with like resolutions, free and projective modules, gradings and simplical homology you’ve got like everything you need for that book

Herzogs book is considerably less gentle, but is also very nice, a touch terse at times but not too much so

[chilly radish]

This is also true in the commutative case

Take a polynomial ring k[x1,...,xn] with k algebraically closed

The quotient by any maximal ideal is isomorphic to k

[hidden wind]

a counterexample with noniso ideals is A = Z/(2) x Z/(4), I = ((1,0)) + ((0,2)) iso Z/(2) x Z/(2) and J = ((0,1)) iso Z/(4), but A/I iso Z/(2) iso A/J

[hidden wind]

oh

[elfin wraith]

Oh god yeah haha the obvious example is really the hardest to spot

[chilly radish]

A direct product also works

R x R with 0 x R and R x 0

[elfin wraith]

Yeah I think the because the example I had in mind for the noncom case was slightly involved I was over complicating that but I also didn’t really think about it much because it’s bed time

[hidden wind]

i have said so many stupid things lately

[elfin wraith]

I think it’s a perfectly reasonable thing to assume would be true

It just unfortunately quite catastrophically isn’t lol

[tough raven]

It's true as R-modules: I = Ann_R(R/I) (for two-sided ideals of a general ring as well).

But once you go to one-sided ideals, it's false: M_n(k) for k a field has many maximal ideals (one for each codimension-1 subspace of k^n) and all the quotients are isomorphic to k^n as M_n(k)-modules.

[glad osprey]

Isn't this an unnecessary use of Wedderburn-Artin? F is a field, and therefore simple, and FxF is isomorphic to F \oplus F, so it is semisimple basically by definition. Have I misunderstood something?

[dull ginkgo]

this is such overkill, the ONLY ideals of this ring are (0,F) and (F,0) lmao

[tough raven]

It's not exactly the same thing. F is a field and hence simple as an F-module, but you also need it to be simple as an (F ⨯ F)-module. Which it is, but note that the first and second F factors are non-isomorphic simple (F ⨯ F)-modules.

But yes, it is very elementary that F ⨯ F is a semisimple ring.

[fierce wind]

$\Bbb R$-linear ring homomorphism $f:\Bbb C$ $\to M_n(\Bbb R)$

[cloud walrus]

spiritedcaveman

[fierce wind]

can i get any hint about the structure or anything

i mean what would be the constraints for the the matrix entries and how

[tawny dune]

C is isomorphic to the ring of matrices of the form

[ a -b ]
[ b a ]

with a,b in R

[fierce wind]

whats the motivation behind such construction ?

[tawny dune]

Well mutliplication by a complex number z = a + bi

Is better thought of in polar coordinates

say z = re^itheta

Then it corresponds to scaling everything by r and rotating by theta

So if you consider multiplication by a complex number as a linear map on C = R^2, it’s a rotation matrix composed with a scalar matrix

Rotation matrices are exactly of the form

[ cos theta, -sin theta ]
[ sin theta, cos theta]

[fierce wind]

for a minute i was thinking of a rotation matrix mutiolied by a scalar

oh yeah

[tawny dune]

That’s exactly right

[fierce wind]

its same, got u

awight

so close, yet so blinded

[tawny dune]

lol yeah it happens

same thing happened to me on a problem i was working on yesterday its just a matter of practice

[fierce wind]

i mean, jesus christ its never the case that i need a material from another galaxy for most of teh hardest(apparently) problems

its something so vivid and you know it, but for christ's sake it will be in your blind sight

[elfin wraith]

This may be a hard question to answer, if its even well posed, but why do we require a sufficiently large sum for the Hilbert polynomial? Looking through the proof in my notes im struggling to see exactly where we use that s>>0 (upper bound on the sum). I have also seen that there is a more general formula that works for all s using the Euler characteristic, but this seems to discuss a lot about sheafs and the like which I've never really looked at, so whats the difference here?

I appreciate this is maybe a bit of a vague and ill posed question which possibly has an answer I'm not equipped to understand but

[rotund aurora]

What is the definition of Hilbert polynomial?

[rotund aurora]

The >>0 condition comes from the d=0 case (r=0 in the screenshot) only (and it "propagates" via induction)

@elfin wraith

like if you take d=0 and a vector space, if it were a polynomial for all s you would have h_M(s)=0 for all s, which need not be true

This is what Eisenbud says for small values, maybe try reading it. Idk this so can't help

[elfin wraith]

Thanks that’s helpful, I’ll maybe take a look at Eisenbud and see what he has to say!

Ah yeah that actually was helpful, the proof via the syzygy theorem is a bit more enlightening imo (but outside the scope of my course so I see why we didn’t go that route!)

[crystal vale]

let f in Z[x] with odd integer coefficients and even degree. Then f(x) has no rational zeroes.

let p/q be the zero of f, then both must be odd since coefficients are odd.

now we have f(p/q) mod 2 = 0, since q is odd and all coefficent are odd and even degree so it gives 1 = 0mod2, which is not possible.

correct?