groups-rings-fields

[latent anvil]

contradiction

[solemn rain]

wtf

thats so cool

yea thats what im writing

tysm

im done with rings in AM 😄

trash structure anyways

now onto vec spaces over rings lmfaao

i know normal G-module

is virtual different?

[oblique river]

usually "virtual" means it's a formal sum of G-modules

with possibly negative coefficients

so if X and Y are G-modules then "X-Y" is a virtual G-module

[solemn rain]

cool

[latent anvil]

could someone help me understand why v is preserved in the successor ordinal case?

okay so M_a and N_a will always intersect trivially, as M and N do, so you just need to check that they span F_a

clearly they span F_b still

and our big process with the Qs involves making sure the generators of each x_{ij} are in F_a

so I think I get it

and if I understand correctly the process terminates because we strictly increase F_a at each successor ordinal step? so if we didn't hit F eventually we'd get an injection from Ord to 2^F which is a no-no

[solemn rain]

i really want to help you as you help me everytime but i know nothing of what ur saying haha

[latent anvil]

i think i got it

[latent anvil]

okay there's this lemma that if $x \in M$ and $M$ is projective we can find a free direct summand of $M$ containing $x$

[cloud walrus]

shamrock:

[latent anvil]

we first find a free module with $F = M \oplus N$ and write $x = \sum_{i = 1}^n a_i u_i$ in a basis, then write $u_i = y_i + z_i$ for $y_i \in M, z_i \in N$, so that $x = \sum_{i = 1}^n a_i y_i$

[cloud walrus]

shamrock:

[latent anvil]

and because we actually chose this basis to be miniml somehow we can replace $u_1,\ldots,u_n$ in the basis with $y_1,\ldots,y_n$.

[cloud walrus]

shamrock:

[latent anvil]

so clearly this means $\langle y_1,\ldots,y_n\rangle$ is a free submodule of $M$ containing $x$ which is a direct summand of $F$

[cloud walrus]

shamrock:

[latent anvil]

but it then goes says "hence also of M"

which I do not get

so let $P = \langle y_1,\ldots,y_n$ and $Q$ be such that $P \oplus Q = F$. can we just say $M = P \oplus (Q \cap M)$?

[cloud walrus]

shamrock:

[latent anvil]

that feels wrong, I don't see why those two would generate $M$....

[cloud walrus]

shamrock:

[latent anvil]

okay so given any element $t$ of $M$ we can write $t = a_1 y_1 + \ldots + a_n y_n + q$ where $q \in Q$. Then uhhh $q = m + n$ for $m \in M, n \in N$

[cloud walrus]

shamrock:

[latent anvil]

oh and in fact $t = a_1 y_1 + \ldots + a_n y_n + m$

[cloud walrus]

shamrock:

[latent anvil]

but there's no reason that $m \in Q$

[cloud walrus]

shamrock:

[latent anvil]

oh i forgot to say up top but this is all over a local ring in case that matters

okay let $Q' = {m \in M : \exists n \in N, m + n \in Q}$. This is clearly a submodule of $M$ and as I showed above $M = \langle y_1,\ldots,y_n\rangle + Q'$. Is this a direct sum?

[cloud walrus]

shamrock:

[latent anvil]

that is, suppose $z = a_1 y_1 + \ldots + a_n y_n$ is an element of $Q'$. Then there's an $n \in N$ such that $a_1 + \ldots + a_n y_n + n \in Q$. The decomposition $F = P \oplus Q$ and the fact that $P \cap N \subseteq M \cap N = 0$ shows that $n \in Q$, so also $(a_1 y_1 + \ldots a_n y_n + n) - n \in Q$. But this clearly implies $a_1 = \ldots = a_n = 0$

[cloud walrus]

shamrock:

[latent anvil]

blech that should not have been hard, I'm missing something obvious

omfg I'm dumb

start at the top

let $P = \langle y_1,\ldots,y_n\rangle$ and choose $Q$ such that $F = P \oplus Q$

[cloud walrus]

shamrock:

[latent anvil]

I claim $M = P \oplus (Q \cap M)$. It's clear that these two intersect trivially, so we just need to show any element of $M$ can be written as a sum of $P$s and $(Q \cap M)$s. Let $x \in M$ and write $x = p + q$ in $F$ for $p \in P, q \in Q$. Then $q = x - p \in M$, so in fact $q \in Q \cap M$ as desired.

[cloud walrus]

shamrock:

[latent anvil]

the proof that a projective module over a local ring is free is whack

[wintry yacht]

If I have a ring R with it's composition law: $x^2-x \in R$, then is it correct to assume that $x^2 = x$ for any $x \in R$ ?

[cloud walrus]

canfly:

[carmine fossil]

No

[radiant briar]

Can you guys help me if C and G are true or false??

[solemn rain]

what did you try @radiant briar

what did you think about

[wintry yacht]

@carmine fossil how can I prove that $x^2 = x$ ?

[cloud walrus]

canfly:

[carmine fossil]

What is the original question?

[radiant briar]

@solemn rain im so confused what's "trivial subgroups" , does it mean {e}

[carmine fossil]

Yes

[solemn rain]

yes

and the group itself

[wintry yacht]

Let (R, +, *) a ring with &x^2 - x \in Z(R)& for any &x \in R&. Then prove R is abelian.

[solemn rain]

whats Z(R)

center?

[wintry yacht]

yes

[solemn rain]

okay

what did you try

[wintry yacht]

@carmine fossil here is the original question

[carmine fossil]

Ok

[wintry yacht]

@solemn rain well I assumed that &(x+y)^2 - (x+y) \in R& for any &x,y \in R&

I feel like there must be something with this center who will help me prove that it's commutative

[solemn rain]

yea

[wintry yacht]

because the center is already commutative

so if I can prove that R is Z(R) then it's commutative

but seems too easy to be just that

[solemn rain]

first say x^2-x is in Z(R) for some x

play around that

u know that (x^2-x)(y)=y(x^2-x) for any y

in the ring

cna u show that xy=yx

[radiant briar]

@solemn rain so as it's e and itself, should be true?

[solemn rain]

yea thos eare the trivial subgroups

[radiant briar]

for (g) i know 1 2 3 4 6 8 12 24 orders, i have no idea how g^2 is identity

[carmine fossil]

Cauchy's theorem

Order 2 will satisfy that

[radiant briar]

oh

so it's true?

damn how u guys

do those thing

[wintry yacht]

@solemn rain I see, but is that enough to prove commutativity?

[solemn rain]

yeda

[wintry yacht]

ok thanks then

[radiant briar]

how about the (f)?? i think su(n) is normal subgroup of U(4)

[solemn rain]

wahts U(4)

[radiant briar]

i meant U(n(

U(n)

like unitary group, but 4x4 matrix

U(4)

[solemn rain]

okay

[radiant briar]

SU(4) is special unitary 4x4

[solemn rain]

okay

so what do u think

[radiant briar]

i think the element of U(4) belongs to SU(4)

so i guessed like it would be normal subgroup

but im doubting muyself

[wintry yacht]

can someone recommend me some good sources from where I can learn abstract algebra?

[solemn rain]

check pinned in #book-recommendations @wintry yacht

[radiant briar]

so from my poor logic, i guessed it'd be normal subgroup

[solemn rain]

so U(4) is the set of all 4x4 unitary matrices

unitary meaning its inverse is its transpose

now for any unitary matrix

the determinant is = to 1

1 or -1 tbh

[radiant briar]

for det = 1, isn't it for SU

[solemn rain]

yea

[radiant briar]

okay so 1 or -1 for U

[solemn rain]

so obv SU is a subgroup

is it normal

how do u see if a subgroup is normal

[radiant briar]

so it's normal iff ghg^-1

[solemn rain]

for a group G and a subgroup H

H is normal iff h in H --> ghg^-1 for any g

in G

[radiant briar]

yes and h in H

[solemn rain]

so go ahead

check it

[radiant briar]

okay thankss

[next obsidian]

Does anyone have ideas for this problem?
Let $A,B$ be noetherian rings and $f\colon A\to B$ a ring homomorphism. Let $M$ be a finite $B$-module, and $f^\colon\text{Spec }B\to \text{Spec }A$ the corresponding map, show that $f^(\text{Ass}_B(M)) = \text{Ass}_A(M)$

[cloud walrus]

Chmonkey:

[next obsidian]

So first thing to note is that one can show that $f^*(\text{Ass}_B(M))\subseteq \text{Ass}_A(M)$ even without the Noetherian or finiteness hypotheses, you can just manually take a prime in the first set and show that it exists in the latter set.

[cloud walrus]

Chmonkey:

[next obsidian]

For the other direction though, I'm really not sure. If you take a prime ideal of the form ann$_A(x) \in \text{Ass}_A(M)$, and then look at $f(\text{ann}_A(x))$ this certainly is a subset of ann$_B(x)$, but it might not be completely equal, and there's not even a reason it should be prime. Finite Noetherianness means I can find a maximal element of the set of annihilators containing ann$_B(x)$ and that this is prime, but I don't see why it should have to pull back to ann$_A(x)$

[cloud walrus]

Chmonkey:

[vital quail]

heh Ass

[next obsidian]

I could try and do some weird localization stuff... but it would complicate things + I don't know neat stuff about modules (finite or otherwise) over local Noetherian rings (although I know there's a lot of stuff about Noetherian local rings, I just don't know them yet and I don't imagine it's actually necessary).

[toxic zephyr]

what is the exact relationship between R^2 and C? it seems like they should be isomorphic but in what context, if any, is that appropriate to state?

[neat ginkgo]

f : C -> R^2
f(x) = (Re(x), Im(x))

idk if thats a hom tho

[next obsidian]

Both are 2 dimensional vector spaces over R, this leads to them being isomorphic as vector spaces, and gives an isomorphism of their underlying additive group

The multiplication of C is special tho

R^2 is not even a field

[neat ginkgo]

(C, *) is not iso to (R^2, *)

but (C, +) is iso to (R^2, +)

[next obsidian]

there's a lot of differences between the two, pretty much all you can say is you can add things inside of them the same way

this is fine for some purposes in complex analysis or something but the two are way different

[knotty mason]

good question

[timid hull]

Hmmmm

[timid hull]

Like using the inclusion $A/\text{ann}_A(x)\rightarrow B/\text{ann}_B(x)$ construct an associated prime $p\subset B$ for $B/\text{ann}_B(x)$ such that $\text{ann}_A(x)=f^{-1}(p)$ (this is where you will need noetherian hypothesis ...) and then using the inclusion $B/\text{ann}_B(x)\subset M$ conclude that $p\in \text{Ass}_B(M)$ .

[next obsidian]

hmmmm

I'll try and trace out the details

[cloud walrus]

WhyamIsohot?:

[next obsidian]

If we're creating an associated prime for B/ann_B(x) why do we need to assume that M is finite as a B-module?

Or is the claim here that we don't actually need M finite?

[timid hull]

Yes M being finite is rebundant

[next obsidian]

okay, I had a suspicion that it might have been

The other part of this is just to say that (then Ass_A(M) is a finite set)

[timid hull]

Even A neednt be noetherian

[next obsidian]

huh, interesting

[timid hull]

All we need is B to be noetherian

[next obsidian]

Cool, that's nice

I guess it makes sense that if we consider "restriction of scalars" that I'd hope a module with finitely many associated primes stays having finitely many associated primes

[latent anvil]

@next obsidian i think thus is the problem I did at Oxford lol

The one which I wrote down on a whiteboard and said to email me if anyone solved it

But then I solved it a couple days later with no email 😎

[chilly ocean]

@solemn rain yo homeboy where did you learn group stuff from

[next obsidian]

Lmao

Here's my final proof'

Wait wut

Conversely, suppose we have ann$A(x)$ an associated prime, then we have an injective map $A/\text{ann}A(x)\to B/\text{ann}B(x)$, and we see that the map on specs must hit all minimal primes of $A/\text{ann}A(x)$, this is because for any injective map $A\to B$, the associated map on specs has this property. If $\mathfrak{p}$ is a minimal prime of $A$, then $A\mathfrak{p}$ has only $\mathfrak{p}$ as its minimal prime, and since localization is exact (thus $B\mathfrak{p}$ is not zero) this implies it is in the image of Spec$B\mathfrak{p}\to \text{Spec }A\mathfrak{p}$, and thus $\mathfrak{p}$ is in the image of Spec $B\to \text{Spec }A$. We see also that any minimal prime in the image of the map Spec $B\to \text{Spec }A$ is mapped to by a minimal prime of $B$, suppose $\mathfrak{p}\subseteq A$ is a minimal prime, and mapped to by $\mathfrak{q}\subseteq B$. Take a minimal prime $\mathfrak{r}\subseteq \mathfrak{q}$, then the image of $\mathfrak{r}$ is a prime that is a subset of $\mathfrak{p}$, and thus is equal.

[cloud walrus]

Chmonkey:

[next obsidian]

To go back to our specific case, since $0$ is a minimal prime of $A/\text{ann}_A(x)$, there exists a minimal prime $\mathfrak{p}\subseteq B/\text{ann}_B(x)$ mapping to it, and we identify this with a prime of $B$ which is minimal among primes containing ann$_B(x)$. We see that this is in Ass$_B(B/\text{ann}_B(X))$, to do so we show that it is a minimal prime of Supp$(B/\text{ann}_B(x))$, remembering that this is a finite module over a Noetherian ring. Consider that any prime being in the support of $B/\text{ann}_B(x)$ is equivalent to containing ann$B(x)$, as this is the same as saying that ann$B(x)B\mathfrak{p} \neq B\mathfrak{p}$, and this is true if and only if ann$B(x)$ does not contain a unit of $B\mathfrak{p}$, if and only if ann$_B(x) \subseteq \mathfrak{p}$. From here it follows that it is a minimal element of the support, and thus a minimal element of the associated primes, and since $B/\text{ann}_B(x)$ injects into $M$ we see that this prime $\mathfrak{p} \in \text{Ass}_B(M)$. We see further that $f^{-1}(\mathfrak{p}) = \text{ann}_A(x)$, due to the fact that $\mathfrak{p}$ when considered in $B/\text{ann}_B(x)$ lies over $0 \subseteq A/\text{ann}_A(x)$ which lies over ann$_A(x) \subseteq A$

[cloud walrus]

Chmonkey:

[timid hull]

whats the source btw is it from vakil ?

[next obsidian]

This is from Matsumura

[solemn rain]

@chilly ocean dummit foote

[latent anvil]

no it isn't enough to do that

do you know the formal definition of a presentation?

@chilly ocean

quotient of free group?

cool

What group and presentation are you thinking of?

For what group?

Oh sorry I just didn't see the D_n message

until now

okay so

Because those relations are satisfied, you get a homomorphism φ : F(a, b)/N -> D_n where N is the normal closure of {a^n b^-2, abab^-1}

Does that make sense?

well, people like to use presentations without actually defining what that means

and if you're doing that there's no way to prove anything

because you don't have a definition

ask them

I can't help you

oof

Okay fine you can maybe do this

show that any word can be broken up into a small number of words

Using those relations

Like you can assume it's a^i b^j using the second one

yes

Yes

It's good to figure out miniquestions like that/sanity check yourself, good job

So I'm using the second relation here

To say any word of as and bs can be written as a^i b^j

why is that true?

I was actually trying to figure it out lol

actually wait

I don't think I believe your presentation

maybe I do but I'm concerned

How were you able to show b^2 = 1 from those relations?

you can't use the fact that b is a reflection

b is just some symbol

yes

oh

That simplifies things lol

okay so

then I have a proof

you can use the second relation

Get a^i b^j

wlog 0 <= 1 < n and 0 <= j < 2

how many possible products are there?

Right

How big is Dn?

Exactly

So these relations are good enough to distinguish all elements of the group

Not sure exactly how to phrase it

wym

oh

well

You have some big word

Of as and bs

but the second one always let's you turn ba into a^-1 b

so I keep swapping the rightmost b forward

this works even if it's b^-1 btw

and then you do the same for the second rightmost b

and so on

yes

Sorry

it's the rightmost among bs

but maybe our word is ba^2

yup

I'm sort of describing an algorithm here

proving that it preserves equality of words under each step

and then when it's done all the bs are to the right of all the as

I wouldn't phrase it like this in an algebra course ofc

so what I'm really doing

Is we have this surjection F(a, b)/N -> D_n

And I've shown the left group is at most as big as the one on the right

so the map is necessarily a bijection

By pigeonhole

Does that make sense?

We showed each element of the left can be represented as a^i b^j, so there's at most 2n elements

Yup, np

I really hate this kind of thing lol

I've seen it in intro algebra classes before

my prof for grad algebra did something like this but more based

he put a presentation on our midterm after never talking about them before

but he was like "cmon yall can read what they are in the book"

and we'd covered free groups

Yeah mine was like that too

It swapped fields for modules and did Galois theory in the second quarter

But it was the same amount of content as ug algebra

Just a slightly different choice of topics

man now I'm nostalgic for 11 months ago 😢

I was a different man...

Lmao

I'm taking advanced comm alg with the prof in spring

Looking forward to it a lot

Hopefully

idk lol

We covered a fair amount of comm alg last spring in grad algebra

[latent anvil]

Suppose $f : A \to C, g : B \to C$ are surjective ring maps and $A, B$ are noetherian. I want to show $A\times_C B$ is noetherian.

Let $I = \ker f$ and let $\pi : A\times_C B \to B$ be projection onto the second coordinate. Then $\pi$ is surjective since if $b \in B$, we can use surjectivity of $f$ to find $a \in A$ with $f(a) = g(b)$, meaning $(a, b) \in A\times_C B$. Clearly $I \times 0 \leq \ker \pi$, and if $(a, b) \in \ker \pi$ then $b = 0$ and $f(a) = g(b) = g(0) = 0$, so we get the reverse inclusion as well. Note that since $B$ is a noetherian ring and a quotient of $A\times_C B$, it is also a noetherian $A\times_C B$-module. By the short exact sequence $0 \to I \times 0 \to A \times_C B \to B \to 0$, in order to show $A \times_C B$ is noetherian it suffices to show $I \times 0$ is (as an $A\times_C B$-module).

We show that submodules of $I \times 0$ are all of the form $J \times 0$ for $J\leq I$ an ideal of $A$. It's clear they have to be of this form for some subgroup $J$, and if $j \in J$ and $a \in A$, we can find $b \in B$ such that $f(a) = g(b)$ (by the same trick as before) and then necessarily $(aj, 0) = (a, b) (j, 0) \in J \times 0$. This shows $J$ is an ideal of $A$.

Now if we have an ascending chain of $A\times_C B$-submodules of $I\times 0$ it's really just a chain of ideals of $A$, and so has to terminate

[cloud walrus]

shamrock:

[latent anvil]

@next obsidian

I couldn't quite get the idea I had earlier about $(A×_C B)/(I\times J) \cong C$ to work, but I realized that $(A ×_C B)/(I \times 0) \cong B$ and submodules of $I \times 0$ are much more constrained so it works out. I did manage to show $I \times J$ was finitely generated but I couldn't quite get noetherianness lol

Is this what you did?

[latent anvil]

ping me if you reply

[next obsidian]

@latent anvil that is not what I did.

I used the previous problem

Also dang he’s gonna do local cohomology in the Spring?? Dope

[solemn rain]

how do i get intuition for algebra

so i can solve more problems

[carmine fossil]

Solve problems

Yea,it's pretty circular

[solemn rain]

how can ii sovle problems if if im just bad with them

90% i just look them up

[carmine fossil]

Stop doing that

[solemn rain]

and be stuck with them forever?

[carmine fossil]

Look up after struggling for a sufficiently long time

[solemn rain]

whats a suifficiently long time

[carmine fossil]

How long do you try before giving up?

[solemn rain]

idk like

20 mins is extreme

maximum

earlier it woudl take me 3 mins to figre i couldnt do it

but shamrock advised me to take longer

haha is this good enough

[carmine fossil]

Try atleast an hr

[solemn rain]

bro listen

solving math problems

is just tricks really

like either you see it or not

tahts it

its so trash honestly what do i do

99% its just doing this 1 trick and thats it

[carmine fossil]

Can say the same about physics

[solemn rain]

lmfao so you agree

[carmine fossil]

Or literally any field in existence

[solemn rain]

what a trash field

'harder' problems are just doing more tricks and having to know more sutff

thats it

trash

[carmine fossil]

99% its just doing this 1 trick and thats it
@solemn rain Definitely not "1" "trick"

[solemn rain]

doesnt matter really just trash

[carmine fossil]

So, what do you think is not trash?

[solemn rain]

idk

[carmine fossil]

Stop generalising

You won't go anywhere

[solemn rain]

im just saying bro

problems are just getting these tricks

thats it

knowing ohh did you know that bla bla bla

no 'creativity' bullshit

false advertisment

anyways this is #groups-rings-fields

so an algebra problem: how can i show that the set of prime ideals has a minimal element

[next obsidian]

wtf 20 mins?

Dude I spend like upwards of 3 hours on a problem before I try and look anything up

Anyway, Zorn's

[paper flint]

problems are just getting these tricks
Being mathematically precise, what do you call a trick?

Is understanding a certain property, and interrelationships between different properties, or their implications a trick in your opinion?

[next obsidian]

I'm also inclined to believe you believe these to be tricks because you just look them up after struggling for not even half of an hour

if you spent serious time on the problems you'd come up with them yourself and see the natural path of thought which led to discovering that method and it wouldn't feel like a trick

(Some things though are just tricks IMO, but they come few and far between)

[paper flint]

I guess what most people think is a trick is actually ingeniousness of someone's persevered efforts.

[next obsidian]

I'm inclined to believe the quote by person I don't remember which says something is a trick if you only ever use it once

and if you use it twice it isn't a trick

[paper flint]

Just the fact that we're not able to come up on our own in a given time period doesn't mean the proof/solution is a trick/farce.

I'm inclined to believe the quote by person I don't remember which says something is a trick if you only ever use it once
I second that opinion. Although that makes tricks more interesting, since they'll require genuinely creative thinking.

[carmine fossil]

He might just be seeing anything as a trick

[paper flint]

People who don't see the depth in math tend to write it off as a bunch of formulas/tricks(my high school physics teacher who had a PhD is one example for me)

[wintry yacht]

Is the normalizer of a group G actually the set of all centralizers of G?

[knotty mason]

no

N_G(G) = G, C_G(G) = Z(G)

[wintry yacht]

alright, thanks @knotty mason

[old hollow]

How are you allowed to just define a “scheme” like this

[oblique river]

that doesn't look like the definition of a scheme to me

it doesn't even say "scheme" anywhere in there

[sharp sonnet]

it does

[oblique river]

oh sorry

[sharp sonnet]

under the following scheme:

[oblique river]

scheme is a math word

which has a technical meaning

which isn't this

[sharp sonnet]

i mean, this is just a regular old map

[oblique river]

so I was confused

yes

that is a map

are you confused about how the map is defined?

if you're asking about the wording, that's not a scheme in the algebraic geometry sense, it's a scheme in the common english usage sense http://prntscr.com/ubomdb

if you're asking about defining that map, the goal of that exposition is to prove that Z_6 is isomorphic to Z_2 x Z_3. To prove that, we just need to find an isomorphism between them. It doesn't matter how we found that isomorphism, just that we can find one. And the one that they wrote down is an isomorphism

(or if you're asking about something else, maybe someone else can answer cuz I need to run)

[next obsidian]

Is any finite length module the sum of length one submodules?

[latent anvil]

okay chmonkey I have a dumb question, does finite length means there's a finite chain up to your module with equal quotient simple?

and length 1 just means simple?

I think Z/4Z over Z is a counterexample

any way to write it as a sum of submodules has to include Z/4Z, which is not itself simple

@next obsidian

[next obsidian]

blech

wait

can't you write it as like

(0,1) + (1,1) + (1,0)

[latent anvil]

What

[next obsidian]

Ugh

oh wait

wtf am I doiong

sdjklfds;lkfjds

[latent anvil]

Are you thinking of klein Four?

[next obsidian]

Z/4Z

right

right

uhhh

so the only submodule is

{0,2}?

[latent anvil]

yup

[next obsidian]

fuck

ugggggggggggggggggggggggg

And this is length 2

Okay so maaaaaybe

let's think

okay what are the associated primes?

is it just (2)?

[latent anvil]

2 probably?

[next obsidian]

and (0)

aha!

[latent anvil]

I forget what an associated prime is lol

[next obsidian]

annihilator of an element

which is pirme

wait no

[latent anvil]

Ah yeah then 2 and

I don't think 0?

[next obsidian]

0 isn't even one

fuck it really is just 2

God damn it

okay wait a second hold up???

ugh

nope

Z/4Z = Z/4Z

and the prime it belongs to is (2) which is maximal

That wasn't meant for you to understand it at all

that was me thinking out loud via discord

[latent anvil]

Z/4Z = Z/4Z
@next obsidian
yes

I figured

[next obsidian]

Okay so this is the thing right

I can sort of dumb this down

Let M be a finite length module <==> noeth and Artinian

suppose it has only 1 associated prime

show that $\sqrt{\text{Ann}(M)}$ is that associated prime

[cloud walrus]

Chmonkey:

[next obsidian]

You can do this on the length 1 case since the module is cyclic

and via some stuff (like induction and some other theorems) if I could write M as a sum of length 1 submodules then I get it for M as well

[latent anvil]

Well the only way I can see getting length 1 stuff is using SESs

Like

Could you maybe inductively prove this by showing it's preserved by SESs?

[next obsidian]

You see, I wished so

and it's really close but

[latent anvil]

Oof

[next obsidian]

the issue is with an SES

you get Ass(M) a subset of the union of the Ass of the two outer ones

[old hollow]

@open torrent yeah, I understand everything else in the image except the part where they map like (0,0) to 0 and (0,1) to 4

I don’t know exactly how to explain why I don’t get this, but it feels like they’re just renaming elements of the set to make it isomorphic

[next obsidian]

and same for Att(M)

where Att(M) is this thing which for our purposes is just {\sqrt{Ann(M)}} due to the work I already did

I put it on SE

but since this is some whacky thing that's really odd no one's responded

If A were Noetherian I could actually be done

But I don't know of any way to like "Noetherianify" a ring

and module lol

[latent anvil]

hmm

[next obsidian]

So one thing is

the associated ideal of M contains sqrt(Ann(M))

but I don't see how I can show the other one

M is f.g. and if I can show that for each generator that ann(x) is prime I can get it

via the like reverse prime avoidance

I tried to show this by taking a minimal set of generators but I couldn't make it work

[latent anvil]

you can't get such a generating set

Look at Z/4Z again

[next obsidian]

what do you mean?

I meant minimal in size

[latent anvil]

Oh sorry I misread ass(x) as ann(x)

[next obsidian]

oh shit

I did mean ann(x)

my b

[latent anvil]

Oh okay

Then what I was saying was

Any generating set for Z/4Z contains 1

[next obsidian]

right

and that's (4)

[latent anvil]

ann(1) = 4Z isn't prime

[next obsidian]

ugh

[latent anvil]

Yup

[next obsidian]

ummm

maybe I can show

blech

[old hollow]

For me it feels like if someone said:

$(\bZ, +)$ is isomorphic to $(\bR, +)$. These structures are isomorphic under addition, under the following scheme (interval notation):
All elements in $[0,1) \mapsto 0$. \\
All elements in $[1,2) \mapsto 1$. \\
All elements in $[2,3) \mapsto 2$. \\
etc.\\

For example, $3.5 + 4.2 = 7.7$, which translates in the other system as $3 + 4 = 7$.

I know I'm wrong here but I tried to make an analogy, I don't quite understand how you can just map numbers

[cloud walrus]

abs_0:

[next obsidian]

They aren't isomorphic under that tho

[old hollow]

no I know

I'm not saying they are

[next obsidian]

if you have a finite set

you can define a map by listing where every elemetn goes

the example you were looking at only had like 6 things to map i think

[old hollow]

yeah there were 6 elements in each group there

[next obsidian]

You could define it on each element, then verify it's a group homomorphism

then verify it's bijective

[old hollow]

ohhhh I think I get it

so it's not about the numbers being equal, it's more that they have the same number of elements and if you map the numbers like that, then addition is preserved

[next obsidian]

yes

[old hollow]

ok that makes sense lol

[next obsidian]

but not that there's the same number of elements

you need it bijective

[upbeat burrow]

couldn't you just compare the multiplication tables?

[next obsidian]

That's essentially what this is doing

[upbeat burrow]

(for something small enough obviously)

[next obsidian]

this is just renaming the elements

[old hollow]

ok I may be dumb

[next obsidian]

a map between two sets of the same size need not be bijective

it has to be surjective and injective

[old hollow]

but in order for two groups to have a bijection don't they have to have the same cardinality

o

[next obsidian]

take {1,2,3} to {1,2,3} mapping all elements to 1

this isn't bijective

[old hollow]

Ah, each element in $\bZ_2 \times \bZ_3$ was mapped to a unique element in $\bZ_6$

[cloud walrus]

abs_0:

[next obsidian]

yes

[old hollow]

ok, wow

[upbeat burrow]

but it's more than that

[next obsidian]

it's also additive

which makes it a group hom

protip also: for finite sets of the same size bijective <==> surjective <==> injective

proof: pigeonhole

[old hollow]

wait does a <==> b <==> c mean a <==> b and b <==> c and a <==> c

[upbeat burrow]

equivalent

Yeah

[old hollow]

??

oh

[latent anvil]

okay so

(a <==> b) <==> C means something different

And maybe we shouldn't use a <==> b <==> c because of ambiguity

But what chmonkey meant above is equivalence

[cloud walrus]

abs_0:

abs_0:

[next obsidian]

@latent anvil I realized I can quotient out by Ann(M) and consider M an A/Ann(M) module

In this case we want to show that the nilradical is an associated prime

We also know that every endomorphism map given by multiplication by a for all a in A is either surjective or nilpotent

[latent anvil]

hmm

Makes sense I think

I looked at the first part of the SE post

Wait if x -> ax is nilpotent and Ann(M) = 0 isn't a nilpotent in A?

[next obsidian]

well this is only after we have quotiented out

so a is really an equivalence class

[latent anvil]

Sure but you're wloging it right?

[next obsidian]

right

oh right

hmm

yeah

[latent anvil]

So assume Ann(M) = 0

[next obsidian]

you're right

[latent anvil]

then every element is either surjective onto M or nilpotent

[next obsidian]

yup

but I need some x in M such that ann(x) is the set of a which aren't surjective

[latent anvil]

wait so you want to show the nilradical is the annihilator of some element?

[next obsidian]

is what you're pointing out

yah

[latent anvil]

and also is prime

yeah

[next obsidian]

no

it's prime already

by assumption

[latent anvil]

oh

uhh

[next obsidian]

so I don't need that bit

[latent anvil]

why?

[next obsidian]

Uhhh

[latent anvil]

Whatever I'll just assume you assumed it

[next obsidian]

because my module is secondary

aka mult by a

[latent anvil]

Oh yeah sorry

[next obsidian]

is either surj or nilpotent

[latent anvil]

Yup

[next obsidian]

you can show then that sqrt(Ann(M)) is prime

[latent anvil]

So the nilradical is prime

[next obsidian]

yup

[latent anvil]

hmm

Okay so this tells us a lot about A

like

There's only one minimal prime

[next obsidian]

yeah

[latent anvil]

there's a minimum prime even

[next obsidian]

can I relate that to like

associated primes somehow

hrmmm

[latent anvil]

Idk I'm just thinking

[next obsidian]

So I have idea

can I embed

A/nilradical into M?

or something

this would give it to you

[latent anvil]

hmmm

[next obsidian]

I feel like

[latent anvil]

Yeah I see what you mean

[next obsidian]

the natural way you embed the ring is as

like take map A -> M by a maps to ax

then mod out by ann(x)

but I don't have a distinguished x to use here

[latent anvil]

lol you're going in circles

[next obsidian]

yeah

[latent anvil]

Can we assume M is finitely generated? I remember you using that above

[next obsidian]

Yes

It's finite lenght

so noeth and artinina

But we don't have assumptions on A

other than, now, that it's reduced

[latent anvil]

Oh lol I totally forgot it was finite length

[next obsidian]

yeah

[latent anvil]

That's important

[next obsidian]

we also have inductive stuff for all submodules

but I don't see how to relate it back up

Hmm okay so

let me think

take the set of nilpotent a

look at like uhh, the a-torsion of each one

we want this intersection to be non-empty

is the idea

[latent anvil]

a-torsion within M?

[next obsidian]

yeah

That isn't even quite enough

if I take something inside the intersection

ann(x) contains the nilradical

Wait hold up

okay so easy thing duh

Umm

there exists one associated prime

[latent anvil]

Any associated prime contains the nilradical

[next obsidian]

so we know at least that for that x that ann(x) contains the nilradical

yeah

[latent anvil]

Oh you said that

oof

Can there be more than one associated prime?

I don't need a proof

[next obsidian]

nope

[latent anvil]

Cool

[next obsidian]

I am assuming there isn't haha

I showed that

one associated prime <==> secondary

for a finite length module

[latent anvil]

Sick

[next obsidian]

But the proof relies on using the fact that the associated prime is sqrt{Ann(M)}

I used that for doing it via induction

So I need to show it now

[latent anvil]

wait I'm confused

do you know M has exactly one associated prime or not?

wait

Isn't there stuff about minimal primes in Ass(M)?

Like from primary decomp?

Idk I barely remember this

[next obsidian]

I do know that

[latent anvil]

Ah I checked stacks and you need noetherian

[next obsidian]

Wtfffff

[latent anvil]

For what I was thinking of

[next obsidian]

Oh okay

Yeah o know what ur thinking of

Minimal primes such that there’s a submodule such that the quotient is iso to A/p

Minimal primes of uhh supp

And minimal primes of Ass

So I did already prove one associated prime

I want to show it’s the radical of the annihilator

[latent anvil]

So like what I was thinking of was whether Ass(M) has to contain some minimal prime of A

Because if so we're done I think?

[next obsidian]

Yeah

[latent anvil]

but that's probably not true in general

[next obsidian]

Hmmm

[latent anvil]

Yeah it's obviously not, let Ann(M) be big

Bigger than all the minimal primes

[next obsidian]

Right

Idk if that’s true even in Noetherian case

[latent anvil]

It isn't

I gave a counterexample

Well a sketch of a counterexample

[next obsidian]

Oh maybe Z/4Z again

Associated prime is (2)

Oh wait but

[latent anvil]

Yeah yeah good catch

[next obsidian]

When we quotient out

And consider it a uhh, Z/4Z module

[latent anvil]

Yeah I mean what we really care about is minimal among primes containing Ann(M)

[next obsidian]

Then the two coincide

[latent anvil]

For our problem

[next obsidian]

Yeah

In this case it’s true

Bruuhhhh

[latent anvil]

and there's no way to like, zorn

[next obsidian]

I can prove this if umm

A is Noetherian

So for Noetherian

[latent anvil]

based

[next obsidian]

It being coprimary (one associated prime)

Is equivalent to the map give by mult by a either injective or nilpotent

But if you remember

[latent anvil]

Yup

[next obsidian]

Surj endomorphisms are bijection a

When the module is finite or w/e

It’s in Chapter 2 haha

[latent anvil]

Oh yeah yeah I remember this

[next obsidian]

Section*

So yeah haha

Blech

And the example u gave with Z/4Z

Z is Noeth

So I wonder if you need a Noetherian hypothesis for this

[latent anvil]

wait dude

[next obsidian]

But I doubt it because I managed to show both Ass(M) and Att(M) are finite

[latent anvil]

hang on

[next obsidian]

Wait hmm

Actually not really

[latent anvil]

We said we could wlog Ann(M) = 0 right?

[next obsidian]

Only for n = 1

Yeah

[latent anvil]

and M is finitely generated and noetherian

[next obsidian]

Yeah

[latent anvil]

doesn't that mean A is wlog noetherian?

[next obsidian]

WAIT

BASED

UR A G

I think so??

[latent anvil]

Lmaoooo

I'm looking at the thing

Yup

Totally works lmao

wlog Ann(M) = 0 and so A is noetherian

[next obsidian]

Bro BASED

[latent anvil]

I'm a god it's true

[next obsidian]

Wooooot

[latent anvil]

You should double check that this is actually wlog btw

[next obsidian]

So now all I need to do is show it’s a direct sum of secondary submodules belonging to maximal ideals

I think so

[latent anvil]

I don't really understand the assumptions

So

It's definitely still finite length

and I think Ass(M) shouldn't change?

[next obsidian]

All I need to do is show it still has 1 associated prime

Or that it’s secondary

I’m pretty sure I convinced myself it’s secondary

If the map mult by a isn’t surj

Then taking a representative that one isn’t

[latent anvil]

yeah if you take a representative of an element of A/ann(M) it's action on M is the same

[next obsidian]

So it’s nilpotent

Yeh

[latent anvil]

So all the maps are either surj or nilp

[next obsidian]

And it’s length doesn’t change

So induction still goes through and crap

[latent anvil]

Well

[next obsidian]

Pretty sure

[latent anvil]

It's also still noetherian and artinian

oh yeah yeah I see

[next obsidian]

Yeah

[latent anvil]

Sure

Okay cool so wlog M is faithful and thus A is noetherian

[next obsidian]

Massively based

Dang I forgot about that

I said I didn’t know a way to Noetherianify a ring and module

Okay wait this is huge tho

I can consider any like finite?? Noetherian Module

As a module over a Noetherian ring

By doing this trick right?

[latent anvil]

Wait

Isn't A also artinian?

[next obsidian]

Uhh

Is it?

[latent anvil]

I feel like this proof also works for artinian

You get that A is a submodule of an artinian module right?

[next obsidian]

Right

Huh

[latent anvil]

I think products of artinian modules are still artinian...

[next obsidian]

Ummm

I feel like

Okay so

[latent anvil]

hmm

[next obsidian]

If it’s also Noetherian absolutely

Cuz finite length still

Right?

Just go in each component one at a time

[latent anvil]

Oh yeha lol

And it is

[next obsidian]

To get your composition series

[latent anvil]

Okay so A is artinian

[next obsidian]

If not Noetherian I hope so

[latent anvil]

And thus noetherian

And M is faithful

This is pog imo

A has finitely many primes

[next obsidian]

Right but... we already knew it was Noetherian haha

[latent anvil]

their intersection is the nilradical

[next obsidian]

Okay so this is nice tho

[latent anvil]

but the nilradical is prime

And all primes are maximal

so A has only one prime

[next obsidian]

Any Noetherian module

[latent anvil]

right?

I think we're done lol

[next obsidian]

Is a module over a Noetherian ring

[latent anvil]

dude

[next obsidian]

Oh wait

Yeah lmao

With A Artinian this becomes even easier

[latent anvil]

Im a god

[next obsidian]

Sham

[latent anvil]

yes

[next obsidian]

Do you want to write the answer for the MSE post?

Lmao

[latent anvil]

oh hell yeah

[next obsidian]

One of us unironically should

[latent anvil]

I will I already up voted it

[next obsidian]

👍

[latent anvil]

Wait so this was showing what exactly?

lol

[next obsidian]

That Ass(M) = Att(M)

[latent anvil]

Yeah I meant how does it show that

Why is the rhs the radical

[next obsidian]

I’ve already shown that Att(M) = {sqrt(Ann(M))}

This is because M is secondary

[latent anvil]

ahh okay

[next obsidian]

So it’s p-secondaey for p = sqrt(Ann(M))

Att(M) is the set of attached primes

= primes for which a quotient of M is p-secondary

But a quotient of a p-secondary module is p-secondary

[latent anvil]

Okay so I get why M is p secondary

but I don't get why a quotient can't be q secondary for a different q

[next obsidian]

I did the details

you manually just showed it

let's see if I wrote it down

[latent anvil]

Basically why is $\sqrt{Ann(M)} = \sqrt{Ann(M/N)}$ for all $N$?

I get why a quotient is still secondary

[next obsidian]

You showed that Ann(M/N) is a subset of the first one there

Is what I did I believe

so if a in Ann(M/N) then aM subseteq N

so it isn't surjective

so then a is nilpotent

[latent anvil]

hmm yes based

[next obsidian]

(as maps)

[latent anvil]

got it

[next obsidian]

I was looking at the chicken scratch I was writing yesterday haha

it took me a sec

[latent anvil]

and in the case N = M the radical isn't prime

So I guess we exclude it from the definition of Att

[next obsidian]

Yeah

I mean

more like

RIght

It's a non-zero quotient

that we look for

[latent anvil]

All primes are maximal+nilradical prime means there's only one prime ideal, right?

Yeah

[next obsidian]

right

let p be any prime, then nil subset p so p = nil'

[latent anvil]

Uee

[next obsidian]

what does nilradical prime really mean?

if a product is nilpotent, one of them is

wonder what that means for the

G E O M E T R Y

[latent anvil]

how did you know M has an associated prime? Is this always true?

Sorry lol I'm not good with this stuff

[next obsidian]

I'm assuming it

first I showed that for length n M, that being coprimary (having one associated prime) is equivalent to being secondary

[latent anvil]

Ah okay right

[next obsidian]

then using both of those I want to show Ass(M) = Att(M)

Showing secondary => coprimary was hard and suuuuper fucking whack dude

[latent anvil]

Right right the coprimaru thing

[next obsidian]

First I had to show it even had an associated prime

I used the fact that maximal annihilator ideals are prime

and the artinian thing

I looked at like Ax_1

then for an x_2 in there looked at Ax_2

assuming ann(x_1) was properly contained in ann(x_2)

[latent anvil]

wait dude

[next obsidian]

I can eventually get an end to this chain

[latent anvil]

A is noetherian in my case

So it has an associated prime

:^)

[next obsidian]

right

I feel like

quotienting out by the annihilator from the start

might've made it easier

but dude it was so whack

So at the end right

I had an associated prime p

and then sqrt(Ann(M)) = q

[latent anvil]

Do you have the direct sum thing?

[next obsidian]

Nope not yet

[latent anvil]

like the other part of your question

Oh okay

[next obsidian]

But dude it was so dumb

[latent anvil]

I'll answer as is

[next obsidian]

I showed that p in Ass(M)

and Ass(M) is a subset of {p,q}

[latent anvil]

Pls proofread my answer

[next obsidian]

so assuming q isn't in there it's coprimary

[latent anvil]

I don't want to get schooled on mse

also upvote it

[next obsidian]

if I assumed that q is in there

I actually showed that Ass(M) = {q}

lmao

sure

[latent anvil]

Oh lmao

I love non contradictions like that

like you do get a contradiction

But like

[next obsidian]

the contradiction is

[latent anvil]

You also just prove it

[next obsidian]

"we assumed it isn't, it is"

so we assume it is

[latent anvil]

Yeah lmao

[next obsidian]

to avoid contradiction

but you can't get that unless you assume it ins't

lmfao

It's like assume RH, true, assume not RH, true

[latent anvil]

(¬p -> p) -> p

[next obsidian]

but on "idk RH" you can't do it

not enough info

[latent anvil]

LEM is op

Pls nerf

Hmm curious my answer doesn't have any upvotes

yo wtf

[next obsidian]

lmfao what

nerd

what were u doing then umm

I don't even remember

this was post Linear ALgebra

was this homological algebra?

[latent anvil]

I guess??

And rep theory

Oh this was around when I was visiting Thomas

[next obsidian]

Oh right

[latent anvil]

okay so

let $A$ be a UFD

[cloud walrus]

shamrock:

[latent anvil]

Suppose $I$ is an invertible fractional ideal of $A$, and also $I \leq A$

[cloud walrus]

shamrock:

[latent anvil]

by the previous problem $I = (f_1, \ldots, f_n)$ and $I^{-1} = (\alpha_1,\ldots,\alpha_n)$ where $1 = \alpha_1 f_1 + \ldots + \alpha_n f_n$

[cloud walrus]

shamrock:

[latent anvil]

we can write $\alpha_i = \frac{p_i}{q_i}$ for each $i$, for $p_i, q_i$ coprime

[cloud walrus]

shamrock:

[latent anvil]

For each $i, j$, we have $\alpha_i f_j \in A$, so $q_i (\alpha_i f_j) = p_i f_j$. Thus $q_i$ divides $p_i f_j$ in $A$, and since $p_i, q_i$ are coprime we get that $q_i$ divides $f_j$, i.e. $(f_j) \subseteq (q_i)$. Thus $I = (f_1,\ldots, f_n) \leq (q_1) \cap \ldots \cap (q_n)$

[cloud walrus]

shamrock:

[latent anvil]

since we're in a UFD, that intersection is principal, generated by the lcm (write every qi as a product of irreducibles and take the max of the powers of said irreducibles)

so we just need to show the reverse containment, i.e. if $x\in A$ is divisible by each $q_i$ then $x \in I$. If this is the case, then $x = x \cdot 1 = x \cdot \left(\sum_{i=1}^n \alpha_i f_i\right) = \sum_{i=1}^n (x \alpha_i) f_i$. The coefficients $x \alpha_i = p_i (x/q_i)$ here are all in $A$ since $q_i$ divides $x$, and so this equality shows $x\in I$

[cloud walrus]

shamrock:

[latent anvil]

Is this what you did @next obsidian ?

[next obsidian]

So what I did

first thing is I supposed $a_i/b_i$ generate $I$

[cloud walrus]

Chmonkey:

[next obsidian]

Then for any $u/v \in I^{-1}$ it has to be that $u$ is a multiple of the $b_i$ and that $v$ is a common divisor of the $a_i$

[cloud walrus]

Chmonkey:

[next obsidian]

(take $u,v$ coprime)

[cloud walrus]

Chmonkey:

[latent anvil]

wait what

ai/bi?

[next obsidian]

Because else you don't have $u/v\cdot a_i/b_i$

[cloud walrus]

Chmonkey:

[latent anvil]

but it's an ideal of A?

[next obsidian]

Oh right haha I guess all $b_i$ can be $1$ then

[cloud walrus]

Chmonkey:

[next obsidian]

I think $v$ has to divide all $a_i$ though

[cloud walrus]

Chmonkey:

[next obsidian]

so it divides the gcd of them all

yeah so in the end I had $I$ generated by the gcd of all the $a_i$ divided by the lcm of the $b_i$

[cloud walrus]

Chmonkey:

[next obsidian]

if all $b_i = 1$ then the lcm is just $1$

[cloud walrus]

Chmonkey:

[latent anvil]

wait that's weird

[next obsidian]

I forgot $I$ is an ideal of $A$

[cloud walrus]

Chmonkey:

[latent anvil]

I got that it was the lcm of all the v's

for u/v generators of I^-1

Interesting

[next obsidian]

yeah so

basically I did exactly that

but then I showed that

$I^{-1}$ is generated by

[cloud walrus]

Chmonkey:

[next obsidian]

the lcm/gcd

then concluded $I$ is generated by gcd/lcm

[cloud walrus]

Chmonkey:

[next obsidian]

So I guess I did mine kinda backwards unnecessarily

[latent anvil]

also I'm on the last problem of section 3

I think I've almost got it

[next obsidian]

NICE

oh i remember this

[latent anvil]

Basically the point of this is problem is that if M is finitely presented

[next obsidian]

solution is really clean IMO

[latent anvil]

And finite generating set has a finite presentation

then you use the fact that A/I is finitely generated

[next obsidian]

Do you have one direction?

[latent anvil]

oh no I forgot to eat

[next obsidian]

oof

[latent anvil]

yeah the finitely generated => finitely presented over a noetherian ring is trivial

the kernel of A^n -> M is finitely generated since A^n is noetherian

so we can get A^m -> A^n -> M -> 0 by picking a generating set for the kernel

[next obsidian]

right

I proved this as a lemma when proving 3.1

Err

[latent anvil]

Oh lol

[next obsidian]

2.5 a)

[latent anvil]

I gotta stop this and get food/go home

[next obsidian]

where r u??

are you still

watching the bees

[latent anvil]

My apartment was like, testing fire alarms/smoke detectors or something

So I've been at the park since noon or so

haha yes

So my sketch of an argument for the other direction

Suppose finitely generated => finitely presented

Let I be an ideal of A

Then A/I is finitely generated, and we have the SES $0\to I \to A \to A/I \to 0$

[cloud walrus]

shamrock:

[next obsidian]

Right

[latent anvil]

thus there's some exact sequence $A^n \to A^m \to A/I \to 0$

[cloud walrus]

shamrock:

[next obsidian]

Wait so you're trying to show what?

Non-Noetherian implies a finite but not finitely presented right?

[latent anvil]

Since $A^m is projective we get a map $A^m \to A$ making a square commute

yeah but the contrapositive

[next obsidian]

Oh