groups-rings-fields

[tardy hedge]

Ok im close

No, i dont get it, i dont know why cosets need to be fized

Fixed

[rotund aurora]

Let A be a G-module, G a finite group, and for each p fix a Sylow p-subgroup G_p. Is it true that if H^1(G_p, A)=0 for all p then H^1(H, A)=0 for all subgroups H of G?

[tawdry plover]

👏

Thanks @rocky cloak for being so patient with me

[tardy hedge]

Still am unsure why cosets of B/psi(A) need to be fixed under B->B

Wait bruh

(Referring back to the actual example i was looking at) The cosets of B/psi(A) need to be fixed under B->B, or else it would send stuff from B to distinct elements in C when it should be sent to the same element in C

Given the diagram needing to commute

I think i actually understand it now

[tawdry plover]

In Boolean rings with unity is it always true that given an two different element there are different prime ideals containing them

Wait nvm

I meant to ask given x not equal to y is there a prime ideal containing X that does not contain y

[dull ginkgo]

,rotate is there an intuitive reason why the isomorphism has to pass through the transpose?

[dull ginkgo]

Is it due to the convention in which we multiply matrices in the first place?

[tardy hedge]

What book is this, and what are you studying?

[dull ginkgo]

Jacobson I

[chilly radish]

This is an entire page of a book

Can you point out the part you're looking at

In the end?

[chilly radish]

Ok found it near the bottom of the page. It's because this is really an isomorphism to M_n(R)^o
and you can go between the 2 using the transpose.

The deeper reason why this is true is that

End_R(M^n) = M_n(End_R(M))
Where M is any R-module.

This means that the matrices that we are multiplying by are matrices of ENDOMORPHISMS.

Now in the case M=R one gets End_R(R)=R^o
(Check that this is true and understand why), so really you get

End_R(R^n)=M_n(End_R(R))=M_n(R^o)

Finally M_n(R^o)=M_n(R)^o in the obvious way

In the commutative case we have that End_R(R)=R so everything works out nicely, but it turns out that generally we need to introduce the opposite ring

It really all follows from this fact that endomorphisms of R compose the opposite way as numbers multiply,

[dull ginkgo]

Oh I see

So it’s like almost over the dual

(Informally speaking)

[chilly radish]

Sure, in a sense

[prisma umbra]

Is this question have any mistake?

Or B right answer?

[marsh scaffold]

What is B even saying

[prisma umbra]

I have no idea😆

[languid trellis]

I imagine it's "the number of left and right cosets of H in G are equal when H is a normal subgroup of G"

or that's what is trying to be said

[marsh scaffold]

No way they forgot to print that many words ;-;

[languid trellis]

I don't see any other reading tbh

[marsh scaffold]

True

So B it is

[languid trellis]

Maybe there was an attempt to paraphrase A with the normality condition

[nimble folio]

If $I$ is an ideal of a ring $R$, is it true that if $n < m$, then $I^n \subseteq I^m$?

[cloud walrus]

clubsoda14

[hollow imp]

yes

[nimble folio]

Sweet thanks

[hollow imp]

in general, I is always a subset of IJ

[nimble folio]

RIGHT

[marsh scaffold]

Does R have a unit?

[nimble folio]

It might....

What happens if R has a unit?

[marsh scaffold]

Wait that's not required

Nvm

[nimble folio]

Let $I$ be an ideal of $R$. Let $M'$ be the subset of elements of $a$ of $M$ that are annihilated by some power, $I^k$, of the ideal $I$, where the power may depend on $a$. Prove that $M'$ is a submodule of $M$.

[cloud walrus]

clubsoda14

[nimble folio]

pretty sure I have to use this to prove that

[dull ginkgo]

In a way is this purely because of the convention we chose for matrix multiplication being in the opposite direction as function composition?

[marsh scaffold]

Wait sorry to bother you again but in Z/2^5Z can't we consider <2> then <2>^5 is 0 but <2>^4 is not?

[nimble folio]

Uhhh

You're asking the wrong guy lol

[marsh scaffold]

Isn't it a counter to your statement

[hollow imp]

no, because 0 is a subideal of any ideal

[marsh scaffold]

Wait i m definitely messing n and m here

[coral spindle]

Let $m = n + k$ where $k>0$. Then $I^m = I^n I^k$ and $I^n$ is an ideal so what do you get?

LOL KET

[cloud walrus]

Boytjie

[coral spindle]

For a counterexample to your claim take the ideal I = (x) of Z[x] or something idk

[marsh scaffold]

Yeah

[coral spindle]

n = 1, m = 2

[marsh scaffold]

It's not true

[nimble folio]

darn

[coral spindle]

In general you do get the reverse inclusion

[coral spindle]

as this demonstrates

[hollow imp]

oh wait I got the n and m mixed up

💀

[tardy hedge]

Hey dude, what textbook r u working thru?

Dummit?

[nimble folio]

Yep lol

[marsh scaffold]

But this is not always true too?

[tardy hedge]

Nice what section r u on rn?

[nimble folio]

what chapter of modules are you on?

[coral spindle]

High powers of ideals get smaller rather than bigger

[tardy hedge]

Im on 10.5

Last section of ch10

[nimble folio]

im barely on 10.1 🥶

are you doing every problem

[tardy hedge]

Dude i was stuck on this one example for days

Nooo no im not lol im too impatient

[coral spindle]

No this is just incorrect

[marsh scaffold]

I see

[coral spindle]

I is always a superset of IJ

[marsh scaffold]

Yeah

[nimble folio]

The later sections look very scary im not gonna lie

[coral spindle]

Like bro I just gave a counterexample

[nimble folio]

Maybe I should stop doing every problem

[hollow imp]

yeah

shoudl've said superset

[coral spindle]

Remember the mantra for ideals

"To Divide is To Contain"

a | b iff (a) contains (b)

so if you like the numbers get 'bigger' but the ideals get smaller

High powers of ideals are divisible by more ideals, i.e., they are contained by more ideals

They're smaller!

Not always but yknow sometimes

[nimble folio]

Wait so does this mean that if $m<n$ then $I^n \subseteq I^m$

[coral spindle]

Yes that's what we proved up here!!!!!!!!!!!!!

[cloud walrus]

clubsoda14

[nimble folio]

Oh lol i missed this there were a couple of convos going on

[coral spindle]

how dare u not pay perfect undivided attention to me and only me

[nimble folio]

[marsh scaffold]

wouldve totally shit my pants after reading this

[dull ginkgo]

Product of (two sided) ideals is between their intersection and the ideals themselves. Ideals absorb multiplication :3

[coral spindle]

Dw mine are permanently shidded

[dull ginkgo]

Pre-shat pants

[coral spindle]

Is this like a presheaf

[dull ginkgo]

Dear god

Gluing axiom

[mighty kiln]

Presheaf

[coral spindle]

That's when the pants are fully shitted

Imma stop now

[dull ginkgo]

LOL

[dull ginkgo]

The one thing you don’t want coherent

[tardy hedge]

Are all the problems in a section actually doable? I always just assumed that the later problems must be just extremely hard but to be fair i never bother doing them lil

Lol

[barren sierra]

but you should still try some harder problems

[nimble folio]

Ehhh

Some of them can be tricky

I was able to get through almost all of the problems in section 10.1 related to "proving things are submodules", problems 1-13

also this might be "cheating" but i always attempt a problem then check my solution online

and if i am completely stuck i ask here or just look it up online

i usually skip the problems that arent the most difficult but the ones that aren't worth the trouble

like im skipping problems 14 and 15, they might be important but i am over it

[still dew]

Well I had to ask, but can dummit just be used for it's problems and like if I am not reading it's theory but from somewhere else is there a chance I can go through the problems

[nimble folio]

Probably

I've read a lot online that dummit and foote does not make a good reference text because a lot of examples are taught through the problems

[shell pilot]

Find Aut(Z). I have looked online and some people say you have to map generators to generators but in the Gallian Abstract Algebra book, I do not see that anywhere in the Isomorphisms chapter. How do people "know" to just go straight to 1 and -1 for Aut(Z)?

I was trying to use only the methods of proving something is an isomorphism. So I was trying to look for a map that is one-to-one, onto, and preserves the operation.

[coral spindle]

Intuition 👍

There is no general method to prove anything, you have to get creative usually

Looking at the generators is a good first step, and Z is particularly nice because it's a so called 'free' group, meaning the homomorphisms out of Z correspond to any choice in the group

That is, you choose where 1 goes, and you have the rest of the map decided for you.

Remember that nobody 'knows' that one proof method will work without trying. You simply have to try things until something works. There is no algorithm to prove a general fact.

[shell pilot]

Ok, well my first "instinct" was to simply say for Z -> Z you just need n so the map is f(n) = n

Does that not work because it's not a finite group?

[coral spindle]

That's just not a proof, I don't get what you mean

It doesn't work because it's not true

[shell pilot]

It just says to find the Aut(Z)

[coral spindle]

Yes

[shell pilot]

So why can't our automorphism be phi(x) = x?

Every element is mapped to itself

Feed it 0, 1, 2, 3, ... you get 0, 1, 2, 3, ...

[coral spindle]

The identity map is in fact an automorphism but there is more than one automorphism of Z

In fact this is not saying very much at all, because the identity map is an automorphism for any group.

[shell pilot]

I don't really understand this answer I guess

[coral spindle]

Can you elaborate on what part of it you don't understand

[shell pilot]

The whole thing. Automorphisms are so easy but so confusing.

[coral spindle]

OK I can't help if you just say you don't understand the whole thing.

What's the very first claim you don't understand.

There are claims written in that screenshot, so read along and tell me the first one where you think you don't know why it's true.

[shell pilot]

I'm trying to match f(n) = nf(1) to the definition of an automorphism

[coral spindle]

OK

What's the definition of an automorphism? Humor me here

[shell pilot]

An isomorphism from a group to itself.

[coral spindle]

OK great, yes, but more specifically what properties does it satisfy

so it's a bijection such that...

There is one very specific property that I am thinking of, the defining property even

[shell pilot]

The only thing I can think of is the operation preservation phi(ab) = phi(a)phi(b)

[coral spindle]

Yes that is the property

We're working in Z so we should write the operation as +

Now here's something. Let's assume n is positive. Then n = 1 + 1 + ... + 1 (n times) right?

So what's f(n)?

[shell pilot]

nf(1)

[coral spindle]

OK how'd you get there.

Explain

[shell pilot]

f(n) = f(1*n) = nf(1)

[coral spindle]

I don't find that satisfying. You haven't explained this in terms of the property we talked about

You haven't used the things I've pointed out to you

[shell pilot]

f(1*n) = f(1)f(n)?

[coral spindle]

I'm not convinced you've understood what I've been pointing out here

[coral spindle]

No, the group operation of Z is +, not *

so we have f(a + b) = f(a) + f(b). We definitely do not have f(a*b) = f(a) * f(b)

Is this clear? This is very important.

[shell pilot]

Ok, f(1+1+1+...+1) = f(1)+f(1)+...+f(1) = nf(1)

[coral spindle]

Great!

Yes

Well done

So for positive numbers n we have f(n) = nf(1)

[tardy hedge]

Why cant the map B->B not fix cosets of B/psi(A), but shift them over, bijectively? I am thinking its because we ALSO need to fix psi(A) under B->B

True?

[coral spindle]

For negative numbers, it's a fact that f(-n) = -f(n) for homomorphisms f so we're done

Clear, soap?

[coral spindle]

I'd like confirmation you're clear on this too. If you're not then we really need to discuss this.

[tardy hedge]

Since we ALSO need to fix psi(A), our only choice is for B->B to restrict to identity on cosets of B/psi(A)?

[coral spindle]

This is as simple as yes or no, I'm not your professor, my job doesn't rely on your understanding

[shell pilot]

Ok, that helps, so basically we only took care of the positive integers with f(n) so we also needed to consider negative integers

[coral spindle]

Indeed, and I covered it up here ^

So is everything we talked about clear?

Or should I clarify

[shell pilot]

I think I need to just read more about automorphisms. I understand the idea but I can't understand Aut(Z) really

[coral spindle]

So is that a 'not clear' then

OK

Well come back here whenever questions arise

[acoustic igloo]

how to you describe / name a finite group that isn't simple?

[coral spindle]

Well what kinds of groups do you know about?

Can you list a few?

[shell pilot]

Ok @coral spindle , So phi maps Z -> Z. So we know that phi(n) must be n*phi(1) because Z = <1> so phi has to map the value to a multiple of 1. However, it could be 1 or -1 multiple, so we have to consider n=1 or n=-1 and phi(1) = 1 and phi(1) = -1. But what is the actual automorphism? Like, what is the "function" that maps Z to Z? Do we just say phi maps n -> n and phi maps n -> -n?

[coral spindle]

We've just discussed how to describe the whole automorphism! If phi(1) = 1, for example, then phi(n) = n*phi(1) = n*1 = n.

N.b. there is going to typically be more than one automorphism! I fear your use of the word 'the'

[acoustic igloo]

alternating, symmetric, permutation, cyclic

[coral spindle]

Great that's really good

So in particular we could talk about a couple of groups here, but let's just talk about a really elementary class

Cyclic groups

They're Abelian, so every subgroup is normal. Are you familiar with this fact?

[acoustic igloo]

no

[coral spindle]

OK

[acoustic igloo]

but fair enough

[coral spindle]

What is the definition of a simple group that you know?

[acoustic igloo]

no normal subgroups

[coral spindle]

Exactly

Except the trivial ones of course :)

And what's the definition of a normal subgroup?

N is a normal subgroup of a group G if and only if what?

[acoustic igloo]

isn't it something like the left and right cosets are the same?

[coral spindle]

Hm this is true but I was looking for something more precise

Do you know the definition or should I remind you?

[acoustic igloo]

i'm not sure

[coral spindle]

OK it's fine just to say you don't know, it's much faster

A subgroup $N$ of $G$ is normal if, for all $g \in G$, we have $gNg^{-1} = N$

[cloud walrus]

Boytjie

[chilly radish]

If by this you mean matrices multiply column vectors from the left instead of row vectors from the right, kinda?

In this case you'd need to consider R as a right R-module though. It's more about the fact that multiplying by r and then s from the left is not the same as mutliplying by rs from the left, but it is from the right.

(which is the same thing as a left R^o-module, which is the same as R as a left R^o module).

[coral spindle]

But what happens when $G$ is Abelian (i.e., commutative)? $gNg^{-1} = gg^{-1}N = eN = N$

[cloud walrus]

Boytjie

[coral spindle]

So in fact every subgroup is going to be normal in an Abelian group

[acoustic igloo]

makes sense

[coral spindle]

Neat huh? This is very helpful in general and this will help us find a nonsimple group

OK so the cyclic groups are Abelian, so all we need to do is find a cyclic group with a nontrivial subgroup

What do you think at this stage, can you name one?

[acoustic igloo]

non-prime ones?

[coral spindle]

Why so?

Can you explain?

[acoustic igloo]

because they have subroups corresponding to the factors

[coral spindle]

That's exactly right

So there you go! You've just found an infinite family of non-simple groups

And in fact as a corollary, you've also found an infinite family of simple groups, namely those of prime order

Those cyclic groups of prime order* to be clear

(In fact every group of prime order is cyclic, but you may not have the tools to prove that yet)

[acoustic igloo]

thanks

i asked a general question but let me make my question more specific, if you don't mind

[coral spindle]

Go ahead

[acoustic igloo]

how can i name/identify this group?

[coral spindle]

Hm this cayley table is a bit nonstandard since we usually include the identity :)

[acoustic igloo]

right, sorry

[coral spindle]

I can see this group is of order 6, yes?

[acoustic igloo]

yes

[coral spindle]

OK so it's not exactly easy, but there are in fact only two groups of order 6, up to isomorphism!

It's either the cyclic group with 6 elements, or the symmetric group on 3 letters (i.e., with 3! = 6 elements)

And I can see it's not Abelian, so...

This is probably the easiest way but I appreciate it requires some knowledge :)

[acoustic igloo]

so it's the symmetric group

so i guess in general this is a hard problem?

[coral spindle]

Yeah in general it's very hard indeed

It's actually not totally known exactly how hard it is, but it is a candidate for an NP-intermediate problem. A similar question for certain infinite groups is impossible.

[acoustic igloo]

okay thanks a lot

oh, that's not sarcastic

haha

but finding subgroups should not be too hard

you can select a member and compose it with itself

[coral spindle]

Yes you can find a cyclic subgroup very easily

But in general finding all the subgroups is extremely hard

I recently attended a conference where several of the talks were about progress in finding subgroups of certain finite groups :)

[marsh scaffold]

i was thinking interms of probability of finding a unit

And any non unit is in a maximal ideal

But i think that the events of an element not being in m and n are not independent

So it fails i guess

So pls help

[dull ginkgo]

I’ll try thinking of this

I am trying to think of a Legrange-esque way of doing this

As commutative rings are abelian group “acting on itself”

[coral spindle]

A unit is something that is contained in no maximal ideals. Use inclusion-exclusion.

[marsh scaffold]

Yeah that's what lead me to believing that 1-1/R_m represents the probability of it not being in that maximal ideal m

But i guess inclusion exclusion is the way

[coral spindle]

Wait I'm suddenly doubting

Give me a sec

A term of the product is going to be 1/some Rms

so typically...

[south patrol]

Yes, the probability of it being 0 in R/m is 1/N(m)

[coral spindle]

Well yes I suppose that does it

[south patrol]

Now multiply etc

[marsh scaffold]

Yeah but these aren't independence events?

Or can we prove independence?

[south patrol]

Hm

[coral spindle]

The trick would presumably be to use linearity of independence somehow right

[marsh scaffold]

Linearity of independence?

[coral spindle]

linearity of independence? wow

I meant linearity of expectation

[south patrol]

I didn't even notice you had mistyped

Lol

[coral spindle]

Perhaps I need sleep

[marsh scaffold]

Don't we all

Expectation is linear for independent and non independent RVs?

[coral spindle]

Yes

[south patrol]

Yes

[marsh scaffold]

So uh

[coral spindle]

I'll leave this to you cool folks

[south patrol]

It is essentially just integration

[marsh scaffold]

Okay fair but what's our plan with the problem

[coral spindle]

The product expands to $|R|\sum_n \sum_{\mathfrak m_1, \dots, \mathfrak m_n} \frac{(-1)^n}{|R|^n} |\mathfrak m_1|\cdots|\mathfrak m_n|$, is there some way to interpret this latter thing as an expectation?

[cloud walrus]

Boytjie

[coral spindle]

I really need to sleep actually tho

[marsh scaffold]

It's looks more like inclusion exclusion lol

[coral spindle]

Well yes

[marsh scaffold]

I hate combinatorics on God 😭

It's the PIE always

[south patrol]

Definitely can't use this but worth pointing out R is actually iso to the product of the R_m (localisations) and then it reduces to the case R has one maximal ideal where this is obvious

[languid trellis]

I wish evan almighty was real so my prayer to destory combinatorics would finally be fulfilled

[delicate orchid]

details, immediately

[delicate orchid]

oh ok nvm maybe later

[cobalt heath]

Me when groups are so difficult

Conceptually they are on easier side compared to other researches, but

[delicate orchid]

I wouldn't be so sure

[cobalt heath]

I am sorry for trivializing something I don't know

I did mean, it is relatively easy to think of groups. But that might not be true as well

[delicate orchid]

yeah groups themselves are easy because they're really general

but they're hard because they're really general

[still dew]

It's easy to think of sets and axiom of choice too but oh boy the consequences of it

[dull ginkgo]

I found a ridiculous overkill way to do this I think

[dull ginkgo]

Let M_n be the nth maximal ideal, let J be the intersection of all maximal ideals

Then R/J is isomorphic to the direct product/sum of R/M_n. An element in this ring is a unit iff the nth term is a unit for each n. These are all fields, so the only non-units in F_n = R/M_n is 0 + M_n (congruence class). Thus the probability of being a unit in this ring R/J is the product over (1 - 1/|F_n|).

[marsh scaffold]

🤯

[tardy hedge]

Soo … a short exact sequence that “splits”

What is going on here

[dull ginkgo]

The harder part is showing each x + J that is a unit as an equivalence class has a single unit within it

I think there must be a better way to do this besides my silly way

[cobalt heath]

Oh this has a number field feel.

[dull ginkgo]

Jacobson

[tardy hedge]

A -> B -> C is a split short exact sequence, this means B is a direct sum of A and C?

[dull ginkgo]

5 lemma ass mf

[dull ginkgo]

There HAS to be an easier way of doing this

[dull ginkgo]

R/Jac(R)

Oh wait

Misread

You can use that I’m pretty sure lmao

[shell pilot]

Why can we just take the 2 out of the function in this answer?

I thought that phi(2r/2) = phi(2)phi(r/2) but we don't know what phi(2) is

[south patrol]

Isn't this a group homomorphisn

[shell pilot]

Thank you very much!

[south patrol]

So phi(1) needn't be 1

It's just this

[dull ginkgo]

Trying to see if it’s true that x + J is a unit iff x - u is in J for one and only unit u

Idk how to do this lmao

[dull ginkgo]

@chilly radish let R be a ring, be viewed as a left-module over itself. Then is End_R(R) isomorphic to R’s opposite ring due to the order reversal due to composition & linearity

[chilly radish]

Yes

[tardy hedge]

Thanks

[dull ginkgo]

Thanks man

[chilly radish]

If you choose a right module structure you get R, but morally this is because this is the same as considering R^o as a left R-module (or R as a left R^o module)

[dull ginkgo]

Oh i see

So once again is it almost like a consequence of convention

[chilly radish]

I mean, kinda

I think of it as a deeper relationship between a ring and its opposite

You wouldn't expect a priori that left modules would be the same as right modules

So why should the endomorphism ring of R as a left and right module be the same

[dull ginkgo]

Same up to (anti)-iso?

[chilly radish]

Like choosing a side is basically thinking of either left or right ideals

Sure

[dull ginkgo]

that’s what i would expect a prior ithough

Just not iso

antiso

[chilly radish]

For the case of R sure. It's more generally true that picking a side gives you different perspectives on a ring tho

[dull ginkgo]

Anti-homomorphisms seem like weirdly behaved objects

[chilly radish]

Like R-mod and mod-R may not be equivalent!

And why would they be?

Simple left R modules are quotients by left ideals, and similarly for right ideals, and generally a ring can have very different left and right ideal structures

[dull ginkgo]

Probably less so due to opposite rings being like tensors in a way where they “turn” anti-homomorphisms into homomorphisms through them, instead of multi-linear morphisms into linear

[chilly radish]

You can phrase it in terms of a universal property if you really want to

[dull ginkgo]

I’d assume for every antihomomorphism A into B it factors through a homomorphism from A^op to B

[chilly radish]

You can just compose with A^op->A, so it's not factoring through in this case, it's being factored through

[dull ginkgo]

Ah true

Not too well versed in category stuff yet

Thank u though!

[chilly radish]

Np. As a final note i'd say that I think the opposite ring is a more fundamental construction that left vs right modules, so to me viewing this as "choosing a different side switches the roles of R and R^o" is a better perspective

[dull ginkgo]

I see

Jacobson doesn’t touch it that much

If at all

[chilly radish]

But I'm not sure how useful this perspective is to you.

Yea makes sense

I think if anything morita theory would cover this

But I have not gone too far into that yet so idk how much more insight there is to be gleamed

[dull ginkgo]

it explained a lot of my confusion

Right R-modules satisfy (xa)b = x(ab) right

[chilly radish]

Yes

[dull ginkgo]

that explains the opposite ring thing

[chilly radish]

Ye

[dull ginkgo]

R^op into End_Z(M)

[chilly radish]

Yea, and the opposite order fixes the associativity problem

[dull ginkgo]

this should be covered in intro alg books imo

I spent like a good 3 hours confused as hell about this, and that cleared it up a lot

[chilly radish]

I'm glad to hear

[chilly radish]

If you introduce it too early it can be confusing

But if you don't mention it at all that could be confusing too

[dull ginkgo]

Also you mentioned how for a right $R$-module $K$ that $\mathrm{End}_R(K^n) \cong M_n\left[\mathrm{End}_R(K)\right]$?

is that what you said?

[cloud walrus]

Arzela-Açaí Theorem

[chilly radish]

Yes

[dull ginkgo]

that's a ring endo right

[chilly radish]

Yes

[nimble folio]

Let $F=\mathbb{R}$, let $V=\mathbb{R}^2$, and let $T$ be a linear transformation from $V$ to $V$ which is rotation clockwise about the origin by $\frac{\pi}{2}$ radians. Show that $V$ and $0$ are the only $F[x]$-submodules for this $T$.

[cloud walrus]

clubsoda14

[nimble folio]

i dont know how to start this problem

[dull ginkgo]

I'll ask you what is the matrix corresponding to that rotation

[nimble folio]

Im guessing I have the find the subspaces $W$ of $\mathbb{R}^2$ such that $T(\mathbb{R}^2) \subseteq \mathbb{R}^2$

[cloud walrus]

clubsoda14

[nimble folio]

{a,-b;b,a}

[dull ginkgo]

geometrically think about it

[nimble folio]

multiplying by i

[dull ginkgo]

every subspace would have to be generated by either 2 vectors or 1 vector

[nimble folio]

Why?

[dull ginkgo]

R^2 is of dimension 2

[nimble folio]

Uhh

[dull ginkgo]

Better question. If W is a 1-d subspace fixed by T, then what does that say about T?

[nimble folio]

No clue

[dull ginkgo]

It's generated by a vector v, so T(v) = sv for some v

ergo, it has an eigenvector

But what about rotations?

[nimble folio]

I'm a bit lost

[chilly radish]

I'll just summarise:

I showed, or rather told you that M_n(End_R(K))=End_R(K^n) and End_R(R)=R^o for left modules. This gives us an isomorphism M_n(R^o) = End_R(R^n)
In the commutative case this gives you the required isomorphism.

Do note that Jacobson's method is a little different. He proves the isomorphism to M_n(R)^o, which is easier if you do it explicitly like he did, but less "natural" than what you get abstractly. The reason there's an ^o is the same as before though, but now in the commutative we need the extra transpose which sends us to M_n(R^o)=M_n(R).

[dull ginkgo]

gonna try to prove that first isomorphism

[chilly radish]

Good

It's really not difficult, follow your nose

[dull ginkgo]

Btw can K be an arbitary module

[dull ginkgo]

immediately, for \mu in End_R(K^N), i can split it into \mu_n maps from the indexed K_n to K^N.

[chilly radish]

Of course

[chilly radish]

First, notice that submodules of V are in particular vector subspaces.
What is the property that makes a vector subspace into an F[x]-module?

[nimble folio]

when a vector subspace W is T stable

[dull ginkgo]

It's a bit mind bendy

Trying to think it through

[chilly radish]

Follow how you'd do it for any vector space

[dull ginkgo]

I haven't done much linear algebra before doing this textbook lo

Anyway

Take \mu endomorphism of K^N, then we can break up \mu into \mu_n sending K_n to K^N. Can we then break up the \mu_n into \mu_(n,m) sending K_n to K_m?

through linearity

and describe \mu as a double sum over them?

[chilly radish]

Exactly. So your question becomes: what are the T-stable subspaces. Since T is a rotation in the plane, it shouldn't be too hard to figure out that no nontrivial subspaces are T-stable.

Now to show this, consider a F[x]-submodule W. If W is not 0, then W is at least one-dimensional. Now show that W=V by showing that you can get to any vector in V by applying an element of F[x] to a vector in W

If you think about it visually it should be more obvious

[chilly radish]

Rather the definition of a direct sum

But yea pretty much

You compose mu_n with each of the projections

Onto each component

[dull ginkgo]

okay kinda neat how it uses the direct product def for the first part and the direct sum/coproduct for the second part

biproduct go brrr

[chilly radish]

This is exatly what you do when finding the matrix of a transformation. You look at the image of a basis vector e_i then take the j-th component (aka projecting to Fe_j)

[chilly radish]

Yup, the beauty of abelian categories!

The univ property going both ways is SUPER fundamental

[dull ginkgo]

salamander lemma my beloved

actually some black magic shit you can do with complexes i figured out

anyway

[dull ginkgo]

i can't wait to do fun homological shit

[chilly radish]

You're almost ready

[opal osprey]

Could you guys give me explicit examples of finite extensions of $\mathbb{Q}$ of the form $\mathbb{Q}(u,v)$ for which $u+v$ is not a primitive element ?

[cloud walrus]

MisterSystem

[chilly radish]

Feels like that shouldn't be possible, but probably is

[opal osprey]

What I know for sure is that when $\mathbb{Q}(u) \cap \mathbb{Q}(v) = \mathbb{Q}$, then $u+v$ is an a primitive element. Also, it is always possible to pick $c \in \mathbb{Q}$ belonging to some cofinite set for which $u+cv$ is primitive, but apparently $c$ doesn't need to equal $1$.

[cloud walrus]

MisterSystem

[dull ginkgo]

Hey so I did some silly shit

Want to hear about it

[chilly radish]

Yea, those 2 facts are what make me think it's probably possible but really hard

[chilly radish]

I'm going to sleep soon, but you can mention me and i'll probably read it tomorrow

[dull ginkgo]

Really quickly, lets say we have like, as an example, a quadruple-complex, indexed by X_(a,b,c,d) with the morphism-"lanes" called A,B,C,D

Then we can create "diagonal" morphisms by composing them and using the commutativity of the complex

We can characterize them by a subset of {A,B,C,D}

and we know composing them is 0 (complex structure) if these subsets intersect?

i.e some "lane" gets composed twice, forcing it to be a 0 map

i wonder if this idea can be generalized

I say lane as in

if f is the "first lane" morphism of X_(a,b,c,d), it sends X_(a,b,c,d) to X_(a+1,b,c,d)

so like the diagonal corresponding to {a,d} would map X_(a,b,c,d) to X_(a+1,b,c,d+1)

but turns out looking at the images and kernels creates this neat elaborate structure

and some wacky diagram chasing schenanigans

if you quotient the images and kernels along with sums of images and intersections of kernels n shit

[coral prawn]

I just worked out some proofs of my own but I’m not entirely sure how correct they are - are these statements true?
(1) Given a free group G and a subgroup H, if G is generated by generators g_a and H by h_a, and if g(h_a)g^-1 is in H for all g in G, then ghg^-1 is in H for all g in G and h in H
(2) If (g_a)h(g_a)^-1 is in H for all h in H, then ghg^-1 is in H for all g in G and h in H
(3) (both (1) and (2)) If (g_a)(h_b)(g_a)^-1 is in H (for all g_a generating G and h_b generating H), then ghg^-1 is in H for all g in G and h in H
And lastly, is there any efficient way to compute quotient groups knowing just the generators of G and H (assuming H is normal)? It would seem there must be if (1)-(3) are true.

[dull ginkgo]

Like for a triple complex, where we have horizontal H (1,0,0) , vertical V (0,1,0) , and forward F (0,0,1) morphisms for the complex.

So if we have X_(a,b,c) and do a {H,V} diagonal map, we map to X_(a + 1,b + 1,c)
But composing a {H,V} and a {F,H} would be a zero map due to commutivity and exactness

You can also create sorta generalized homology groups from these sets

Like

Say we have a {H,V} map in and out of X_(a,b,c), then we can like, take the kernel of the outgoing {H,V} one over the sum of the images of the H and V ones going in

[chilly radish]

The idea sounds kinda similar to spectral sequences, although not exactly. Seems like this might be useful for something, although I can't think of any uses off the top of my head (and I'm not sure how interesting this is in isolation). Your comment about looking at the images and kernels does make me think of spectral sequences, although the structure is different

[dull ginkgo]

I don’t know what spectral sequences are

[chilly radish]

Taking quotients of images and kernels is what you do yo pass to the next page in spectral sequences

It's kind of really bad to explain and it's even worse if you don't have any examples

[dull ginkgo]

That was just my way to diagram Chase

[chilly radish]

So I wouldn't worry too much about it at this point

[dull ginkgo]

I defined a sorta “substructure” of kernel/image (vice versa) quotients so i could just weave exact sequences through them

[chilly radish]

The way I learned SS proper was 2% weibel exercises 98% CHURNING serre-leray SS computations after an initial example

[chilly radish]

Sounds like (co)homology

[dull ginkgo]

It is

I think

[chilly radish]

Should be

[dull ginkgo]

Well they are the only examples of reasonably big complexes i could make lmao

[chilly radish]

Homology algebraically is the failure of a complex to be exact

[dull ginkgo]

I suppose so

I don’t know much use of like

The nine lemma and shit

[chilly radish]

I used this once

Couldn't tell you where tho

[dull ginkgo]

Just seems like practice for diagram chasing

[chilly radish]

These kinds of technical results sometimes matter more to the people setting the foundations than it is to the people using it

[dull ginkgo]

This whole exact sequence schenanigans was more of out of interest than use

Just was neat how it existed

[chilly radish]

It might be used in some abstract proof of some property in a general abelian category

[dull ginkgo]

Maybe idk

It just requires like

HUGE complexes in the first place

[chilly radish]

It's even cooler when you have good motivation from algebra/geometry/topology

You should revisit once you get around to any one those (for motivation from algebra i'd say something like group extensions is a good jumping off point for motivation)

[dull ginkgo]

I mean like

Complexes of Double complexes are triple complexes I guess

So that can make “big ones” in a sense

[chilly radish]

Sounds awful

[dull ginkgo]

The only homologies i know the bare bones def of are hom and tensor prod’s Ext and Tor derived functors

That’s about it

And not much at that

[dull ginkgo]

Jacobson II covers more category stuff

Might encounter it in AG considering whatever the rhyme in the living fuck this def of multiplicity is by Serre

[chilly radish]

Tor can show up pretty often in commutative algebra

[still axle]

not even gonna try to understand that one

[chilly radish]

This just looks like some variation on the euler characteristic/generalisation

[marsh scaffold]

What you've just showed that probability of finding elements that aren't in any those ideals is that product, and those are precisely the units of R cause for every non unit we can get a maximal ideal containing that

[marsh scaffold]

So you're basically done here

And I believe this same procedure can be done to show that probability of element not being in either M and probability not being in N is independent

So thanks @dull ginkgo

[prisma ibex]

this is a very intuitive definition of intersection multiplicity actually

[prisma umbra]

Can anyone help me

I want to learn and get answer along

[paper flint]

Can you try to describe what the quotient group is going to be?

[prisma umbra]

{0,4,8,12...} Negatives too

It is asking for order 2

2^2=4
2^3=8
2^4=16

[paper flint]

Is this really a group?

Ah okay, you said negatives too

But

Why do you think it's the quotient group?

What would 4Z be?

[tribal moss]

Do you know what "quotient group" means?

[prisma umbra]

{...-8,-4,0,4,8,12,16}

[prisma umbra]

Actually no

Even i trued to look up but I don't understand

My native english is hindi

And all the stuffs are English

[prisma umbra]

Is this 4z?

[tribal moss]

Yes.

[prisma umbra]

Phew

Hmm

So what should we do next?

I know modulo operations

[tribal moss]

Next is to understand "quotient group" (but I'm not sure how much help I can be because all the explanations I can give would be in English).

[paper flint]

I can try to offer help in Hindi

[prisma umbra]

Do you understand hindi sir?@paper flint

Yeah i can understand simple english

[paper flint]

No "sir" please, but I do know Hindi

[prisma umbra]

But standard English word out of mind

[paper flint]

Hahaha, it's my native language

[prisma umbra]

Wowow

Never thought that a senior moderators will be Indian

Bhut bhut khushi hua

[paper flint]

Aapko quotient groups ki definition pata hai?

[prisma umbra]

Definition pata hain lekin mujhe jyada samjh ni aayi

[paper flint]

Integers aur unke subgroups model case hai ye idea describe karne ke liye

Given a group G and a normal subgroup H of G (agar "normal subgroup" ki definition nahi pata, to abhi ke liye ignore kar do)

[prisma umbra]

Left coset=right coset

[paper flint]

For now, just assume that G is abelian

[paper flint]

Careful, this is only true when the group is abelian

So it works for the above case (integers), but is just false in general

[prisma umbra]

Acha

[paper flint]

So the quotient group (as a set) consists of all these cosets

[prisma umbra]

Integers follow commutative

[paper flint]

Can you try to find the cosets of 4Z in Z?

For starters, what is 4Z?

[paper flint]

Okay, you answered that here

[prisma umbra]

.

[paper flint]

Right

[prisma umbra]

Haan

[paper flint]

What is 1+4Z going to be?

Remember, a+H is defined as {a+h: h in H}

[prisma umbra]

{.....-7,-3,1,5,9...}

[paper flint]

Correct

Now describe 2+4Z, 3+4Z, 4+4Z, 5+4Z

[prisma umbra]

2+4z={...-6,-2,2,6,10...}

3+4z={...-5,-1,3,7,11...}

Same like i will do

[paper flint]

4+4Z is important to see

[prisma umbra]

Opps thikk

4+4z={....-4,0,4,8,12...}

[paper flint]

This is indistinguishable from 0+4Z

[prisma umbra]

5+4z={....-3,-1,5,9...}

[paper flint]

And similarly if you just continue with 5+4Z, it ends up looking like 1+4Z again

[prisma umbra]

Ha dono same hi hain

[prisma umbra]

Ha ji

[paper flint]

To humare pas kitne distinct cosets hain yahan par?

[prisma umbra]

Both looks same

[paper flint]

Quotient group as a set would consist of these distinct cosets

[prisma umbra]

Four

1+4z,2+4z,3+4z,0+4z

Should I include 0+4z?

[paper flint]

Yes, and why include 5+4Z?

[prisma umbra]

Ohh it is same 1+4z

[paper flint]

Yep

[prisma umbra]

So four😋

Aapne toh itna simply bata diye😑😑

[paper flint]

Yes, and these are {0,1,2,3} (I'm omitting the coset notation for brevity)

[prisma umbra]

Bhut hi shanadar

[paper flint]

Now you must determine the elements of order 2

[prisma umbra]

Ye kaise karna hain bhai?

Modulo 2?

[paper flint]

Instead of jumping to modular arithmetic (it is basically that, lekin terminology laane se pehle use samajhna zaruri hai), let's just work directly with the definition of "order" of an element

What is the order of 1+4Z?

[prisma umbra]

1?

Ohh hang on

-7,-3,1,5,9 so i will get multiple of 2 when i add 2 times

So order is 2

[paper flint]

[prisma umbra]

Odd numners odd numbers

1+1 will be 0 remainder

Mod2

Ese karna haina?

5+5mod2

[paper flint]

Isliye modular arithmetic use karne se rok raha tha

Aise nahi karna

[prisma umbra]

Fer power karke?

🙈

[paper flint]

Z/4Z as a set {0+4Z, 1+4Z, 2+4Z, 3+4Z} include karta hai

Hum is coset me addition aise define karte hai: (a+4Z)+(b+4Z)=(a+b)+4Z

Quotient group ke liye identity element kya hoga?

[sly frost]

do you know what the order of 1 in Z is btw?

or the order of 4, 10, etc?

[prisma umbra]

Matlb ki (0+4z)+(x+4z)=(0+x)+4z mil jaye

Ye toh (0+4z) hi hoga identity

[sly frost]

do you know what order is in general?

and what is the result of (2+4Z)+(3+4Z) for example?

[prisma umbra]

Yeah

Addition and get element

Idenity

[paper flint]

You should try to answer this

[prisma umbra]

(1+4z)

[sly frost]

great

[prisma umbra]

??

Sorry if i am wrong

[sly frost]

and that's because if you add an integer whose remainder of division by 4 is 2 and another whose remainder is 3 you get remainder 1

[prisma umbra]

(2+3)+4z=(5+4z) and it is same as (1+4z)

[sly frost]

for example 6 + 11 = 17 = 1 mod 4

[sly frost]

yes

[prisma umbra]

Ha ye toh aata hain sir🥰

[prisma umbra]

Bas itna hi aata bas

[sly frost]

cool idk what that means but happy for you

[prisma umbra]

I am lost

What was my focus to find?

Idemity?

[sly frost]

I wanted to know if you understand the operation on cosets

[prisma umbra]

I see

If someone helps and i give simple example

Then i can understand the concept and follow it

[sly frost]

the thing about operation on cosets is that you basically take one representative element from both cosets, compose them and get an element from another coset

[prisma umbra]

What is operation here mod?

1+1+1+1+1+mod?

[sly frost]

no, I mean the additive group of Z

[prisma umbra]

Order of 1 is 1

[sly frost]

no

what's the identity in Z?

[prisma umbra]

0

[sly frost]

that's the normal group of integers

without any modular arithmetic, just normal addition

[sly frost]

why 1?

[prisma umbra]

If we add -1

[sly frost]

you don't understand order then

[prisma umbra]

It is additive inverse

[sly frost]

you have to compose the element WITH ITSELF

[prisma umbra]

order is we apply the Operation. Multiple times

And gets identity

[sly frost]

right, but 1 + (-1) doesn't apply to order

[prisma umbra]

Yes

1+1+1+1??

[sly frost]

in this case you have to add 1 with itself

[prisma umbra]

I do not think it will return to 1

Bigger

[sly frost]

not only that, it will never return to 0

this is what you need

which means that the order of 1 is infinity

[prisma umbra]

Hmm

[sly frost]

∞

because you can add 1 to itself infinite number of times theoretically and you'll never get 0

[prisma umbra]

Yes

So my previous question complete?

[sly frost]

well, I wanted to know if you understand order

but your exercise was to get elements of order 2 in the group Z/4Z

and we've not done that yet I think

[prisma umbra]

Order 2 in the quotient group?

Four we found

[sly frost]

wait, what?

[prisma umbra]

{(0+4z),(1+4z),(2+4z),(3+4z)}

Order 2 in it?

Am I right?

(1+4z) order 4 right?

(1+1+1+1)+4z gives (0+4z)?

Ahha yoo

[sly frost]

yes

[prisma umbra]

And we want order 2

(2+4z)

Order 2

[sly frost]

yes

[prisma umbra]

Nice nice

Things are easy to learn

If people like you guys surrounded ☺️☺️

Why maths is written so difficult?

Lots of symbols and difficult definition 😭

[sly frost]

well, sometimes you need people to explain things to you

[prisma umbra]

(0+4z) order?

0?

[paper flint]

Yeah, these ideas take time to get used to. Even historically, these did not originate in a day or two, but often over the course of decades if not centuries.

[paper flint]

Every element trivially gives you the identity if you operate 0 times

[sly frost]

not quite, it's 1

[prisma umbra]

I was not able to understand modular arithmetic for a monyh

And by help of a guy named a lonely bean he taught me so nicely and now i feel so good when i see modular

[prisma umbra]

Power 1

Once

Added

Even i tried to look at youtube and websites

But due to english

🫣🫣🫣

[sly frost]

0+4Z is itself the identity, so you just take this element once and you have the identity

[prisma umbra]

Order of identity is 1

[sly frost]

yes

[prisma umbra]

Are you indian too?

I meant understand hindi

[sly frost]

no, I'm polish

[prisma umbra]

Why my color changed?

[sly frost]

you received an "active" label for some reason

[prisma umbra]

I am using discord all the day and this server 🥰

[sly frost]

but if you don't understand english well and try to learn from english sources then it will be challenging

[prisma umbra]

Learning many things which i didn't learn at school and university

[sly frost]

you should either find good sources in your native language or understand english better

where did you get that exercise from?

[prisma umbra]

I can communicate in simple English 😑

I am learning english too from discord after my graduation

[sly frost]

that's great, you should

[prisma umbra]

There is a mathematical telegram group

Where indian giys post question

And i am preparing for a teaching exam DSSSB

But i am learning all the stuffs first so no student complain for me🤣

[paper flint]

You should try to unpack definitions, etc., work with simple examples wherever possible

It's difficult but you get a hang of it as time goes by

The English used in math is generally accessible enough and avoids unnecessary jargon (math itself has a lot of it, so authors tend to be kind enough not to make things worse)

[dull ginkgo]

Well, that’s not the end of it

We’ve shown the probability of finding a unit in R/J, not R

[marsh scaffold]

I mean to say what you've actually shown is a unit in R/J CAN'T BE IN ANY OF THE MAXIMAL IDEALS

[dull ginkgo]

I’m trying to think that if x + J is a unit in R/J then x is a unit in R

And because all the equivalence classes are the same size, means the probabilities are equal

The main thing here is that if x is in J, then 1 + rx is a unit for each r

[marsh scaffold]

But you've already done the problem if I am understanding ur argument correctly

A element not in any maximal ideal has to be a unit

[quartz wind]

is there such a thing as an "order homorphism" between ordered sets

[south patrol]

Oh yeah nice

[south patrol]

Yes

If you have two ordered sets then an order preserving map is defined in the obvious way

[quartz wind]

was reading a blog post that gave an explicit map from $\mathbb{R}$ to a chain in $\mathcal{P}(\mathbb{N})$ under set inclusion, does that count? since $\mathcal{P}(\mathbb{N})$ is a poset not a toset

[tribal moss]

More precisely, we usually require x <= y implies f(x) <= f(y), but not the other way around.

[cloud walrus]

esca (@ with reply)

[south patrol]

Well you can just generalise it to posets easily :)

[quartz wind]

alright thanks

[dull ginkgo]

@marsh scaffold here’s my whole proof. Assume we are working over finite commutative ring R

Lemma 1: assume K is a proper ideal of R. If x is an element of K, then 1 + x cannot lie in K.
Proof: 1 + x ( - x) = 1 must lie in K, implying r(1) = r is in K for each r in R. Then K = R, a contradiction

Let J (Jacobson Radical) be the intersection of all maximal ideals of R.

Assume x + J is a unit in R/J. Let y + J be its inverse in R/J.
Then xy - 1 is in J.
Thus by lemma 1: (xy - 1) + 1 = xy is a unit of R.

Let (xy)^-1 = u. Then xyu = x(yu) = 1, thus x^-1 = yu, therefore x is a unit.

Since this holds for each x in the congruence class, that means that if x + J is a unit in R/J, then each element in that congruence class in R (|J|-many) is a unit in R

Now by Chinese remainder theorem, R/J is isomorphic to a direct product/sum over R/m, for maximal ideals m. An element in this sum is a unit if and only if the nth component in R/m_n is a unit. But R/m_n is a field, so it’s a unit iff it’s not 0. Thus the probability of being a unit in R/J is the product over (1 - 1/|R/m_n|)

However as shown, if x + J is a unit, then all |J|-many elements in that congruence class are units in R, and every congruence class in R/J has |J|-many elements. Thus the probability of being a unit in R/J is EQUAL to the probability of being a unit in R. This proves your formula

[marsh scaffold]

Okay this definitely slaps but hear me out

Once you've R/J

Consider projecting an element to each component of the product that it splits into

Now what is the probability that it's non zero in each component

[dull ginkgo]

That’s basically what I am using

In the “component” part

That it’s a unit in the product iff each of the projections gives a unit

Of which the only chance it is a nonunit(as the projections are into fields) is 0

[marsh scaffold]

Yeah it's okay but a unit can also be characterized as that it's not a part of any maximal ideal

[dull ginkgo]

So it’s the probability (1-1/|F_n|)

[marsh scaffold]

And that it's projection is non zero in each component

[dull ginkgo]

Note the projection is on R/J

Not R

You have to lift the probability to R

[marsh scaffold]

Yes I know

You don't

Wait let me write down what I am trying to say

So R/J splits right

Now do u agree that an element is a unit iff it's not in any maximal ideal

@dull ginkgo

[dull ginkgo]

yeah

[marsh scaffold]

Okay so now R/J is same as prod R/m

Now define a map from R to prod R/m

Projecting x onto each component

Now each component is independent in the product right

So the probability that it projects onto sth non zero in each component is just the multiplication that you obtain

So 'this' element is 'not' in any maximal ideal

Cause all its projections are non zero

So it's a unit

[marsh scaffold]

Cause for each component there can be only R_m -1 equivalence classes it maps to

[dull ginkgo]

That’s what i was saying

That’s a proof of the iff i said

[marsh scaffold]

Ah okay but i believe the iff can also be shown using the zorn's lemma for arbitrary commutative unital rings

[dull ginkgo]

I don’t think that’s necessary

[marsh scaffold]

The set of ideals if x is not an unit containing it is non empty

xR is in that set

Bound that by union in a chain

[dull ginkgo]

It should be intuitive. Image of a unit under a homomorphisms is a unit

mu(xy) = mu(x)mu(y) = mu(1) = 1

[marsh scaffold]

I agree

Okay anyways thanks again

But i still believe apart from the lifting that you're doing (which is completely fine)

[marsh scaffold]

You solved the problem here

[dull ginkgo]

Once again

You need to go from R/J to R

[marsh scaffold]

We could actually using the same method that you did but now only on two maximal ideals show that an probability of an element not in them is independent

[dull ginkgo]

This is an overkill way to do it

[dull ginkgo]

Maybe, unsure if it’s independent without doing some trickery (e.g Chinese remainder theorem)

[marsh scaffold]

Yeah as I and J are maximal here so I n J is IJ

So after CRT

[dull ginkgo]

Chinese remainder theorem makes it independent MODULO J which is what i was saying

[marsh scaffold]

R/InJ is R/I x R/j

[dull ginkgo]

So yeah

It’s just not an obvious problem

Unless you know about the Jacobson radical

[marsh scaffold]

Now an element is not in I and not in J iff it's projections are non zero in both components

[dull ginkgo]

Yes that works

[marsh scaffold]

And the probability of the projection being non zero in each component is (1-1/R_I)

[dull ginkgo]

yes

[marsh scaffold]

And rest follows

[dull ginkgo]

Because they’re fields

1 - 1/R_i yes

This problem is not something I’d expect to see in an exercise

The 1 + rx for x in J being a unit is not obvious

Plus i looked it up and i didn’t see a proof using this otherwise actually lmao

Usually it uses prime ideals apparently

[marsh scaffold]

Could you reference it then

[dull ginkgo]

What do you mean

[marsh scaffold]

Nvm

[dull ginkgo]

The Jacobson radical (intersection of maximal ideals) is actually really cool

If you want, you can show for an arbitrary ring:

Show that the set of elements x such that 1 + axb is a unit for any a and b in R is an ideal, and is the intersection of maximal ideals

[marsh scaffold]

Seems non trivial to me at first

I'll let you know if I get stuck with the proof

[dull ginkgo]

It uses Lemma 1 in the proof I did

Namely ideals are additive subgroups, and units cannot lie in any proper ideals

So if i is a unit, K is a proper ideal containing x, then x + i cannot lie in the ideal

[tardy hedge]

mu gives an injection into B right? Or else phi(mu) could not be identity

I also dont understand how the proof is like one line, how does this show that B = A(+)C

I am thinking it has something to do with the kernel being A?

[rocky cloak]

C' is a submodule of B that doesn't intersect A and for which B = A+C'. Hence B is the direct sum A(+)C', and since C' is isomorphic to C, that's it.

[tardy hedge]

C' does not intersect A exactly because phi(mu) is identity right

Since A is kernel

[dull ginkgo]

5 lemma blush

[tardy hedge]

yea i saw in hungerford they used that

[tardy hedge]

Why does B = A+C'?

something to do with A being kernel and some quotient thing maybe?

yea i feel like im close to understanding but need an extra push in the right direction

this section has been tough for me

[rocky cloak]

Consider b in B.

Then b = mu(phi(b)) + b-mu(phi(b))
Note b-mu(phi(b)) is in the kernel of phi, i.e. in A

[tardy hedge]

what element is in C'?

oh lol dumb

thx

Im going to have to think about this now

[tardy hedge]

Oh ok i think im getting this a bit better now

[tardy hedge]

Ok cool this blurb was helpful actually

[dull ginkgo]

I think i got this btw

the linearly independent elements e_{n,m} of matrices (1 in n,m, 0 elsewhere) generate M_n(R) and naturally correspond to endomorphism sending a module element to the nth component in the mth position (0 elsewhere)

They satisfy the same relations

[chilly radish]

This works in the case of K=R (Although the "1" is actually an endomorphism of R, in this case the identity), but otherwise I don't see why e_{n,m} should generate M_n(End(K)).

Take an endomorphism f of K that is not given by multiplication by an element of R (For example take K=R^2 and swap the two entries).

Define a map on K^n by

(a1,..,an)->(f(a1),a2,...,an)
As a matrix, this would have to be diagonal, but
You can't write this as a linear combination of the e_{n,m}, since that would mean f= r e_{1,1}, i.e. f is multiplication by an element of r.

It works over R since every element of End_R(R) is mult by an element of R

[dull ginkgo]

Well.. they generate M_n(End(R)) as an End(R) algebra

[chilly radish]

Sure

In the case of R they also generate it as an R-algebra

I thought you meant that this should generally generate the matrices as an R-algebra which is ofc not true

[dull ginkgo]

No, it’s a weird argument to word out

[chilly radish]

Oh btw

The indices should be backwards

[dull ginkgo]

As in the described e_{n,m} endomorphisms are linearly independent over End(M) and additively generate End(M^N) as a module, and you can show abide the same relations as the matrix ring M_n(End(M))

[dull ginkgo]

Oh shit yeah

I’m trying to type this out on my phone without any paper or anything

Excuse small errors or big ones lol

[chilly radish]

The map sending the n-th copy to the m-th copy is represented by e_{m,n}

[dull ginkgo]

Yeah

[chilly radish]

Yea, I get what you mean.

[dull ginkgo]

But yeah, End(M^n) is an End(M)-algebra, generated linearly by linearly independent e_{n,m} using the relation e_{n,m}e_{m,k} = e_{n,k} ande_{n,m}e_{a,b} = 0 iff m ≠ a

And that gives the isomorphism

That’s what i was getting at

,rotate @chilly radish do you think it would be acceptable to short cut these problems using that idea

[dull ginkgo]

Skip a lot of tedious verification work

And one more thing. If we have a free (projective) object A mapping onto (epi) B, then does every endomorphism of B extend to an endomorphism of A

[chilly radish]

Not sure what kind of shortcut you're thinking of

[chilly radish]

What if B=A

I think you can choose preimages of the generators of B, and then preimages of their images and freely define an endomorphism of A such that the induced map on the quotient is the endomorphism of B

Don't think it's true for projective non free

[chilly radish]

Sorry wait

This is only if a basis of A maps to generators of B

If you can choose the surjection this is true but idt it should be for a general surjection

Even for a free module

[tardy hedge]

HomR(D,L) is an abelian group?

Oh, under addition not function composition

[dull ginkgo]

R module so yes

[tardy hedge]

Im on something that gives a homomorphism between Hom’s, i dont think ive seen that before

[dull ginkgo]

I was a bit confused about a part of a proof, that’s why

Because it was about endomorphisms of a finitely generated module

And it used matrices over the generating set but it confused be because it wasn’t free, just finitely generated

So there should be a matrix extension on the free monoid endomorphism into the finitely generated module

[chilly radish]

Yea but you're looking at "vectors" of elements of a module

Unless you're talking about smth else

[dull ginkgo]

Proving Cayley Hamilton

For finitely generated modules

Uses matrices and determinants

Over the basis but for the matrix to be a thing I’d assume you’d need to extend the endomorphism

Just hurts my brain a bit

[chilly radish]

The matrix is on M^n

Not on

M

[dull ginkgo]

What do you mean

[chilly radish]

Show me the text

[dull ginkgo]

I’d have to find it again

It’s on my computer, I’m using my phone rn

Like here’s my whole confusion esque sorta thing going on

So let’s say

We have a finitely generated module M, rank N

[chilly radish]

Well i'm not sure what you're confused about without seeing the proof for myself. This is the proof I know (R commutative obvs so no need to worry about direction stuff w matrices)

[dull ginkgo]

Just kinda processing matrices and shit here

So

M, easier to phrase here as we have an epimorphic valuation map from R^N to M, sending e_n to x_n

And we have an endomorphism of M, f

We can make a matrix “representing it”

[chilly radish]

Not sure why you need the free module

[dull ginkgo]

Not done typing

[chilly radish]

I'm saying that the standard proof (above) has no reference to free modules

[dull ginkgo]

This is more about matrices and endomorphisms over finitely generated modules

[chilly radish]

Just ping me when you can take a pic of the text, it's hard trying to parse your confusion without knowing what caused it

[dull ginkgo]

Well it’s a small part of the text that is more of an overarching thing

Like how to handle matrices over finitely generated modules

Because we can look at the image of the generators, represent them as scaled sums over x_n and make a matrix out of that

But in general it’s not a unique choice

There’s multiple matrices representing the same thing

So the idea that we are working with determinants of those matrices just feels not well defined

[chilly radish]

So as in the picture I sent you can make a matrix, but it doesn't act on M, rather on M^n. It just represents the endomorphism phi I in End(M^n)

[dull ginkgo]

I’d expect it to be a congruence class of matrices

[chilly radish]

We just need a polynomial

A priori you're right that you might get different "characteristic polynomials"

If you choose different generators

[dull ginkgo]

I mean we have the same generators, but the choice of coefficients is not unique

That’s my point

[chilly radish]

Oh sure

[dull ginkgo]

We get different matrices that “induce the same endomorphism”

[chilly radish]

Sure

[dull ginkgo]

I’d assume it’s a congruence class

And we are taking the determinant of a matrix that doesn’t have a canonical choice

We are just choosing some “representative” matrix

I’d assume it’s modulo the kernel of Hom(R^N, M) into End(M)

Just processing this

How matrices over the finitely generated module work

Actually let me just sorta use an example

Let’s say we have rank-2 module M generated by (x,y)

Then if we have an endomorphism, we can observe for example f(x)

But let’s say f(x) = ax + by = cx + dy, f(y) = ux + vy

Then we have two different choices of matrices

That when acting on M give the same result

[chilly radish]

The matrices don't act on M

There's no reason why a matrix of coefficients for arbitrary generators should give you an endomorphism of M

[dull ginkgo]

Good point nvm

It’s just the jump to a matrix here

But i see it a bit more clearly now

[chilly radish]

So the important thing to note here that x I and A are not the same endomorphism

They only act the same way on the vector of generators (m1,...,mn)

And this will be true for any choice of coefficients

A priori, they only have to agree on this vector

I think this is where your confusion stems from

[dull ginkgo]

So our matrices are acting on M^N as you said right

[chilly radish]

Yea, but also you only need it to be 0 on a specific vector

You don't care what it does to the rest of M^N and because of what you said about non uniqueness of coefficients there's no reason it should

Idt the "characteristic polynomial" is ever unique in this scenario

In part because you can do the same thing for more generators and get a degree N+1 polynomial

[dull ginkgo]

I’d assume it’s a congruence class of polynomials again but still

[chilly radish]

Possibly

[dull ginkgo]

Pondering the matrix part of it

[chilly radish]

It's just using the fact that End(M^N)= M_N(End(M)) and the fact you can define the adjugate over any ring

It is a trick tho

[dull ginkgo]

i like that

I was about to say

If we have an R-module M then we can let the R-matrix ring act on M^n

[chilly radish]

Yea this is basically what's going on

[dull ginkgo]

And that seems like the embedding of M_N(R) into M_N(End(R)) functorially

[chilly radish]

Not sure what you mean by the embedding is functorial.

[dull ginkgo]

M_n(R) for a ring R is an endofunctor i think

[chilly radish]

Yea

Maybe you meant natural

[dull ginkgo]

And R embeds into End(M)

Yeah

[chilly radish]

I see what you mean, but also M also varies

[dull ginkgo]

R embeds into End(M) by left-multiplication so M_n(R) into M_n(End(M)) ~= End(M^n)

[chilly radish]

Ye

[dull ginkgo]

That’s what i meant yeah

Okay so

Does Cramer’s rule generalize to this case

So like

Well i guess it does