groups-rings-fields

[coral spindle]

I suggest you go back and check the proper definition.

[coral shale]

u seem to be wanting to prove uniqueness of identity, but i dont think that falls under the scope of the q. Just proving existence is enough?

[coral spindle]

The identity is necessarily unique!

Rather, the question is just asking you to show that phi(1) is an identity.

[coral shale]

yh

[pastel cliff]

ahhhh okie

technically not necessarily a commutative ring but w/e

a unit should just be an element with an inverse

[coral spindle]

We're not talking about units, we're talking about the identity

We're getting off track here!

I am going to remind you of the definition of the identity.

The identity 1 of a ring R is the (unique) element such that, for all x in R, we have x1 = 1x = x.

Exercise: prove that this is indeed the unique element with this property.

Note that you only assumed that one side works, but the identity requires both sides.

[pastel cliff]

ok fair i wasnt being formal about that

[coral spindle]

This is why I asked you to go back and check the definition

[pastel cliff]

proving that 1 is unique is just simple computation i have it in my notes so i'll take that much for granted

im definitely overcomplicating this

[coral shale]

yh isnt this like half the proof

or almost the whole proof, in terms of ideas

[pastel cliff]

that should hold for r \neq 0 and the 0 case is simple enough to address?

[coral shale]

it holds for r = 0 doesnt it?

e is mult identity iff forall a, ae = ea = a, and you've 'found' an e that works

[coral spindle]

You do not need to distinguish the 0 case.

Start from the definition of the identity, and use the tools you have.

[south patrol]

Okay so I've realised I'm a bit stuck on something which should be kinda easy lol.

Let $F$ be some (infinite) field and $K = F(\alpha,\beta) / F$ some separable algebraic extension; let $\Lambda$ be a formal variable and $\tilde F = F(\Lambda), \tilde K = K(\Lambda)$. This is an algebraic extension so we can consider the minimal polynomial $m(X) \in \tilde F(X)$ of $\alpha + \beta \Lambda$. The coefficients of $m(X)$ are just rational functions in $\Lambda$, so we can take the gcd $A(\Lambda)$ of the denominators of these rational functions. Then $G(\Lambda, X) = A(\Lambda) m(X) \in F[X,\lambda]$ and $H(\Lambda):= G(\Lambda, \alpha + \beta \Lambda) \in K[\Lambda]$ is the zero polynomial.

So we can take $H$'s derivative. But this is where I get confused - how can we calculate this? If I try to use the definition as a product, I end up with nonsense

[cloud walrus]

potato

[south patrol]

(Obviously H is just 0, as its derivative, but expressing it in terms of m and A is key to the proof I'm working on)

[chilly radish]

hmm 1 sec

I don't like this proof btw

this is for separable implies primitive element right? in the case of infinite fields

[south patrol]

Zariski-Samuel takes the partials of G with respect to lambda and X basically and I don't quite get it lol

Yeah

[chilly radish]

there's a much nicer proof

imo

[south patrol]

which one do you like

[chilly radish]

which uses infinite pigeonhole principle

[south patrol]

I mean there's the one i know with like

oh maybe it's that lol

but like

[chilly radish]

you take a+cb for all c in F and then there's two that coincide

[south patrol]

Take some splitting field in which min polys of alpha,beta split, then pick c such that like if a_i and b_j are the conjugates then the a_i + c b_j are all distinct

yeah

[chilly radish]

ok lemme see

what do you end up with

if you try to differentiate

[south patrol]

So like the naive thing to do is use the product and chain rules but then you wind up with like $0 = A'(\Lambda) m(\alpha + \Lambda \beta) + A(\Lambda) \beta m'(\alpha + \Lambda \beta)$ lol

[cloud walrus]

potato

[south patrol]

which is nonsense because the first term on the right is zero but the second term can't be zero by separability and things

I guess here the issue is that we're basically evaluating at yet another formal variable so this sort of thing is messed up

Zariski-Samuel still does the sum of partials type thing and winds up with the same second term as me but a different first term (which they do not write out - all you need is it's a poly in lambda, alpha, beta really)

[chilly radish]

why

you're only differentiating by the first variable aren't you

like wrt say Y (Which you plug lambda into) m is a constant

or uhh

wait

[south patrol]

well it's the deirvative of H though

[chilly radish]

do you wanna differentiate wrt gamma here

[south patrol]

gamma lol

Yes, differentiate H(Lambda) with respect to lambda

to get 0 = stuff

[chilly radish]

oh ok

oops

yea

that's starnge

[south patrol]

ye

i am confusion

like i don't even need this proof, it's just very curious to me what that derivative even is lol

maybe i should just write out explicit A and m and compute

Okay yeah I mean my calculation seems to give what I'd expect again so that's still weird

[chilly radish]

Maybe the proof is wrong?

[south patrol]

Oof

Well I mean

Even if we ignore everything else and just focus on what I've written I'm confused

Lol

I can't see where it breaks

(I did adapt it v slightly from the original but only to optimise it for this case)

[chilly radish]

I'm not sure what the aim of the proof even if

Is

What is differentiation supposed to give us

[lethal dune]

R is a jacobson radical and m being the unique maximal ideal.

How do I claim that ker\pi = m ker\pi?

what I get is $(r_1, \cdots, r_n) \in \ker\pi$ then $r_i\in \mathfrak{m}$

[south patrol]

Basically the idea is that because of the βX term, we wind up with [non-zero poly] = β * [another non-zero poly] and then we can find some point where neither polynomial vanishes to give us the λ we'd get from the other proof

But yes I imagine it's just the same as the pigeonhole proof but using the fact that polys have only finitely many roots rather than something more specific

ultimately just feels like a more tedious version though lol

i guess to me the most interesting thing about it technically is that it you don't need to reference conjugates / any splitting fields, plus it explicitly uses derivatives which makes it immediately obvious where the proof fails if we don't assume separability

¯_(ツ)_/¯

[next obsidian]

Can you show the whole problem

[next obsidian]

Do you know about flatness yet?

Gah

The solution I want to use needs flatness / Tor stud

Oh!

[lethal dune]

well yes

but I don't want to use it

can do it the elementary way?

[next obsidian]

Nice okay you assumed the map F -> M is an iso mod m

Yeah so I mean the intended way is like

Because M is flat,
0 -> ker pi -> F -> M -> 0 is exact even after reducing mod m

Since F/mF -> M/mM is an iso (by choice of picking the map to map to a basis, you end up with

0 -> ker pi/m ker pi -> F/mF -> M/mM where the last map is injective

[lethal dune]

well that's what I did but wasn't justify the exactness

[next obsidian]

Right so

The issue is how to show you have a 0 on the left

This actually follows

Because ker pi is a summand

Namely, there’s a splitting

0 -> ker pi -> F

Right?

A map F -> ker pi such that the composition is identity on ker pi

[lethal dune]

one moment let me check

[next obsidian]

I mean you know F = ker pi (+) M

[lethal dune]

yes

[next obsidian]

So this could be summed up saying

ker pi -> F -> ker pi is identity

But this holds true even after reducing mod m

Aka, ker pi/m ker pi -> F/mF splits

You just take the original splitting then reduce that map mod m

So actually
0 -> ker pi/m ker pi -> F/mF -> M/mM -> 0 is exact

Since the last map is an iso, this says ker pi/m ker pi = 0

[lethal dune]

okay so we can claim it from there

[next obsidian]

In general likr

Split exact sequences remain exact after applying like

Additive functors

Because of this argument

[lethal dune]

so what proof were you talking abt flatness?

flat => prj => free for locals

[next obsidian]

Well M is flat so you can put Tor^1(M,A/m) instead of the 0 on the left

Tor -> ker pi/m ker pi -> F/mF

But since M is flat that Tor is 0

[lethal dune]

oh

[next obsidian]

Alternatively you can show using snake lemma and some diagrams that

[lethal dune]

I see

[next obsidian]

0 -> A -> B -> C -> 0 remains exact after tensoring with anything

If C is flat

It obviously follows by using Tor

But you can prove it without that

It’s just annoying

[lethal dune]

agree

[next obsidian]

This is true btw if the flat module is fg

And generally, flat + finitely presented will be projective even not over a local ring

Also it’s true that M flat over (R,m) with m nilpotent => M free

So this is of interest mainly when R is Artinian

[lethal dune]

and how do you prove that?

any ref?

[next obsidian]

If M is finitely presented then Hom(M,N) commutes with flat base change

Ie tensoring with a flat algebra let’s you put that inside

Upshot: hom commutes with localization if M is finitelynpresented

The proof is an easy use of 5-lemma

[lethal dune]

I'll try

[next obsidian]

Now you show for finitely presented M, being projective is equivalent to being locally free ie localizations at all primes (or maximal) ideals are free

[next obsidian]

It actually isn’t that obvious

But it’s easy

[lethal dune]

okay any ref then?

[next obsidian]

The idea is literally, there’s a natural map Hom(M,N)_p -> Hom(M_p,N_p)

As in its a natural transformation

It’s obviously an iso when M is finite free

Because then both sides are just copies of N_p

So you pick a presentation

F_2 -> F_1 -> M

Look at

[lethal dune]

not very familiar with AG terms

[next obsidian]

Hom(F_2,N)_p -> Hom(F_1,N)_p -> Hom(M,N)_p -> 0 -> 0

And you have maps going down making a commutative diagram to the same stuff but the _p is inside the Hom

The four outside maps are iso, the ones with 0 because uh… 0

The ones with the finite free modules because it’s obvious

So the middle map involving M is an iso

Anyway, if M is finitely presented and projective then clearly M_p is free

Conversely, if you know M_p is free for all p, consider a surjevtion
N -> N’, we want to show Hom(M,N) -> Hom(M,N’) is surjective

Write C for the cokernel of this, then look at
Hom(M,N) -> Hom(M,N’) -> C -> 0

Localize at arbitrary primes p, then because M is finitely presented we can write this as

Hom(M_p,N_p) -> Hom(M_p,N’_p) -> C_p -> 0

M_p is frre, so projective, so that first map is surjective ie C_p = 0

This holds for all p, so C = 0

I.e. Hom(M,N) -> Hom(M,N’) is surjevtivr

Ie M is projective

Now the final part is that finitely presented plus flat is projective, but this follows because we know finite flat over a local ring is frre, so a finitely presented flat module is locally free

Thus it’s projective

This is at the end of section 7 in commutative ring theory by Matsumura

[wooden ember]

Solved a neat problem with some friends today but wondered if there was an easier way to do it

We wanted to show that a finite Boolean ring is isomorphic to the ring of functions from some finite set X to F_2

This is easy as vector spaces but the question became finding a basis x_1,…,x_n with x_ix_j=0

And for that we considered the annihilator of an element a_1

And showed we could make it grow till it was of size 2^(n-1) by changing a_1

We could then take the final a_1 as our first basis vector

Then repeat by starting with the annihilator of a_1 + 1

And in general we produce a_k by making the annihilator of (a_1+1)…(a_(k-1)+1) grow

Then we managed to show that this forms a basis as a vector space, and that a_1 + … + a_n = 1 (this required the Chinese remainder theorem which I feel might be overkill I’m not sure), and of course a_ia_j = 0 so then we had a simple isomorphism and we were done

But it took us quite a while to come up with this so I was wondering if there was a simpler argument

Say for example quotienting by some principal ideal and proceeding by induction?

[chilly radish]

The proof I know doesn't reference conjugates but does reference galois closure, and the use of separability is still clear. Ig it's the same thing just implicit, it just uses the fact that a finite separable extension has finitely many intermediate fields, because the galois closure does by the fundamental theorem

[tribal moss]

Since we know what the result is, perhaps start by picking the elements that generate an ideal of size 2, and try to show they must be a basis?

[wooden ember]

Im not sure what you mean here

[south patrol]

yeah sure

I think the finitely many intermediate fields thing is probs the nicest to me

[tribal moss]

I'm saying the basis you're so laboriously constructing can be defined in one go as
B = { x in R | x != 0, and {0,x} is an ideal }

It's easy to see that all those elements must annihilate each other, and I think it ought to be possible to show that every r in R is a sum of finitely many of them, by induction on the size of |<r>|.

[wooden ember]

Ah i see

[next obsidian]

Hehe

Idk if it’s “simpler but”

There’s a lot of math here, but a lot of it is not that hard

Actually, the last part is not needed

The point is, a Boolean ring which isn’t 0 or Z/2Z splits apart as A x B for nontrivial rings A,B

Those are also again Boolean

[lavish gull]

those images are too small to read

[next obsidian]

If R is finite, then inductively you can keep breaking apart R into a bunch of Boolean rings until they’re all Z/2Z

[lavish gull]

so that F_4 = F_2 x F_2 and so on?

[next obsidian]

Yes

Well no

That’s not true

F_4 isn’t a Boolean ring

[lavish gull]

i meant a finite field. how are you defining Boolean ring?

[next obsidian]

Where x^2 = x for all x

[lavish gull]

again, images are too small to read

[next obsidian]

Lol

[lavish gull]

what's funny?

[next obsidian]

That you assumed Boolean ring meant finite field

Also I’m not even responding to you, I was responding to Little Narwhal

[lavish gull]

ok, that is a bit immature

[next obsidian]

Lol

[coral spindle]

Very strange interaction lmao

[next obsidian]

Anyway @wooden ember all you need are Lemmas 3 and 5 of my first image

This gives a kind of conceptual idea about why this should be true

[coral spindle]

Can you remind me of the definition of a boolean ring, I'm curious?

[chilly ocean]

A Boolean ring is a ring coming from the operations of symmetric difference and intersection on a family of sets, sort of

[lavish gull]

can it be defined in terms of extensions of Z/2Z?

i.e. using abstract algebra

[next obsidian]

You just need to note that the algebra of functions X -> F_2 is isomorphic to F_2^|X|

[chilly ocean]

it can be defined as subdirect product of copies of Z/2Z iirc, if that's what you mean

but it's just a commutative ring for which x^2 = x or something

[next obsidian]

.

[lavish gull]

how does it differ from F_(2^n)[x]

[next obsidian]

In like every way

[coral spindle]

Thanks chmonk

[chilly ocean]

am I even correct here. I'm not sure

[mighty spade]

fun fact, Spec (boolean ring) is hausdorff!

[lavish gull]

well what is it at all?

[next obsidian]

For finite ones it’s a direct product of Z/2Z

[coral spindle]

Looks like a fun problem, I think I'll give it a shot without looking at the answers here

[next obsidian]

I imagine you can probably do a transfinite induction to prove it for all cardinalities

[chilly ocean]

I've said that because that's what I know but it holds for Boolean algebras

[next obsidian]

Also wtf is a subdirect product

[chilly ocean]

a subspace of a direct product with the projections being

good

surjective iirc

[next obsidian]

But

[lavish gull]

that's the additive structure of F_(2^n). there must be other differences?

[next obsidian]

Dude that’s a ring

[coral spindle]

Uh yes the multiplicative structure

[next obsidian]

F_2^n is not a product of Z/2Z as a ring

[wooden ember]

Lmao chmonkey what is this

[next obsidian]

Homework from sophomore year

[coral spindle]

Yeah, feels like a job for Zorn

[wooden ember]

Localization

[next obsidian]

[wooden ember]

Ahhh okay

[next obsidian]

You can skip that!

You only need lemma 3 and 5

[lavish gull]

you certainly know what it is not

[wooden ember]

I thought you proved this for me with localization I was confused

[next obsidian]

Dude you’re so annoying

Get blocked

Nah

[wooden ember]

Ahhhh i learned lemma 1 before too

I totally forgot about it

[next obsidian]

That was later, I showed a local Boolean ring is Z/2Z

[wooden ember]

I see that’s very handy

[lavish gull]

do you study Boolean functions in Boolean rings?

[wooden ember]

Yeah alright I see how it works

[next obsidian]

The problem was to make a Noetherian ring such that each localization at a prime is Noetherian, but the ring isn’t Noetherian

I used an infinite product of F_2

Cuz that’s Boolean

[wooden ember]

Nifty

[next obsidian]

And so the localizations are Boolean and local

So F_2

So Noetherian

But an infinite product of F_2 is not Noetherian

[wooden ember]

Smort

[next obsidian]

Anyway does the proof make sense?

You just CHOP CHOP

[wooden ember]

Yeah definitely

[next obsidian]

Swag

[wooden ember]

Pretty easy following from lemma 1

[next obsidian]

Yeh

[wooden ember]

I just kinda forgot that was a thing

[next obsidian]

That’s fair

Also I think Shamrock came up with most of this proof

[wooden ember]

Im pretty happy with our solution though it took us a really fun collab to come up with it

[next obsidian]

But we worked on it together

It took us a while to solve this problem about the Noeth ring thing

And he big brained this solution

Lmao

[wooden ember]

🧠

[chilly ocean]

https://en.wikipedia.org/wiki/Subdirect_product
this is what you do in universal algebra because using subdirects products you can generate whole equational classes
iirc you take something that's subdirectly irreducible
for definition of subdirect products in ring theory there is Lambek, and for universal algebra stuff there is Sankappanavar

[next obsidian]

Okay so Blitz I’m not reading that cuz idc about universal algebra but

If you’re talking about some like subring of a direct product or F_2

The only such things are direct products of F_2

I think

Idk I will actually read this now

[chilly ocean]

direct sum of infinite copies of Z/2Z?

[next obsidian]

Cuz I doubt if that’s true

Not a direct sum

Infinite direct sums aren’t rings

You need to allow infinitely many non-zero things

To get 1 in the ring

[chilly ocean]

hmm... true

[wooden ember]

Wait that’s true

Fuck

[next obsidian]

Yes

[wooden ember]

I used that as a counterexample to smth recently

[next obsidian]

But you just use direct products

So it’s okay

[lavish gull]

are you still talking about Z/2Z

[next obsidian]

Okay maybe it isn’t okay

If you used one

[wooden ember]

Thinking in terms of modules has messed up my thinking of rings lately

[next obsidian]

Yeh

I did the same before

Someone had to remind me

And I was like ah shoot

[wooden ember]

Wait why is it not a ring though

Am I being dumb

[next obsidian]

So

You need to have a 1

[wooden ember]

But like if f and g have finite support so does fg right

[next obsidian]

It necessarily is (1,1,1,…)

[wooden ember]

Ahhh right

[lavish gull]

where do you get x^2 = x from

[next obsidian]

But this isn’t in the direct sum

[tribal moss]

It's not true for all cardinalities. There's a Boolean algebra consisting of the finite and cofinite subsets of N, but it is countable so it cannot be a product of F_2's.

[next obsidian]

Wow!

Is it Boolean in the sense that x^2 = x?

If it’s given by symmetric difference and intersection then it will

[tribal moss]

Yes.

[next obsidian]

:O

[lavish gull]

i mean where does x^2 = x appear in the construction?

[tribal moss]

(Boolean algebras and Booleans rings being equivalent concepts).

[next obsidian]

That’s wild

[wooden ember]

So the product is intersection?

[next obsidian]

Yup

You should prove this gives a ring

It’s not fun to prove distributicity

I just drew a picture and drew the stuff and went

“Okay yup”

[wooden ember]

Had to TA people trying to prove symmetric difference is associative so I’m past that lmao

[next obsidian]

Ooofffff

[lavish gull]

I showed that Boolean OR was given by x^3 in F_4[x]

[wooden ember]

Damn yeah that’s pretty cool

[tribal moss]

Ah, the right generalization to higher cardinalities might be that a Boolean ring is a direct sum of F_2's, rather than a product.
But I'm not completely sure.

[lavish gull]

f(x)=x^3 has f(0)=0 and f(a)=1 everywhere else

[wooden ember]

But what we were just saying is that this is not a ring in general

But yeah the additive group has to be that

[lavish gull]

f(x)=x^(2^n-1) has the same property in F_(2^n)

[wooden ember]

Since it’s a vector space

[lavish gull]

definitions...

[next obsidian]

Right but it can’t be a direct sum of F_2 as a ring because this isn’t a ring

[chilly ocean]

but the argument is that it's not unital

[tribal moss]

Hmm, right, there's no identity.

[wooden ember]

Wait but then does the algebra from above work

Or is N considered cofinite

[tribal moss]

N is cofinite because its complement is {}, which is manifestly finite :-)

[wooden ember]

Yeah okay just in case you wouldn’t consider empty as finite

[lavish gull]

direct sum of F_2 where the operation is component-wise addition?

you can just use F_(2^n) additive group for that and be pleased to be in a field

[coral spindle]

Why are you still on about this finite field stuff, we're past that

Yes, clearly the additive group of F_(2^n) is the same. This is abundantly clear

[lavish gull]

i guess i missed your messages? i don't think i understand how this channel works then

it's not clear. you still need to show it's distributive, associative, .etc. it follow from the properties of F_2, but it's not given

[tribal moss]

I withdrew my suggestion about direct sums, after it was pointed out it wouldn't have a 1.

[chilly ocean]

It's interesting to me that there is a countable Boolean algebra, but maybe I shouldn't be surprised by the fact

[lavish gull]

the identity of F_(2^n)^+ is 0

[lavish gull]

why do you want a multiplicative identity now?

[wooden ember]

Because we’re talking about rings

[tribal moss]

Because that's required by the definition of "ring" we're using.

[lavish gull]

how are you constructing your ring?

[tribal moss]

It's being given to us.

[lavish gull]

please tell

[tribal moss]

The question doesn't make sense. We're talking about what can be proved about an arbitrary Boolean ring.

[lavish gull]

how are you defining that?

[tribal moss]

So even if we didn't require rings to have 1, it would still shoot down my proposal that some infinite Boolean rings do have 1, but my construction would not make those.

[lavish gull]

are you constructing the ring from polynomials over some base structure?

F_2[x] is a ring and as far as i know Boolean logic is compatible with the operations of F_2. but i'm guessing this isn't what you are talking about

[tribal moss]

As I said, I'm not constructing any rings. I'm discussing what can be said about an arbitrary Boolean ring.

[lavish gull]

how can you say anything about them without some concrete examples of what rings looks like

[wooden ember]

You can say things about rings without discussing examples

But we have plenty of examples of Boolean rings

[lavish gull]

and, generally, what do they look like?

[wooden ember]

Namely all direct products of F_2

That’s not all of them though as tropo pointed out

[chilly ocean]

like lattices of sets if they were made into a ring

[tribal moss]

So I suppose I did constuct one ring. :-)

[lavish gull]

ok, so you don't know yourself?

[wooden ember]

I literally just told you an infinity of examples

[lavish gull]

i like Boolean algebra and logic, but we don't appear to be able to communicate. something has failed somewhere.

[coral shale]

what defn of bool alg are we doing here

[lavish gull]

ok thanks. i don't know about lattices

[coral shale]

i dont think its the compsci one

[wooden ember]

To be clear

Here we are talking about Boolean rings

ie a ring such that x^2 = x for all x in the ring

[coral shale]

ah ic. does this correspond in anyway to the computer boolean alg or just same name

[wooden ember]

Or Boolean algebras if you prefer since those are the same thing

[lavish gull]

well you get those from polynomials over F_2, i.e. F_2[x]

[chilly ocean]

Boolean algebra is a certain lattice, it's complemented etc.

[lavish gull]

is it because you want multiple variables? F_2[x,y,z,...]

[wooden ember]

I should’ve said equivalent

[lavish gull]

to enable expressions like: f(x,y) = x*y + x + y

[tribal moss]

A boolean algebra can be defined equivalently as

1. a set with operations AND, OR, NOT that satisfy certain axioms,
2. a partially ordered set that satisfies certain axioms,
3. a Boolean ring

[chilly ocean]

it's a good point, this is probably the source of confusion
in computer science what they mean by Boolean algebra is the two-element Boolean algebra we have in math

[tribal moss]

The connection between those three definitions is not obvious, but standard enough not to care about once you know it.. :_)

[lavish gull]

is AND defined component-wise?

[tribal moss]

AND is just an abstract binary operation, as long as it satisfies the axioms.

[lavish gull]

obviously you get a product in F_(2^n) but that might not satisfy the axioms

[chilly ocean]

a Boolean algebra can also be defined as a topological space

[tribal moss]

F_2[x] is not a Boolean ring.

[lavish gull]

yes, i've been told a lot about what it is not. but what is it?

symbolically, how are you constructing it

[delicate bloom]

wot

[elder wave]

You take polynomials with coefficients in F_2

[lavish gull]

F_2[x], ok

maybe you want more than a single indeterminate?

F_2[x,y,z,...]?

[delicate bloom]

no

[wooden ember]

Im not sure what more you can want

[tribal moss]

Timo told you how to construct F_2[x], not how to construct a Boolean ring.

[delicate bloom]

x != x^2 in F_2[x]

[wooden ember]

We have given a few examples of Boolean rings

[lavish gull]

ok, i'm not going to guess any more. i can give you a construct based on F_2[x] with the property that x^2=x

do you know what that might be?

[delicate bloom]

it's like you're asserting a banana is a boolean ring cause we're talking about boolean rings, and then you're like "ok what's a banana?"

who cares

[wooden ember]

Namely the direct product $\prod_{i\in I}\mathbb{F}_2$ is a Boolean ring

[lavish gull]

it's your conversation. it doesn't interest me really

[cloud walrus]

𝓛ittle ℕarwhal ✓

[lavish gull]

the product has component-wise AND

[tribal moss]

Well, if you quotient out <x>, you're left back with F_2 itself, which is a Boolean ring.

[lavish gull]

why didn't you say that 30 minutes ago

[wooden ember]

Because that’s irrelevant of polynomials

Like you don’t need to go through polynomials to talk about F_2 in and of itself

[lavish gull]

i don't think you agree between you about the definition

[tribal moss]

It's been mentioned several times over the last hour that the direct product of any number of copies of F_2 is a Boolean ring.

[wooden ember]

Including a single copy

ie F_2

[lavish gull]

ok

[chilly ocean]

I'm surprised you have patience for this. I'd just give some reference and run away

[elder wave]

Oops didn’t read enough for proper context

[lavish gull]

i thought maybe you'd like to talk about some interesting things about Boolean algebra once we got the construction out of the way

[elder wave]

[lavish gull]

but we never got there

[wooden ember]

Lmao okay I’m out

Thanks for the help though @next obsidian and @tribal moss

Thought it was a fun problem

[chilly ocean]

google Sankappanavar universal algebra and go read it. It has a chapter about Boolean algebras in it

free online

[tribal moss]

This all started by Narwhal talking about the interesting property of Boolean rings that if they're finite then they're direct products of F_2s.

[lavish gull]

ok thanks but i'm reading other things about F_(p^n)[x] and it's quotients

[chilly ocean]

it even has multiple chapters about it characterizing it as different objects (Boolean rings, Stone spaces)

[coral spindle]

Remember to exercise self-care lmao

[lavish gull]

for example, because it is a vector space, every matrix can be written as polynomial

[chilly ocean]

I'm sure there is a very extensive amount of literature dedicated to Boolean algebras alone

[lavish gull]

there are some interesting matrices and the properties of those carry over to the polynomials that represent them

for example, if Ax=0 so that x is in the null space, then x is a root of the polynomial equivalent to A

@wooden ember is that woke?

i guess you can read about it in your book

[tribal moss]

Cool it with the hostile pinging, okay

[lavish gull]

it wasn't hostile. i wanted to know what he thought of my theorem

i got the impression that he thought what i said was a trivial matter

[coral spindle]

I came up with a somewhat simpler argument in the same vein, you interested? I'll DM

[formal ermine]

let $H \normal G$ then $|C(H)| = 1$ and $|N_G(H)| = |G|$ then by orbit stabilizer $|G| = |H|$ what's wrong here

[cloud walrus]

illuminator3 👻(#eric4honorable)

[coral spindle]

Not sure how you're getting this is = |H|, by my reckoning you'd get |G| = |G| from that

it is G acting on H, after all

[formal ermine]

ahhhh, right. thanks

[coral spindle]

No worries

[formal ermine]

why is $3 \not \in \bZ_{21}$?

[cloud walrus]

illuminator3 👻(#eric4honorable)

[tribal moss]

Who says it isn't?

[coral spindle]

This is (Z_21)*, not Z_21.

[formal ermine]

how does removing the zero also remove other elements?

[coral spindle]

Do you remember the definition of (Z_n)*?

Have you looked at it?

[formal ermine]

Z_n \ { 0 }?

[coral spindle]

OK, that's not it

[tribal moss]

(...)* removes all the non-invertible elements.

[coral spindle]

^

[formal ermine]

ohhhh

makes sense

[coral spindle]

So R* = R\{0} exactly when R is a division ring

[formal ermine]

yeah

thanks

[lapis trail]

The kernel of a group action G acting on A is in the stabilizer of every element of A, is that right?

[coral spindle]

That's right

[lapis trail]

ok thanks

[coral spindle]

Or more specifically, it is the intersection of the stabilisers

[lapis trail]

I was confused on stabilizer vs kernel

[coral spindle]

OK well

The stabilizer only fixes a single element; the kernel fixes all of them.

[lapis trail]

got it

[coral spindle]

👍

[lapis trail]

that helps thanks

[formal ermine]

how's the kernel of a group action defined? like do we have a notion of a neutral element in the output set?

[coral spindle]

A group action is a homomorphism G -> Sym(X)

so we just define it as the kernel of that.

[formal ermine]

oh wait I confused group action with group operation

mb

[chilly ocean]

is this a group action or a permutation representation of a group action?

[pastel cliff]

it's an equivalent definition to the map G x X -> X

[chilly ocean]

"equivalent" meaning...?

[pastel cliff]

meaning... equivalent

it's two ways of defining a group action

[chilly ocean]

Meaning a 1-1 correspondence

i believe that if we have one we can get the other. say, if we use the latter, we can construct a homomorphism that meets the specifications of the former.

i believe we can also go the other way.

this is just done through currying, basically.

however, that doesn't mean they're equivalent. and so i think it's a fair question to ask what a kernel is in the context of the GxA -> A, rather than G -> Sym(A)

Not a homomorphism

[coral shale]

its a matter of notation...

[chilly ocean]

Unless you treat it just as a map

[coral shale]

g . a := phi(g)(a) and ur good to go

[chilly ocean]

In which case kernel is a set of pairs ((g, a), (h, b)) with ga = hb

[chilly ocean]

it's not... both maps have different domains, codomains, and relationships.

there's a correspondence between them yeah. but that's not what I'm asking.

[next obsidian]

Bro they’re equivalent. You have a bijevtion between both things. They describe the same thing, they’re just different ways of phrasing them. Both are useful

[pastel cliff]

that correspondence makes them the same

potato potato

[chilly ocean]

thanks bro makes sense now

[coral shale]

im sure theyre isomorphic or something in a categoric sense

[chilly ocean]

i believe they are. but then it's like... okay what's the kernel.

[coral shale]

exactly the same structure

[next obsidian]

It would make sense if you pulled your head out of your ass and didn’t purposefully be so pedantic

[pastel cliff]

honorable

[chilly ocean]

We can cut it off from here.
Let's cease communications.

Thank you.

[chilly ocean]

is there a name for the "kernel" in the context of GxA -> A ?

[coral shale]

theres this suggestion which im trying to comprehend rn

[agile burrow]

It's still called the kernel

[next obsidian]

Yeah it’s the kernel of the corresponding map into the symmetry group

So it makes sense to call it that

[chilly ocean]

I don't think that definition matches kernels.

I think ...

{ g in G s.t. g.a = a } is the kernel.

i think they're maybe commenting on faithfulness of actions.

[coral shale]

yh i dont get this lol. Isnt the kernel everything in G that leaves everything in A invariant?

yh

[chilly ocean]

Why would it be a subset of G

[coral shale]

cus thats how it is in the original?

{g | forall a, ga = a} makes sense to me

[chilly ocean]

Here the map is from G x A

[coral shale]

oh

[chilly ocean]

right.

G x A to A is just a map in Set since A is a set and we want it to interpret it as a morphism in Set

So we take kernel with this in mind

And that's just a set of pairs, a subset of (G x A)^2

kernel of a homomorphism is a subgroup only because of how nice groups are

We could try to simplify it to something like in the group case

But why are we having this consideration in the first place

If you want to just use kernel under the "equivalence" then why the commotion about the whole equivalence

[coral shale]

oh i missed this. ok makes sense

[chilly ocean]

I didn't create the commotion, so I cannot answer.
I asked by what means we were equating the two definitions.
I think that may have aroused some short circuits somewhere.

If I follow your categoric approach, then I don't have much to say: it's resolved.

Bijection, I already said that

[chilly ocean]

Here

[coral shale]

,,g ._\varphi a \coloneqq \varphi(g, a) \coloneqq (\varphi(g))(a)

[coral shale]

something like this maybe

[chilly ocean]

Again, merely saying "1-1 correspondence" doesn't seem equipped to highlight where the kernel is.

[pastel cliff]

why not

the point of having such a correspondence is to extend understanding of one thing to the way we look at another

[chilly ocean]

Please understand ... I'm not saying we cannot use correspondences.

[pastel cliff]

the homomorphism G -> Sym(X) can inform how we think about G x X -> X and vice versa

i think we're just having a hard time understanding what your question/confusion is

or at least i am

[coral shale]

phi : G -> Sym(X)
Sym(X) : X -> X

So we have 2 options to combine these?

ig

[pastel cliff]

wdym combine

like map composition?

[coral shale]

in (very) loose symbol ideas

phi : G -> (X -> X)
phi : (G -> X) -> X

cus of associativity of functions and all

[pastel cliff]

i mean sure?

[coral shale]

cant quite put my finger on it formally

[chilly ocean]

The second one is wrong

[coral shale]

yh sure

[chilly ocean]

It's not wrong... but it's not complete if we're talking about group actions.

You get G x X like before, not an arrow

[coral shale]

needs to be GxX which is not G -> X

[pastel cliff]

to me the fact that a group action of G on X can be thought of as a homomorphisms G -> Sym(X) is just saying that group actions of G on X are subsets of all possible permutations of the elements of X

G x X -> X to me is a more "concrete" way of seeing the same thing, but in such a way that you can write a proof with this defn more easily somtimess

blitz might wanna fact check that, i learned all this recently, but that's how i see htings

[chilly ocean]

I think it's good that you're giving your own interpretations and all... nothing to say

[pastel cliff]

i meant the correctness of it but i'll assume it's good then

how the turns have tabled

[coral shale]

If we use precise notation; sets of ordered pairs for functions, im sure the bijection will be explicit

to express what i wanted to express

[chilly ocean]

You're currying, yeah.
I mentioned this.
https://en.wikipedia.org/wiki/Currying

I think they are different functions though as I noted above.
What we do, however, is correspond them with a bijection - which I think everyone has agreed with at this point.

When I learned this, however, the G -> Sym(X) was called a "permutation representation".
And so the kernel was understood only in terms of it's relationship to that group homomorphism.

I feel like there's motivation not to refer to whatever corresponds with the kernel as "a kernel" in terms of that definition. Ofc, in general discussion, I'm going to likely follow fine what someone means by "kernel" once it's established we're dealing with a group homomorphism.

[pastel cliff]

currying never heard of that

[chilly ocean]

Getting hungry here

[pastel cliff]

possibly dumb question of my own, but in a field F, 0 has an inverse right?

[chilly ocean]

No, for example Q and R, fields, do not have an inverse of 0.

[coral shale]

what

[pastel cliff]

[coral shale]

R is a field

[pastel cliff]

trying to show that if $F$ is a field, $F[x]$ is never a field.

[south patrol]

additive inverse sure

[cloud walrus]

stμ₂dying

[coral shale]

yh go ahead

with inverse idea

of some sort

[pastel cliff]

i mean if it's a ring R, R[x] can't be a field bc the zero polynomial wouldn't have an inverse right

[south patrol]

try to pick an example of a non-zero element with an inverse lol

[chilly ocean]

Um.... I'm studying that for an exam.
I think F[x] is um... it might be a Euclidean domain or maybe a UFD.
idr.

If you find out what it is in the hierachy, then that might help with finding out what it lacks to keep it from going further down.

[coral shale]

idk

[chilly ocean]

Because units are non-zero elements of F

[south patrol]

the zero element never needs to have an inverse...

[coral shale]

polynomials in single variable with coeffs in F

[pastel cliff]

kek i knew that

[south patrol]

but there is a more obvious example of an element which has no inverse

[chilly ocean]

I know that part.
I'm saying idk where it lands in the hierarchy of classes of rings.

[south patrol]

Yes, it is a Euclidean domain (you can use polynomial division) and hence is a PID hence UFD among other things

[pastel cliff]

ufd?

[coral shale]

oh sry

[south patrol]

unique factorisation domain

[chilly ocean]

Unique Factorization Domain

[coral shale]

yes its an ed for a field lol

For F ufd it might be a ufd... not so sure here

[pastel cliff]

ufd <=> integral domain? havent seen that term before

[coral shale]

no

[chilly ocean]

No.

[south patrol]

no

unique factorisation into primes essentially

[chilly ocean]

Are you studying from a text?

[south patrol]

very different beasts

[pastel cliff]

naw

[chilly ocean]

The definition of UFD is a bit tricky because like... we have to be a bit cautious about "prime" vs "irreducible"

[next obsidian]

Yes

[chilly ocean]

Where's semirings

[next obsidian]

This is basically Gauss’s lemma

[south patrol]

fr

[coral shale]

its been half a yr

[chilly ocean]

Dummit & Foote, apparently, is appreciated for how it handles this hierarchy.
In Ch.7 they introduce rings.
In Ch.8 they make more clear some distinctions and relationships among PIDs, EDs, and UFDs.

[coral shale]

since ive thought about this stuff lol

[next obsidian]

I’m just saying it is true

[coral shale]

yh yh im saying i forgot

sad

[pastel cliff]

dummy and foot fetish

[pastel cliff]

noot quite seeing it

[next obsidian]

[south patrol]

what is an element of F[x] which is not in F

lol

[pastel cliff]

anything with an x

[south patrol]

indeed

[coral shale]

x

[chilly ocean]

I think the big math mafia has merced structures deemed too unaesthetically pleasing with regards to their lack of "natural " properties like commutativity and associativity. Watch out for big math.

[south patrol]

x is the simplest possible such example

[next obsidian]

Actually if F isn’t a field that’s not true

[south patrol]

can x have an inverse

[coral shale]

oh no

[next obsidian]

Things which have an x can be invertible

[south patrol]

F is clearly meant to be a field

[next obsidian]

Sure, but it’s a bad thing to say anything with an x

Well I guess it doesn’t matter but

[south patrol]

Oh sorry

[pastel cliff]

field coefficients of an x

[next obsidian]

You can quickly say “polynomial rings are not a field”

[coral shale]

can i have an explicit example, not following.

[south patrol]

Yeah anything with an x opens up a can of worms

[next obsidian]

By pointing at x

[south patrol]

but x is what i was trying to point at

[coral shale]

R[x], x has an inverse...?

[next obsidian]

A polynomial is invertible iff the constant is invertible and all other coefficients are nulpotent

So x is never invertible

[delicate bloom]

1/x my favorite polynomial

[south patrol]

atiyah macdonald chap 1 exercises

[next obsidian]

But something like 1 + ax could be

[south patrol]

😩

[coral shale]

oh ok uh

[south patrol]

ye e.g. if A = R[ε] where ε^2 = 0 then (1 + εx)(1 - εx) = 1

[next obsidian]

Consider Z/4Z[x]

[coral shale]

not thinking clearly rn

[next obsidian]

And 1 + 2x

[coral shale]

ok lol

[next obsidian]

Its inverse is 1-2x

[south patrol]

lol

same example

up to iso

nice

[coral shale]

so x always works

[next obsidian]

Yes

[coral shale]

given F is a ring

yh ok

[next obsidian]

That’s why I think it’s a better thing to say “x is not invertible” because this is true regardless of F

[south patrol]

x always works because e.g. ||evaluation||

Oh i mean agreed

[chilly ocean]

You mean degree?

[south patrol]

no, i did mean evaluation

degree needn't play nicely in general rings anyway because of the example I gave above

[next obsidian]

Evaluating a poly at 0 is a homomorphism and sends x to 0. Units always get sent to units so x can’t be a unit

[chilly ocean]

Oh

[south patrol]

or more concretely: if xf(x) = 1 then plugging in x =0 gives 0 = 1 lol

[chilly ocean]

I was thinking of valuations and how it's an Euclidean domain in field case

[coral shale]

Next, F(x) is always a field

[next obsidian]

If F is

[south patrol]

is there a good notion of R(x) for a more general ring lol

[next obsidian]

Can you even make sense of it without F being an integral domain?

I don’t think so

[coral shale]

Smallest field containing R and x

D:

[next obsidian]

Yeah so

[chilly ocean]

x = 0 = 1 when our ring is trivial

[next obsidian]

That makes sense iff R is an integral domain

[south patrol]

ig you'd pass to quotient field if integral innit

but otherwise like

[next obsidian]

Well potato I was thinking

Define it as rational functions

[south patrol]

big no no if you invert a zero divisor lol

[next obsidian]

With coefficients in R

[south patrol]

ye

[next obsidian]

This is gonna be exactly

[south patrol]

that works

[next obsidian]

Frac(R)(x)

[south patrol]

ye

[next obsidian]

And this will be a smallest field containing R and x

[south patrol]

indeed

[next obsidian]

And this can only exist if R is an ID

Or else u can never embed in a field

[south patrol]

yh lol

öwö

algebra is fun

[chilly ocean]

Embed into division algebra

[south patrol]

everything is commutative

hm

[chilly ocean]

Especially in Lie algebras. They're all trivial

[next obsidian]

You can’t

[south patrol]

i want to give a tangentially related question on like - when is R[x] a PID

[next obsidian]

Zero divisors prevent this

[south patrol]

that one is cool

[next obsidian]

That’s iff R is a field no?

[south patrol]

like it all kinda suggests the general intuition that like adjoining an indeterminant kinda makes it slightly worse

yes but i intended it to be an interesting q lol

[next obsidian]

Oh

I thought it was a question

Seeking an answer

[south patrol]

no

lol

[next obsidian]

[south patrol]

you underestimate my power obi wan

[chilly ocean]

But xy = 0 implies x = 0 or y = 0 in a division algebra

[south patrol]

yes, so if you can find non-zero x and y such that xy = 0, then you cannot embed in a division algebra

[next obsidian]

Sorry I don’t know what you were saying

My point is if you had a zero divisor this prevents you from embedding into a division algebra

[south patrol]

o

[chilly ocean]

Oh. I was saying, take something non-commutative and then try to embedd it

[next obsidian]

Sure, but zero divisors prevent this, the issue about non-integral domain I was bringing up wasn’t being noncommutative lol

It was that zero divisors obstruct it

[chilly ocean]

I think it doesn't always work to embedd it into division algebra in the non-commutative case (assuming no zero divisors)

[next obsidian]

It’s tricky because of left and right inverses?

I don’t see why you can’t do it like this thoigh

Take R<R> the noncommutative polynomial ring with generators indexed from R

Quotient by the two-sided ideal generated by expressions of the form r•x_r = 1, x_r•r = 1

Now everything in R<R> was a product of a bunch of variabeles x_r and elements r, each of these now have a two-sided inverse

And the hope I guess is that the natural map from R into this ring is still injective

This feels like it should be clear

[pastel cliff]

gonna interject one more time - is this valid or

[chilly ocean]

Maybe it does work. I remember this had some obstructions, or maybe that was for embedding a semigroup into a group. Either way going sleep, its past my bedtime

[pastel cliff]

that secont to last = shoud be \iff

[next obsidian]

There’s no way that first thing could be a biconditional without extra assumptions (which I think are present in the setup for your problem)

You’re claiming that every idempotent is the identity

Let y = xr, then you’re basically saying y^2 = y => y = 1

But that’s not true

In the particular case you’re assuming every element has a left inverse then multiplying on the left is injective, so avtually you can cancel a y

But you need to explain this

[pastel cliff]

i see the problem but when would that implication you wrote be false

[next obsidian]

I’m not sure

Frankly

In this particular setup assuming everything has a left inverse and that xr = 1

Well if you assume xr = 1

Then yeah obviously it’s true that xrxr =xr => xr = 1

Both are tautologies

But anyway, you only need to go one direction

that biconditional in general isn’t true, you can’t get it just from the axioms of a ring so I think you should remove it

But also, this proof won’t work

You need to use that every element has an inverse, your argument as written only uses the assumption that x has a left inverse

I think you can have elements with a left inverse and no right inverse

[pastel cliff]

i thought that would follow from x being arbitrary

[next obsidian]

The right inverses only come when you assume everything has a left inverse

x is not arbitrary

You have fixed r, then x came into existence because you assumed r is left invertible

I think you have to use that x is also left invertible, I’ll try to run an argument in my head really quick

Yeah

You need to use that x is left invertible

[pastel cliff]

ok then how abt "let r be an arbitrary nonzero element of R. by assumption, it is left invertible so there exists x \in R such that xr = 1 ..."

[next obsidian]

Yeah I mean that’s what your argument already did

But your argument after that

You never used that x is left invertible

[pastel cliff]

oh i think i see what you mean

[next obsidian]

ie you never pulled in a y such that yx = 1

[pastel cliff]

one ssec

[pastel cliff]

im being dense chmonkey but why do we need this

if r is an arbitrary element of R, the same argument applies to all elements of R

[next obsidian]

Because you need to again use that r’s left inverse has a left inverse

It doesn’t matter if you’ve quantified over all r, as you’ve written it your proof would be able to prove “if r has a left inverse, it has a right inverse” and this is false

Your proof ultimately will be able to prove the statement “if r has a left inverse with a left inverse, then r has a right inverse”

But you need to actually use that r’s left inverse has a left inverse in the argument that r has a right inverse

If this still doesn’t make sense, from what you had written there with biconditionals, extract a proof that xr = 1 => rx = 1 and I’ll point out where it fails

And okay, maybe you feel that in writing “let r be arbitrary and nonzero, let x be such that xr = 1” you’ve used that every element has a left inverse. This is true, but my point is if you modify this to just say “let r be nonzero, and have a left inverse. Let x be such that xr = 1” now you haven’t used that every element has a left inverse, but the rest of your argument won’t be changed at all. Thus this proof could show “if r has a left inverse it has a right inverse”

[pastel cliff]

this made sense, will try again

[pastel cliff]

do right inverses and left inverses have to be equal if they both exist for some r

[delicate bloom]

simplify by associativity in 2 ways $a^{-1}_Laa^{-1}_R$

[pastel cliff]

asking bc of this but this is probably also wrong

[cloud walrus]

Merosity

[next obsidian]

That doesn’t prove that r has a right inverse

Oh wait

Yeah it does

And yeah, what mero said

[tribal moss]

If the claim is true at all, then the proof needs to use distribuitivity somehow. As long as you're only looking at multiplication, it's easy to construct a semigroup where everything has left inverses but some things don't have right inverses.

[next obsidian]

Maybe u can just expand on why r has a right inverse, but this proof works

No it doesn’t tropo, this works

Here’s a proof that a monoid with everything left invertible is a group

[tribal moss]

Hmm.

[next obsidian]

|| let x in M, let y such that yx = 1, let z such that zy = 1. Then zyxy = xy, but also is z(1)y = zy = 1, so xy = 1||

[tribal moss]

Oh right, I see why the thing I had in mind actually won't work.

[pastel cliff]

very vague question out of curiosity - are there any results that hold for non-abelian groups that dont hold for abelian groups

or rings for that matter

[next obsidian]

Yes, in non-abelian groups there exist x,y such that xy ≠ yx

[pastel cliff]

im

im not sure what i expected

[tribal moss]

Perhaps less facetious, conjugation only really makes sense to study in a non-abelian setting.

[next obsidian]

Hungerford will the the gentlest

Don’t do Lang first by any means

[pastel cliff]

why so against lang?

[next obsidian]

Have u read it

[pastel cliff]

no that's why im curious

[next obsidian]

Then go try to read it

[pastel cliff]

[gloomy night]

Lang is dense and merciless. Even if you know algebra, there will be parts of it that are hard to understand.

[coral spindle]

Lang wrote books for himself

[wooden ember]

dont we already have terminology for this? I thought a section was called a right splitting

(not asking for help on the exercise, just terminology check)

[chilly ocean]

https://math.stackexchange.com/questions/336893/can-we-embed-cancellative-rings-into-division-rings

Yep it doesn't

Conditions for when it embeds seem interesting but I'm not going to dive deeper

[hidden haven]

A section of a map is a pre inverse for it. A retraction of a map is a post inverse. A right splitting of a short exact sequence is a section of its second map, and a left splitting is a retraction of its first map.

ie sections and retractions are more general

[wooden ember]

right okay

though in this exercise they only do it in the context of an s.e.s

and make the mistake of calling a section of i a morphism q such that q o i = id (should be a retraction if i understood correctly)

[hidden haven]

Ye section should be on the right

On the right of ∘

[lethal dune]

is arbitrary direct sum of injective modules is injective? If yes can you give an elementary proof

like without using derived functors

[wooden ember]

I think that's false? take the map k[x]->k[x] sending p |-> xp. This is a k-linear map but if you extend this to an exact sequence 0->k[x]->k[x]->k[x]/(x)->0 this doesn't left split right?

Sounds true for direct products though

not too familiar with injective modules though so better to wait for someone else

[wooden ember]

I feel i may be missing something key here... Nowhere in proving this have i used the fact that im working with projective modules which is worrying

i just defined maps phi_i from Hom(P_i,N) to Hom(P_i,N) by post compisiton by psi which is easily seen as a cochain morphism

and the induced map on cohomology sends a + imHom(alpha_(i-1),N) to phi_i(a) + imHom(alpha_(i-1),N) = psi o a + imHom(alpha_(i-1),N) = r(a+imHom(alpha_(i-1), N)) so im not sure where i need to use that things are projective

seems kinda direct to me

additionally im working with R an integral domain that isn't a field and this wasn't used either (though i suspect this is only needed in the exercises that follow)

[next obsidian]

No

It’s true over Noetherian rings because of Baer’s criterion, proof is not that difficult

It’s true over a PID even easier sort of again by Baer’s criterion because injective <==> divisible

Actually, if R is such that arbitrary direct sums of injectives are injective, then R is Noetherian. So it characterizes them

[lethal dune]

I see

[untold basin]

Hello, can someone give me a tips for this ?
Let K,L be fields. Show that if there exists a field morphism between K and L, then car(K) = car(L)
(where car is the characteristic)

I found many things but I don't find some contradiction (I supposed car(K) =/= car(L))

[lapis trail]

I just got DF and like it

[untold basin]

I just realised that my morphism is injective ^^^^^^

[hidden haven]

When you pass to cohomology, you get Ext only if the resolutions are projective

[wooden ember]

yeah i do get that i just thought there might be some more things to it but after working on the other exercises i concluded the same

Wondering if someone can help with this exercise that I've been stuck in for a bit. I want to show that for an integral domain that isn't a field R and an R-module N with Ann(N)=/= 0, then Ext^i(Frac(R),N)=0 for all i. The idea was to use some results shown before, namely that for a fixed r in R, the multiplication map by r phi:M->M will induce the map Ext^i(phi, N): Ext^i(M,N)->Ext^i(M,N) which will also be multiplication by r. Additionally if we have a multiplication map by r psi:N->N we get an induced cochain complex morphism on Hom(P, N) for P a projective resolution of M, that when passed to cohomology is also multiplication by r. Since Ext^i(phi, N) is unique, it must be equal to this second map. But now taking r in the annihilator of N this means Ext^i(phi,N) is 0

The idea would be that precomposing a representative by the multiplication map gives 0, and we'd like to show that this implies the representative itself is 0. This is true for the 0-th cohomology group since the multiplication map is surjective on Frac(R) (since it is a field), but im not sure how to do this for higher groups

id somehow want to say that the multiplication map induces a surjection on the Ext groups so that precomposing by multiplication being 0 still implies representatives are 0

[pliant raptor]

How would you prove that $\mathbb{C}\otimes_{\mathbb{R}}\mathbb{C}$ is a left $\mathbb{R}$-module?

[cloud walrus]

ImHackingXD

[chilly ocean]

C is an R-bimodule here, so of course

If $M$ is a $S-R$-bimodule, $N$ is a left $R$-module, then $M\otimes N$ is a left $S$-module

[cloud walrus]

Blitz

[chilly ocean]

a (very) known property of tensor product

[pliant raptor]

I see, thank you

[toxic zephyr]

if a=xy is an irreducible element of an integral domain R, then (by definition) either x is a unit or y is a unit. If we say WLOG that x is a unit, then y cannot be a unit, correct? since a cannot be invertible?

[next obsidian]

Slight caveat, this is for i > 0

Oh huh wait

If Ann(N) ≠ 0 then maybe this is true for all i

[wooden ember]

according to the exercise it's true for all i>=0

and ive already proved it for i=0

[next obsidian]

Yeah

I misread the problem

@wooden ember I think I have the solution

your idea is almost correct, but we want to see that multiplying by an element of R is injective

and then we get that the Ext has to be 0 since actually we can see directly that map is 0

So we do this by induction, you handled the base case

[wooden ember]

why injective 🤔

oh right i see what youre saying

yeah okay

[next obsidian]

FUCK

my argument almost worked god damn it

[wooden ember]

sadge

[next obsidian]

My idea

was to use a LES in Ext

where we got something like this

[wooden ember]

we havent covered LES yet so there's something simpler anyways

but im all ears

[next obsidian]

Ext^i-1(Frac(R), some shit) -> Ext^i(Frac(R),M) -> Ext^i(Frac(R),M) where the last map was induced by multiplying by x where x in Ann(M), so it's 0

some shit was something which had a nontrivial annihilator

so by induction that Ext is 0

[wooden ember]

right

[next obsidian]

damn it

this is almost the opposite of what you do for local cohomology

[wooden ember]

why doesnt it work 🤔

[next obsidian]

that proof works basically like this but you're dealing with regular elements

[next obsidian]

I don't think so

I'm gonna think about it

harder

Oh lol

I thouht you asked "does it work"

[wooden ember]

here's the full exercise in case a piece of information jumps at you

[next obsidian]

the reason it doesn't work is that I couldn't get the right SES

to then derive an LES in Ext

[wooden ember]

i see

[next obsidian]

what the fuck

am I supposed to take a projective resolution of Frac(R)????

I was trying to do this with injective resolutions of N

since that seemed most natural

[wooden ember]

uhu that's what i was trying

we havent covered injective resolutions

[next obsidian]

bruh

okay

hmm

Oh

hurb

I think I got it

lmfao

$0\to \mathrm{Frac}(R)\xrightarrow{\cdot x} \mathrm{Frac}(R)\to 0\to 0$ is a short exact sequence because multiplying by $x$ is an isomorphism

[cloud walrus]

Dᵇ(onkeyy)

[wooden ember]

yup

[next obsidian]

We can apply $\hom(-,M)$ and get a LES in Ext

[cloud walrus]

Dᵇ(onkeyy)

[next obsidian]

oh yeah

whatever

I'll just write this out first and then

think about how tf to do it without

[wooden ember]

yeah ofc im still listening haha

[next obsidian]

but it gives us $\mathrm{Ext}^i(0,M)\to \mathrm{Ext}^i(\mathrm{Frac}(R),M)\to \mathrm{Ext}^i(\mathrm{Frac}(R),M)\to \mathrm{Ext}^i(0,M)$ exact

[cloud walrus]

Dᵇ(onkeyy)

[next obsidian]

and the only non-trivial map there is multiplying by x

which is 0

okay I see how to do this without the LES

you only have to use that Ext is a functor

you know that multiplying by x gives an iso on Frac(R)

so it induces an isomorphism on Ext^i(Frac(R),M) for all i (by functoriality, the induced map from the inverse will be an inverse for the induced map. This is true for all functors)

but we know that's 0

[wooden ember]

oh god

yeah

bruh

that's a hack i refuse

[next obsidian]

[wooden ember]

thanks

[next obsidian]

ChmonkaS

it took some more machinery to get there

but I did get there eventually

[wooden ember]

lmao

[next obsidian]

man I'm so dumb

I literally used the LES to prove that an isomorphism induces an iso

after applying a functor

[wooden ember]

it happens

sometimes it feels like the more tools you have, the more you miss the obvious little things

[rotund aurora]

Let n=pq where p and q are prime and put R=Z/nZ. Consider the R-module R^2. Then, (p,0) and (0,q) are both torsion-elements, but (p,q) is not a torsion element, therefore the set of torsion elements of R^2 including zero is not a submodule of R^2.

Is this example correct?

[next obsidian]

not true

R^2 is a finite group, everything is torsion

(p,q) just has order pq

[rotund aurora]

but pq=0

[next obsidian]

no?

pq is a number

[rotund aurora]

pq=n which is 0

[next obsidian]

doens't matter

pq is the order, as a number in N

if you add (p,q) with itself pq times it becomes 0

or do you mean torsion over R?

maybe that's where the confusion is coming from, I was considering torsion as an abelian group

[next obsidian]

yes, so are you considering this as an abelian group or as an R-module?

[rotund aurora]

as an R-module

[next obsidian]

then yes

[rotund aurora]

I did say it was a module tho

[next obsidian]

well, to me I read it as Z-module

which is maybe just

my own brain worms

lol

anyway, so this is kind of a deficieny of this definition of torsion

My personal preference for the definition of torsion is as follows

It's the set of m in M such that there's an r in R which is nonzero and not a zero divisor

such that rm = 0

then M_tor is actually a submodule because if m and n are annihilated by r,s then m + n is annihilated by rs and rs is not 0

[rotund aurora]

So in that case, M_tor is always a submodule?

[next obsidian]

Yeah it should be