groups-rings-fields

[scarlet estuary]

in analogy with F_2

[kindred mist]

it sorta looks like F_2[x] almost is what I was thinking

except modulo some polynomial

[scarlet estuary]

indeed, F_8 is actually the field formed by the eight elements of Z_2[x] with degree at most 2

this generalizes

for all F_2^[maximal degree + 1]

and an analogous statement holds for different "base fields"

(ie F_3^p, F_5^p, etc)

[kindred mist]

Ok this is neat, I like how elegant that relationship works out

makes me want to study more Galois & finite fields tbh

[scarlet estuary]

i should clarify that im not stating this

exactly correctly

the "generalization" isnt as nice as im implying

[stone fulcrum]

It's still very friendly though

[scarlet estuary]

yeah, it's "properly" stated by taking about irreducible polynomials

[kindred mist]

like modding out Z_2[x] by irreducible polynomials?

or just Z_p[x] in general?

[scarlet estuary]

basically yes

wikipedia probably has a nicer writeup

[kindred mist]

wiki is so nice

always to the point

thanks again btw

[timid hull]

Or you can think of $\mathbb{F}_{p^n}$ as the splitting field of the polynomial $X^{p^n}-X$ over $\mathbb{F}_p$ .

[cloud walrus]

WhyamIsohot?:

[scarlet estuary]

this is not abstract algebra

try #prealg-and-algebra or a generic questions room

[kindred mist]

Ah like definitionally as that splitting field? neat

Going to think about why those are equivalent

ty for the info

[mild laurel]

Going that route is a bit more tricky I think, you have to talk about why all the roots of that polynomial are distinct and so have to talk about separability issues

[kindred mist]

hmm, one thing I wanted to ask about though btw

Is there somewhere in particular I can read about why "alg. closure is equivalent to not being an ordered field"? (I think I heard this is true from somewhere, it seems kinda subtle, but idk)

(I tried searching for some pdf/wiki/other documentation on this ^^^ but no avail)

[upbeat juniper]

from what I've read, to define F_p^n as a quotient it first needs to be shown that irreducible polynomials of degree n do exist

and that's shown indirectly by constructing degree n extensions via splitting fields

[latent anvil]

@kindred mist that doesn't sound right

Ordered field => char 0

but there are non-algebraically closed fields of positive characteristic

[kindred mist]

Ok yeah it sounded to crazy to me to be true

ty for the heads up

[scarlet estuary]

that would be a wild theorem though

ngl

[mild laurel]

does alg closed imply not being ordered field

[scarlet estuary]

there is no algebraically closed ordered field.

[mild laurel]

yeah okay, maybe that's what they meant

[scarlet estuary]

suppose we have such a field and pick an a < 0

x^2 = a has a solution

but uh oh

that means x^2 = a < 0

😦

clearly this cant be true

since no matter whether x > 0, x = 0, x < 0 is true

we must have x^2 >= 0

this problem goes away if solutions to x^2 = a need not exist

(such as in the case of R)

[scarlet estuary]

(mod) 2 is quite an unusual way to write it

anyway, do you know what properties you need to verify to show something is a subgroup?

[charred pewter]

the subgroup criterion, which is just showing closure under multiplication and inversion

[scarlet estuary]

yep; have you tried showing those for E_n?

[charred pewter]

I'm confused on what the elements of E_n would be

[scarlet estuary]

some sigma is an element of E_n iff, for all i, sigma(i) = i (mod 2)

it might be best to do this with a concrete example

lets say we have S_4, so we take E_2

this is the set of permutations such that each i is mapped to something it's congruent to mod 2

so for example, if your permutation was the cycle (1 2)

it would not be an element of E_2

since sigma(1) = 2 which is not = 1 (mod 2)

however, the permutation (1 3) IS in E_2

since sigma(1) = 3 which is congruent to 1 (mod 2) and sigma(3) = 1 which is congruent to 3 (mod 2) [and obviously 2, 4 are unchanged, so sigma(2) is definitely congruent to 2 mod 2, as is sigma(4)]

similarly, (1 3)(2 4) is in E_2

basically, the idea is that elements of E_n only permute things "by even amounts"

3 wont be mapped to 2, 4, 6, 8, ...

but 3 might be mapped to 1, 3, 5, 7, ...

[charred pewter]

Okay, I think I'm starting to understand a little better

And so I can use the elements like (1 3) and (2 4) to prove using the subgroup criterion? But I would have to use variables to do it?

[scarlet estuary]

well a better thing to do is

suppose you have two elements (say sigma, tau) in E_n

show then that tau(sigma(i)) = i (mod 2)

i.e. $\tau \circ \sigma$ is also in $E_n$ (so it's closed under permutation multiplication/composition)

[cloud walrus]

Namington:

[scarlet estuary]

you also have to show that, given any sigma in E_n, there exists an inverse sigma also in E_n

but this isnt too hard

since, if sigma(i) = i (mod 2), and if sigma(sigma^-1(i)) = i, then this means sigma^-1(i) = i (mod 2)

hold on let me write that with fancy latex

if $\sigma(i) = i\pmod{2}$, then since $\sigma(\sigma^{-1}(i)) = i$, we have [\sigma^{-1}(i) = i\pmod{2}]

[cloud walrus]

Namington:

[scarlet estuary]

since we know $\sigma(i) = i\pmod{2}$ for all $\sigma$ and all $i$

[cloud walrus]

Namington:

[scarlet estuary]

so since $\sigma^{-1}(i) = i\pmod{2}$, we know that $\sigma^{-1}(i) \in E_n$, and therefore $E_n$ is closed under taking inverses (taking $\sigma^{-1}$ for any $\sigma \in E_n$)

[cloud walrus]

Namington:

[scarlet estuary]

combine this with a proof of closure under group multiplication (which is, in this case, permutation composition)

and you've proven E_n is a subgroup of S_2n

hint: congruence is an equivalence relation so $\tau(\sigma(i)) = \sigma(i) \pmod{2} = i \pmod{2}$

[cloud walrus]

Namington:

[scarlet estuary]

hence $(\tau \circ \sigma)(i) = i\pmod{2}$

[cloud walrus]

Namington:

[scarlet estuary]

but youll have to give a quick argument for why you can do this

[since right now im kinda abusing the = sign]

[charred pewter]

All those = signs are to be read as congruence signs right?

[scarlet estuary]

yeah

sorry

im just being lazy

well the $=$ in $\sigma(\sigma^{-1}(i)) = i$ is an actual $=$ sign

[cloud walrus]

Namington:

[scarlet estuary]

but besides that yeah

they're all congruences

[charred pewter]

I'm going to attempt the proof now, thank you so much for your help

[frank solar]

i'm going to assume that this channel is also open for discussion on representation theory lol

so i was going through fulton and harris and there's this exercise:

[bleak abyss]

Is algebra related to representation theory? 🤔

[frank solar]

(1) is there a proof of this that doesn't resort to finding a k[G]-module homomorphism? fulton and harris don't introduce the idea of representations as k[G]-modules until the next section, so there should be a way to do so without resorting to that, and
(2) is the remark actually useful for such a solution?

ahaha are any of the other channels more related to representation theory?

[bleak abyss]

I'm kidding lol

Perfect place

[frank solar]

LOL alright

[bleak abyss]

Uh, Frobenius reciprocity maybe??

That's my knee jerk when I see induced anything

[frank solar]

yeah i think that a claener proof of it goes throug frobenius reciprocity

[bleak abyss]

Would it suffice to do it for irreps?

[frank solar]

unfortunately f+h don't introduce it until like 2 pages down but honestly f+h's organization/ordering has been a little questionable so far so i think i'll just use it

[bleak abyss]

What tech do you have so far?

That might be a hint at the intended solution

[frank solar]

um so far f+h has mainly just shown the defn of an induced representation as well as existence + uniqueness

[bleak abyss]

Gotcha, I'll pull it up one sec

Okay yeah fwiw you can assume that U and W are irreducible

Since everything in sight distributes over direct sums

Eh doesn't feel like it helps

Welllll

[frank solar]

i'm not really familiar with the description of representations of G as k[G]-modules, but could a proof using k[G]-modules like https://math.stackexchange.com/questions/66323/why-is-u-otimes-operatornameindw-operatornameind-operatornameres logically become a proof that doesn't involve the idea of k[g]-modules?

[bleak abyss]

Tensor product distributes over the direct sum

So you can say U\otimes Ind(W) = \bigoplus_{\sigma \in G/H} U\otimes W^{\sigma}

Err wait

W^{\sigma} isn't a G-rep nvm

So you can't do this naively

[cloud walrus]

octave:

octave:

[frank solar]

oops that should not say egg' lol

[bleak abyss]

dem eggs doe

[frank solar]

lmfao 🥚

[cloud walrus]

Daminark:

[bleak abyss]

This using FH notation, I think this is right

[frank solar]

hm ok i see

[bleak abyss]

There's a chance I'm wrong since this is off the top of my head but okay I think this is a bilinear map, descends to a map on the tensor product

Then the idea is to show that the induced map on tensor products is actually a morphism of G-reps

And either show injectivity (probably easier lmao) or surjectivity, then it's just dimension counting

That's I think your roadmap

[frank solar]

ah ok sure sure

alright thank you so much!

[bleak abyss]

I will have to go now but yeah good luck fam

[frank solar]

lol thankss

[oak perch]

Um, I'm peddling free Cayley tables. They are cool because they've been made colorful when your mouse over things (tested in Desktop Firefox). http://mathx.world/demo/ There are three tables total and refreshing picks a random one of the three. (If I cannot peddle here, I apologize and I will self delete.)

[kindred rivet]

Does not work on iPhone

Nvm

[oak perch]

which browser?

[kindred rivet]

It does

It works on iPhone

[oak perch]

yay

[kindred rivet]

But

The display size is wrong

[thin flume]

I can't open it on Chrome?

[oak perch]

You can try Chrome but I'm sure Mac Firefox works.

[thin flume]

🤔

[scenic sage]

i opened

[oak perch]

oh yeah I hadn't gotten around to https yet just http so I think Chrome might not like the security.

hmm ...

[kindred rivet]

do u have an ssl liscence

[oak perch]

yeah I only have http://mathx.world not https so if a browser automatically chooses https or if it bans http then there is an error. I hadn't gotten around to it.

[kindred rivet]

it works on chrome for me

lmao

[oak perch]

But yeah it's a fun little thing. I take Cayley requests! I'm working on randomly jumbling the top and left bar to make it more interesting. Looking to highlight the center and other things I'm hoping I have the skills to code.

[kindred rivet]

uh

theres lots of empty space

[oak perch]

What is the resolution of your monitor?

[kindred rivet]

its not a square.

the page displays a square

hence there is empty space

[oak perch]

yes, but the table itself must be a square.

[kindred rivet]

i understand but

u can add text and shit

on the side

[oak perch]

are you just kidding? 🙂 my sarcasm sensor hasn't been operating at full today. 🙂

I might just make two tables or more. Same tables but displaying different info.

I have noted your observation but I really want to be minimalist text-wise.

anyway we'll see.

[golden pasture]

idk why would someone need a website specially for cayley tables ngl i would just write some script<or ask here for help on how to write a suitable script> to generate them myself

[oak perch]

Well, the cool thing about the comp is it's easy to jumble the elements of the group. It's also theoretically faster for the computer to find inverses, the center, etc.

but I must go. Thank you all .

!!

[vestal snow]

Can someone show me what to do in v) => i) after taking M = A/a?\

[solemn rain]

why do we need the order of the group Aut(K/F) where K is an extension of F to be of order equal to the degree of the extension

can some1 illustrate with examples for me?

[upbeat juniper]

that's only if K is Galois

e.g. Q(cbrt(2)) is degree 3 but its automorphism group is trivial

[solemn rain]

yea ^ iam having trouble with the computations big time

how did u know that

is it because it's a root minimal polynoimal is of deg 3 --> x^3-2 ?

[elder valley]

Yeah, the min poly is degree 3, so the extension is degree 3

[solemn rain]

yeaa

okay

ty

[frank solar]

Can someone show me what to do in v) => i) after taking M = A/a?
@vestal snow consider the map $M\rightarrow M_B$ defined by $A/\mathfrak{a} \mapsto B/\mathfrak{a}^e$?

[cloud walrus]

octave:

[shy bluff]

Can somebody explain to me how, in part b, you're suppoes dot use f(x) -> (f(sqrt(c)), f(-sqrt(c))? Like c is independent of x, so how could you possibly cover all of R x R that way?

[next obsidian]

@shy bluff what problem?

[shy bluff]

Oh it did'nt send

Sorry

[next obsidian]

Oh this is surjective

since you're ranging over all f(x)

this is a function which sends polynomials to points in R^2

Say for concreteness that c = 1

Then if you want to hit (0,0) just evaluate f(x) = 0

aaaaaaa this is harder than I thought oof

I want to tell you to just use the chinese remainder theorem

For (2,-1) you can use f(x) = 1.5x + 0.5

So in general if c > 0, let d = sqrt(c)

Then if you want to hit the point (a,b) I think you can do the following process

find the midpoint, so I'll just assume b > a since the other case is almost the same

you can do (b - a)/2 + a to find the midpoint of b and a

from there you want to sort of renormalize d so that it spans the distance from the midpoint to the two points, so let m = (b - a)/2d

Then f(x) = mx + (b - a)/2 will satisfy what you want

If you plug in d then you end up with (b-a)/2d * d + (b - a)/2 + a

= b

[latent anvil]

It's a map of R algebras, so it suffices to hit (1,0) and (0,1). I think you can write this down more simply

[next obsidian]

and if you plug in -d then you get -(b-a)/2d * d + (b - a)/2 + a = a

Fubb you brendan

[latent anvil]

lol

f(x) = (x-sqrt(c))/(-2sqrt(c))
g(x) = (x+sqrt(c))/(2sqrt(c))

ez

I think (a, b) = φ(ag + bf)

right?

[shy bluff]

Wut

We haven't learnt chinese remainder theorem

[latent anvil]

you don't actually need to

It would give you the result pretty immediately but you can prove the result immediately

So you want to show that some quotient of R[x] is iso to R×R and you have a map R[x] -> R×R. How do you show quotient rings are isomorphic to something?

[shy bluff]

First isomorphism theorem

Find a homomorphism from R[x] to RxR with kernel equal to x^2 - c

Right?

[woven delta]

Yes

Surjective homomorphism

[shy bluff]

Right

[woven delta]

Obviously 1 should get mapped to (1, 1)

[shy bluff]

And they've given us f(x) -> (f(sqrt(c)), f(sqrt(-c))) right

Yea, that's oen of the conditions of it being a homomorphism

[woven delta]

Oh lol I was going to do this by generators

But yeah the map they gave you works

[shy bluff]

Yea but how is that map surjecitve?

[woven delta]

So you have to verify that 1) this is a homomorphism 2) it is surjective

It's pretty easy to show it's surjective

Where does x get mapped to?

[shy bluff]

Like it satisfies the properties of being a homomorphism, I can see that off the bat

[woven delta]

Also the way they wrote it is a bit wrong

[shy bluff]

Uh it gets mapped to (f(sqrt(c)), f(sqrt(-c))?

[woven delta]

No -sqrt(c)

[shy bluff]

Errr sorry yea

How is that surjective? c is just a constant right?

[woven delta]

Okay the way I understand it is that we have to map 1 to (1, 1) and we want to map x to somewhere so that f(x^2-c) = 0

So we must have f(x)^2 = c

[shy bluff]

Yes

Ok

That makes sense I guess

[woven delta]

But we also want f(x) to be linearly independent of f(1)

[shy bluff]

Yea

[woven delta]

So a good choice for f(x) would be (1, -1) (if c was 1)

And you can modify that if you get a different c

So you would map x to (sqrt(c), -sqrt(c))

[shy bluff]

wait question

We want 1 to get mapped to (1, 1)

[woven delta]

Yes

[shy bluff]

But we map x to (sqrt(c), -sqrt(c))

[woven delta]

Yes

[shy bluff]

Oh ok so you just special case the 1?

[woven delta]

1 is the identity in R[x]

So it has to get mapped to the identity in RxR

Which is (1,1)

[shy bluff]

Like "If x not equal to identity, then map it to (sqrt(c), -sqrt(c)), else map it to identity"?

[woven delta]

No x is an element of R[x]

[shy bluff]

yea

I'm confused so as to what x we're talking about

[woven delta]

Okay so we are working with R[x]

And R[x] is generated by 1 and x as an R algebra (ie every element of R[x] can be written as a linear combination of elements of {1, x, x^2, ...}

[shy bluff]

Yes

[woven delta]

x here means the polynomial x

The element of R[x]

Okay so once we determine how our homomorphism f acts on 1 and x we will determine how it acts on the whole R[x]

We know that 1 gets mapped to (1, 1) in RxR and we want to map x to an element f(x) so that f(x)^2 = (1, 1)

[shy bluff]

hrm

ok

[woven delta]

This gives us exactly 2 choices which f(x) can be

Namely (sqrt(c), -sqrt(c)) and (-sqrt(c), sqrt(c))

We pick one of the choices

[shy bluff]

Ok

[woven delta]

And that's our homomorphism

Now we want to show that this homomorphism is surjective

How would we do that?

(this is actually just linear algebra)

[shy bluff]

uhhh we see that it a(1, 1) + b(sqrt(c), -sqrt(c)) = 0 if and only if a = b = 0

[woven delta]

Yeah we know that (1,1) and (sqrt(c), -sqrt(c)) are linearly independent, so they form a basis

By linear algebra facts

And that gives us surjectivity

Okay so we're done

I guess you should show that the kernel is exactly (x^2-c)

Actually

[shy bluff]

Oh right and we only care about 1 and x because it's R[x]/(x^2 - c)

[woven delta]

No we only care about 1 and x for R[x] in general

Okay how do you show that (x^2-c) is exactly the kernel?

(hint: use the division algorithm)

[shy bluff]

Wait I dont' think that we learnt that

[woven delta]

You haven't learned about expressing a polynomial uniquely in terms of another polynomial?

[shy bluff]

I dont' think so, not in the context of rings

Let me doeuble check, we may have but not called it such

[woven delta]

This is just given 2 polynomials p and q being able to write p = qr+s uniquely where s has strictly lesser degree than q

[shy bluff]

Yea no I dont' think that we actually learned that?

[woven delta]

You should have learned that in high school or something

It's very basic

[shy bluff]

Oh I mean we learned it in high school yes

But we can use that here?

[woven delta]

Yes

[shy bluff]

oh ok

[woven delta]

Have you never learned the Euclidean algorithm in other contexts?

[shy bluff]

I learnt it but not in the context of rings?

[woven delta]

This is in the context of fields not rings

[shy bluff]

I just thougtt we couldn't use it here lol

[woven delta]

What book are you using?

You should definitely read the section on the division algorithm in whatever book you're using

[shy bluff]

Dunnin and Foote, but the prof never refers to it

[woven delta]

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

These are the rings where you have the division algorithm

The degree function on R[x] is a Euclidean function as defined in the article

It's not hard to prove the division algorithm for polynomial rings (in a single variable) over a field

[shy bluff]

Ok, but say that we can't use the division algortihm

Can't we say that because we have S = R[x]/(x^2 - c), that all its elements are generated by 1 and x?

[woven delta]

What do you mean?

[shy bluff]

Well we have that S is a quotient ring right

[woven delta]

Sure

We know (x^2-c) is contained inside the kernel

[shy bluff]

I thinkt hat I'm ont understanding quotient rings then?

[woven delta]

Now we want to show that it is the kernel

So another thing we can do is show that x^2-c is uniquely factorizable into (x-sqrt(c))(x+sqrt(c)) (up to units in R)

[shy bluff]

Why?

[woven delta]

Why does it factor as that?

[shy bluff]

No, I seew hy it factors

But why does that help you show the kernel?

But also like I'm just trying to understand where my understanding is breaking down here

S is a polynomial quotient ring

[woven delta]

Yes

[shy bluff]

So every element of S takes on the form of s + (x^2 - c)

Yes?

[woven delta]

We don't know that though

We don't know x^2-c generates the kernel

There are other possible generators

Namely x-sqrt(c) and x+sqrt(c)

[shy bluff]

Why dont' we

Is'nt that the definition of S?

[woven delta]

No

[woven delta]

We are defining S via our map

Actually you are right

We are starting with S defined that way

And we are constructing an isomorphism between S and RxR

We are doing that by defining a surjective map out of R[x]

And showing it has kernel exactly (x^2-c)

We defined our homomorphism and showed it is surjective

[shy bluff]

oh ok

I see

[woven delta]

No we need to show the kernel of the homomorphism is exactly (x^2-c)

[shy bluff]

So here you already know the image and you're showing that where it's coming from is equal to S

Ok that makes sense

[woven delta]

Alright so we know (x^2-c) is inside our kernel

And we know that every ideal is of the form (p(x)) for some polynomial p

[shy bluff]

Ok following

[woven delta]

So if (p(x)) is an ideal that contains (x^2-c) which is not (x^2-c) then p(x) must properly divide x^2-c

So there are 2 choices you have to rule out

Namely f(x-sqrt(c))=0 and f(x+sqrt(c))=0

Why can we rule those out?

[shy bluff]

Because we have the restriction that c > 0

[woven delta]

Well we actually know what the image is

Because we defined f already

[shy bluff]

Yes

[woven delta]

And we know the image of x and the image of 1 are linearly independent

[shy bluff]

Following so far

[woven delta]

So neither of f(x) - f(sqrt(c)) = 0 or f(x) + f(sqrt(c)) = 0 are true

[shy bluff]

yea

[woven delta]

So therefore the kernel must be (x^2-c)

[shy bluff]

Oh wait

Ok now I think I get it

Nhat took a while lol

[latent anvil]

I'd never thought about it but it's super easy to prove nilradical = intersection of all primes with a little bit of scheme theory

f is in all primes iff D(f) empty iff Spec A_f empty iff A_f is the zero ring iff f is nilpotent

@next obsidian you may appreciate this/have already thought of it

I think all this uses is that D(f) and Spec A_f are isomorphic

I guess you really just need to know they're in bijection, which is what the usual proof uses

[vestal snow]

@frank solar I tried that but I wasn't able to justify how the "obvious" map between these two is also the same as the map obtained by considering the exact sequence 0 -> a (x) B -> A (x) B -> A/a (x) B -> 0

using which we get that (A (x) B)/(a (x) B) is isomorphic to A/a (x) B

[next obsidian]

Yeah Sham I did that one

I was like oh huh

I know exactly what problem initiated it too haha

[vestal snow]

Nevermind, I just realized that I defined one of the maps incorrectly

[next obsidian]

Tbh sham, the fact that the nilradical is the intersection of all primes feels sooooo fucking fake to me I’m sorry

[vestal snow]

Doesn't that rely on Zorn's Lemma?

[next obsidian]

... fuck

@latent anvil maybe you're interested in this, but I looked it up and apparently you can prove it using the compactness theorem from logic

[latent anvil]

I would be interested in that

I thought you said you skipped the Proj question?

You can see where I came up with this live on Twitter :P

[latent anvil]

@next obsidian

[next obsidian]

I did skip the proj question, I'm now curious as to why you were thinking of that haha

But uh, look at II.2.18 a) haha

That's what I thought you were looking at when thinking about nilpotence and schemes lmao

[latent anvil]

Look at II.2.14(a)

I have a solution on Twitter

It comes down to showing that if a positive degree homogenous element is in all homogenous primes, it's nilpotent

[steady axle]

Look at II.2.14(a)

[latent anvil]

?

[steady axle]

from which book?

[latent anvil]

Hartshorne

[steady axle]

ohk

[latent anvil]

@next obsidian

and you can prove that the same way as the nilradical thing I posted above

just for Proj and not Spec

[smoky cypress]

Let K be a field, then are K[X][[Y]] and K[[Y]][X] isomorphic? Where K[[Y]] is the set of all formal power series with coefficients in K.

My guess is that they are not isomorphic, but doesn't seem obvious to me to prove

[golden pasture]

they look isomorphic to me

hm

maybe not

cuz like you could get some degen f_n(x)y^n where degree of f explodes with n

[steady axle]

$$K[X][[Y]]\cong \oplus _{\mathbb Z} K[X] \cong \oplus _{\mathbb Z} \oplus _{\mathbb N} K $$

[cloud walrus]

bertwit:

[latent anvil]

I would've thought they were iso but maybe not

[steady axle]

so they are isomorphic

[latent anvil]

No bertwit

[smoky cypress]

Isomorphic as rings

[golden pasture]

uh i think you are thinking of laurent series

[latent anvil]

You might not want that whoever

[smoky cypress]

Also idk what \oplus_\bZ mean

[golden pasture]

here is formal power series

[latent anvil]

Isomorphic as k algebras could be more tractable

[smoky cypress]

idk k algebra xD

[golden pasture]

cuz like you could get some degen f_n(x)y^n where degree of f explodes with n, but x must have finite degree in the other ring

[latent anvil]

oh a ring map is a map of k algebras if it's k linear

Both of these rings are k vector spaces, right?

[smoky cypress]

Yeah

[latent anvil]

Also what bertwit wrote isn't even right

For k vector spaces

Power series isn't a direct sum

[golden pasture]

ok ive convinced myself they arent cuz K[X][[Y]] can have a unbounded degree in X

[latent anvil]

okay more general question

How is $A \otimes_k k[[t]]$ related to $A[[t]]$ if $A$ is a $k$-algebra?

[cloud walrus]

shamrock:

[latent anvil]

I think there's a map one way

[smoky cypress]

Um are they isomorphic as rings?

[latent anvil]

That's kind of my question

But I don't think they are, so I don't want to ask that

[smoky cypress]

Ok

[latent anvil]

so we have a map $\varphi : A \otimes_k k[[t]] \to A[[t]]$ given by $\varphi(a \otimes f(t)) = a f(t)$

[cloud walrus]

shamrock:

[latent anvil]

My current question is whether this is an iso

If it is, the rings you started with are isomorphic

Since $k[[y]][x] \cong k[[y]] \otimes_k k[x]$

[cloud walrus]

shamrock:

[latent anvil]

ahh yeah I think I see the issue

Why this will fail to be surjective

anything in the image of this map will eventually have all coefficients in k

It's probably injective but I don't want to prove that lol

hmm I have no idea how to actually prove these rings are different. It sounds super hard

Hmm what does Spec k[x][[y]] look like

[vestal snow]

This might not work but is there a universal property of k[[x]]?

[latent anvil]

Probably

I don't know it though

Maps out are awkward though

and I can't imagine maps in are better really

[vestal snow]

If there is, maybe you can kind of combine the universal property of polynomial ring and formal power series to get a universal property of k[[y]] (x) in terms of k

[latent anvil]

Doesn't k[[t]] only have one prime or something

It's a local ring

@next obsidian do you remember how this works?

[vestal snow]

I'm pretty sure it has more than one prime

[latent anvil]

hmm yeah you're right

It's just local

[mild laurel]

it has one prime up to units

[latent anvil]

Oh I was talking about prime ideals

[vestal snow]

Aren't 0 and the set of series with constant term 0 both prime?

[latent anvil]

Oh yeah, but I think that (t) is the only nonzero prime?

this would mean dim k[[y]] = 1

and in fact k[[y]] is noetherian because all prime ideals are finitely generated

so dim k[[y]][x] = 2

and the same is true for k[x][[y]], so this wasn't very helpful...

so my thinking is that k[x][[y]] is bigger than k[[y]][x], but I'm not sure how to quantify this

clearly not by dimension lol

Yeah what am I talking about, isn't k[[t]] a dvr???

hmm that makes me feel like I should be able to describe all the primes of k[[y]][x] explicitly

[next obsidian]

Holy shit lmao

I feel like these can't be the same

[vestal snow]

I'm trying if taking both of them to their field of fractions and showing those aren't isomorphic works

[latent anvil]

hmm weird

[golden pasture]

k[x][[y]] is a local ring i think and k[[y]][x] isnt right?

[latent anvil]

It's not

A local ring

m + (y) will be maximal for any maximal ideal m of k[x] I think

[golden pasture]

oops right

[latent anvil]

@vestal snow the fof of k[x][[y]] extends that of k[[y]][x]

Maybe you could understand that extension?

Idk the field of fractions seem pretty scary

[vestal snow]

Yeah that got nowhere

[vestal snow]

For vii), how do I show $V(a+Ann(M)) \subseteq Supp(M/aM)$?

[cloud walrus]

Have a Banana Bitch:

[next obsidian]

I feel like I did this in Matsumura. Try using Nakayama’s lemma maybe?

[vestal snow]

I tried doing contradiction by saying that suppose $p$ is in $V(a+Ann(M))$. Then $a+Ann(M) \subseteq p$. Assume that $M/aM)_p=0$. Then for every $m$ there exists a $u \in A-p$ such that $um\in aM$. For each generator, take the $u_i$ and multiply it together to get a $u$. Then $um\in aM$ for all $m \in M$

[cloud walrus]

Have a Banana Bitch:

[vestal snow]

I need to somehow show that $u$ is in $a+Ann(M)$ to get a contradiction

[cloud walrus]

Have a Banana Bitch:

[next obsidian]

I don’t see how this could work. The only assumption you started with is that a + Ann(M) is a subset of p and I don’t see how that could possibly relate to how you constructed u since this would just hold true for any prime such that (M/aM)_p = 0

Like maybe there’s something really really subtle but I really don’t see it

[vestal snow]

Oh okay

Also which matsamura book did you use?

There's two IIRC

[next obsidian]

The newer one

So here’s something

Suppose $p \in V(a)$ just at all

[cloud walrus]

Mathemagician:

[next obsidian]

then if $(M/aM)_p = 0$ then you know that $M_p = a_pM_p$

[cloud walrus]

Mathemagician:

[next obsidian]

Then since this is a local ring this says that $M_p = 0$ right?

[cloud walrus]

Mathemagician:

[next obsidian]

I think

Maybe I'm just being stupid because it's 3 am lol

but if $p \in V(a + Ann(M))$ and what I said is true that's something you can try and leverage I guess

[cloud walrus]

Mathemagician:

[vestal snow]

Alright I'll try

[next obsidian]

I don't actually see how that would be useful tho tbh

Now that I think about it more

[vestal snow]

perhaps we need to use the earlier parts

[next obsidian]

I mean $(M/aM)_p \neq 0$ says that $M_p \neq a_pM_p$

[cloud walrus]

Mathemagician:

[vestal snow]

I guess yeah

[next obsidian]

Idk, I feel like there's some cooler way to see this with some exact sequence or some bs idk

[vestal snow]

Wait but how does what you said prove the claim

[next obsidian]

Oh wait

Use parts v) and vi) I think

From v) you know that Supp(M) = V(Ann(M)) and from vi) you know that Supp(M/aM = M (x)_A A/a) = Supp(M) \cap Supp(A/a) = V(Ann(M)) \cap V(a) = V(a + Ann(M))

Lol

@vestal snow

[vestal snow]

Yup!

That was easy

Thanks

[next obsidian]

Assuming you've already done the other parts haha

[vestal snow]

I did

I have a bad habit of forgetting the previous exercise/part when I move on the the next one

[next obsidian]

Oof. For a multi part one like this that usually ends up being key

[vestal snow]

Are you an undergraduate or graduate student?

[next obsidian]

Undergrad

[vestal snow]

Just a small thing that's bothering me about that solution

Oh wait nvm

yeah I'm probably just tired af

Cool! I'm an undergraduate too

Rising sophomore

[next obsidian]

Oh yeah I’m a rising junior

[autumn plover]

If S is a simple R-module (R not necessarily commutative), and $n>k$, how do I show that there's no surjective R-module map $S^k -> S^n$?

[cloud walrus]

Ohad:

Ohad:

[vestal snow]

If f: A -> B is a homomorphism and every prime ideal of A is a contracted ideal, then f*: Spec(B) -> Spec(A) is surjective?

How would I prove this?

any prime ideal p of A can be written as q^c, but we don't know if there is a way to write it where q itself is also prime

[vestal snow]

The book claims that it is obvious, but I don't see how

[next obsidian]

Yeah I don't see how this is true in general

You need some sort of going up type theorem I feel

I don't know any way to like fatten up any ideal into a minimal prime ideal containing it, you can get a minimal prime but the fact you have choices would lead me to believe there's no reason it would pullback to the same ideal as the original

[woven delta]

is the set ${\sin(q)|q\in (0,\pi)\cap \mathbb{Q}}$ linearly independent over $\mathbb{Q}$?

[cloud walrus]

Liquid:

[next obsidian]

...

Evil

My gut says yes but I have no idea how I’d show that

[oblique river]

the question is much easier if you replace ${\sin(q)|q\in (0,\pi)\cap \mathbb{Q}}$ with ${\sin(q)|q\in (0,\pi)} \cap \mathbb{Q}$

[cloud walrus]

Buncho Bananas:

[woven delta]

lmao

[oblique river]

:^)

[kindred mist]

Am I correct that if we take a noncommutative unital domain D, then Frac(D) will just be a noncommutative (unital) DVR?

[cloud walrus]

𝓒𝓸𝓾𝓷𝓽𝓪𝓫𝓵𝓮:

[latent anvil]

I'm not sure that Frac makes sense here

Like, won't you need a left vs right division?

[kindred mist]

hmm ok yeah idk I suppose so

[latent anvil]

I also have no idea what a noncommutative DVR is

I would've assumed Frac(R) would be a division ring

[kindred mist]

I'm just using DVR to mean division ring

my bad

[latent anvil]

Oh, DVR means something else

Discrete valuation ring

[kindred mist]

ack... rip my notes >_<

There does exist noncommutative division rings I think, I guess we just might not be able to form them with Frac, I was having trouble finding an example with google though

(I thought matrices but GL isn't a ring, and M_n isnt a division ring)

[latent anvil]

there are noncommutative division rings, but I don't think there's a good notion of frac for a noncommutative domain

Or maybe there is, but it's significantly uglier

[kindred mist]

maybe quaterion-valued rational functions (for noncommutative division ring)?

[latent anvil]

What do you mean?

Do you just want an example of a division ring which isn't a field?

[cloud walrus]

𝓒𝓸𝓾𝓷𝓽𝓪𝓫𝓵𝓮:

[kindred mist]

sorry about that, thanks

[latent anvil]

if a subring of a ring is zero, the whole ring is zero

That's kind of woke

[latent anvil]

someone gave me homework on Twitter so I'm going to try and work it out here

I am a little confused but hopefully I can work it out on my own

Let A be a k algebra. I'm being told to prove that a Z grading on A is the same thing as an action of G_m = Spec k[x, x¯¹] on Spec A

Note that G_m ×_k Spec A = Spec(k[x, x¯¹] (×)_k A) = Spec(A[x, x¯¹]), so the action map G_m ×_k Spec A -> Spec A is the same as a k-algebra map A -> A[x, x¯¹] satisfying some conditions

Okay so what are those conditions?

Let's let ρ : G_m ×_k Spec A -> Spec A be the action map and φ : A -> A[x, x¯¹] the corresponding k-algebra map

Hmm I think it'll be pretty nice to write down. I'm gonna have some diagrams in k-Sch expressing that ρ is a group action, defined in terms of the fiber product and the unit Spec k, but everything's affine so those diagrams are exactly dual to something in k-Alg

[latent anvil]

so one thing I want to understand is the counit and comultiplication maps in k[x, x¯¹]

I think the counit just sends x to 1

On schemes the unit map should send the single point of Spec k to the point (x-1) in G_m, right???

I have elected not to do this problem because thinking about group schemes is making my brain unhappy

[charred pewter]

If two disjoint cycles sigma and tau have order r, then does (sigma x tau)^r=e (identity)

just making sure

[delicate bloom]

yep

[vestal snow]

@next obsidian Is this problem incorrect then? Any idea how to fix it? We haven't done integral extensions yet so I think there might some other additional condition you can add to make it work

[next obsidian]

I mean it might be correct?

I just have no idea how you would show it

Perhaps somehow the fact that every prime is a contracted ideal implies some sort of going up type thing?

I think actually it must be equivalent since I’m pretty sure f* surjective is the same as the going up property, but I might be mixing some stuff up

I guess you need to figure out some way to figure out there’s some sort of prime. Try Zorn’s I guess?

Oh wait I think I see how to do this actually

Let P be a prime of A, and let S be A\P, then S is a multiplicative system. It follows that T = f(S) is a multiplicative system, so consider the poset of ideals disjoint from f(S). This is nonempty since there is an ideal I of B such that f^-1(I) = P, but if I intersected T then f^-1(I) would intersect S, a contradiction. By Zorn’s there’s a maximal ideal in this Poset, and you should know that maximal ideals which are disjoint from multiplicative systems are prime

I think you have to prove it in order to complete the proof that nilradical = intersection of all primes

@vestal snow

So apparently this is equivalent to the going up condition @dapper nebula you might be interested in this too

Since going up is equivalent to the map on Specs being surjective I’m pretty sure, I know it implies it, and I’m pretty sure the other direction holds as well

[vestal snow]

@next obsidian Here by maximal ideals you mean the maximal element of the poset, right?

[next obsidian]

Yeah maximal wrt the poset

[vestal snow]

Why are maximal ideals disjoint from multiplicative systems prime?

Nevermind I figured it out

Is this a commonly known result?

Or did you come up with it while doing this problem?

[shy bluff]

So for this question, where K is a field, how do I start it? I'm looking at the definition of a field and it seems to be the set of units of K?

[vestal snow]

(ab) must be a maximal ideal

So that gets rid of the possibility of both a and b being units

[shy bluff]

Wait we haven't learnt about maximal ideals yet

[vestal snow]

I guess you can still do it using quotient rings instead of maximal ideals, but its a bit of a pain

[shy bluff]

Yea I feel like taht's how they want us to solve it

[elder valley]

Hint: fields don't have zero divisors

[shy bluff]

oh wait this is true

wait

If both of them are units, then there exist c, d such that ac = 1, bd = 1

Then every element k in K[x]/(ab) takes on the form of k + ab right, where k is an element of K[x]?

wait actually, another question, if K[x]/(ab) is a field, then it must satisfy this right?

In this case R^x is K[x]/(ab) \ {0}?

[woven delta]

No

[shy bluff]

Oh

[woven delta]

Yes

[shy bluff]

wait what

[woven delta]

This just means that every element is a unit

You don't have to be fancy

[shy bluff]

Oh so every element of a fiield is a unit

[woven delta]

Every nonzero element

Yeah

[shy bluff]

and we can't have 0 elements in it?

[woven delta]

What do you mean?

0 divisors?

[shy bluff]

We can't have 0 itself in the field

rIght?

[woven delta]

No I said every nonzero element

[shy bluff]

Oh wait no

[woven delta]

Is a unit

[shy bluff]

Ok I see what you mean by zero divisors

Ok I see

[woven delta]

Units can't be zero divisors

[shy bluff]

Yea

Ok sorry I derped

[woven delta]

Anyway if a and b are both units

[shy bluff]

Ok so we have to have that every k in there is a unit

[woven delta]

Then ab is also a unit, right?

[shy bluff]

yea

[woven delta]

So why can't k[x]/(ab) be a field?

(if a and b are both units)

[shy bluff]

I don't know, I dont' really see how that affects K[x] is the thing? Like all that I really understand is that

• elements of K[x]/(ab) take on the form of k + ab
• ab is a unit, and so is k?

[woven delta]

What is the ideal generated by a unit?

[shy bluff]

hrm

If we have a unit i, then the ideal generated by it must contain all ir, ri for all r in R, and so the identity is included and it ends up generating the entire ring?

[woven delta]

Okay so what is R/R?

[shy bluff]

I... don't know

[woven delta]

Think about it

[shy bluff]

Oh wait

It'd all just go to 0

right?

[woven delta]

Yep

So you just get the 0 ring

Can the 0 ring be a field?

[shy bluff]

no

Oh it's because everything is in the same equivalence class

[woven delta]

Yeah

Part of the requirements for a field is 1 \neq 0

So the 0 ring is not a field

[shy bluff]

Yea

oooh ok

So if we have that both a and b are unit,s then we have that the ideal generated by ab is the entire ring

And thus we have that K[x]/(ab) becomes the 0 ring, which it can't be because we've said tha t it has to be a field

oh

[woven delta]

Okay so that's why both can't be units

Why does one have to be a unit?

[shy bluff]

If we have that just one is a unit, then their product is not a unit, thus it does not generate the entire ring, and thus it can be a field?

[woven delta]

This doesn't quite give us what we want

[shy bluff]

hrm

Wait but is the part about their product not being a unit correct

[woven delta]

Yeah their product won't be a unit

[shy bluff]

ok thank god that was a guess

[woven delta]

Do you know about irreducible elements?

[shy bluff]

oh wait that makes sense, we have a group of units

[woven delta]

Yeah

[shy bluff]

No, we have'nt learnt about irreducible elements

[woven delta]

Hmm

Okay suppose both elements are not units

[shy bluff]

Sure

[woven delta]

You want to show that k[x]/(ab) is not a field

What can go wrong?

[shy bluff]

well, if it's a field then we need that everything in it is a unit, and also that 1 is not 0 in it?

[woven delta]

Yeah

All nonzero elements are units

[shy bluff]

And if it's a unit then there has to be an inverse element

[woven delta]

Yes

[shy bluff]

hrm but ab isn't a unit though right?

[woven delta]

In k[x]/(ab) we have ab=0

Aureva gave a hint earlier re zero divisors

They said that fields can't have zero divisors

[shy bluff]

wait why do we have ab = 0

oh right

Yea it gets sent to the equivalence class of 0 right?

[woven delta]

Yes

What zero divisors can exist in k[x]/(ab)?

[shy bluff]

None, because it's a field right?

[woven delta]

Well we want to show it's not a field

Assuming a and b are both not units

[shy bluff]

We would have that ab = 0

which would imply that we have a zero divisor?

[woven delta]

What would your 0 divisors be?

[shy bluff]

wouldn't it be any multiple of ab?

[woven delta]

No that would just be 0

You want a pair of nonzero elements whose product is 0

[shy bluff]

Oh

[woven delta]

In other words you want c+(ab) and d+(ab) with cd\in (ab)

[shy bluff]

ok, but why does both a and b not being units amke that happen?

[woven delta]

Well because a and b are not in (ab)

And ab is

[shy bluff]

wait

(ab) is the ideal generated by ab right?

[woven delta]

Yes

So you have to use a fact here

[shy bluff]

so it's, for all r in R, rab, abr in (ab)

[woven delta]

About k[x] being the polynomial ring over k

Where k is a field

To show that a and b cannot be in (ab)

Namely you have to use the degree function

[shy bluff]

Hrm

degree(ab) = deg(a) + deg(b)

[woven delta]

Yes

[shy bluff]

but it's not a unit so its degree can't be 0?

[woven delta]

As long as neither of them is 0

But that's a case that's not too hard to dismiss

[shy bluff]

oh right, because neither a nor b are units, their degrees are both non-zero, and thus the degree of deg(ab) must be 0?

[woven delta]

deg(ab) must be strictly greater than deg(a) and deg(b)

So that tells us a and b can't be in (ab)

[shy bluff]

why must it be strictly greater

Because ti's a product of two positive integers?

[woven delta]

Because (ab) is the set pab where p is an element of k[x]

And so the degree of every nonzero element in (ab) is greater than or equal to deg(ab)

[shy bluff]

The degree of every nonzero element in (ab) is greater than or equal to degree (ab).... I'm a bit lost, can you come up with an example of a concrete ring and values of a and b?

[woven delta]

Sorry we are talking about polynomials

So say $\mathbb{R}[x]$ as an example

[cloud walrus]

Liquid:

[shy bluff]

sure

[woven delta]

Let $a=x^2+1$ and $b=x^3+3$

[cloud walrus]

Liquid:

[shy bluff]

sure

Those are both irreducible polynomials over R, which means that they are not units

Right?

[woven delta]

Then the leading term of ab is x^5

[shy bluff]

yes

[woven delta]

And any nonzero element of (ab) must have degree at least 5

[shy bluff]

ok sure

So then what does (ab) look like?

Is it just all polynomials with degree greater than 5?

[woven delta]

It looks like $p(x) (x^5+x^3+3x^2+3)$

[cloud walrus]

Liquid:

[woven delta]

Where p(x) is any polynomial

[shy bluff]

Oh so it's just the product of any polynomial and that

[woven delta]

Yeah

You can use this idea of degree to show that a and b are not in (ab)

And then you can use that to show that $a+(ab)$ and $b+(ab)$ are zero divisors in $k[x]/(ab)$

[cloud walrus]

Liquid:

[woven delta]

And I think that you should be able to work out things from here

[shy bluff]

hrmmm I'll sleep on it... I think that it's just too late lol

[woven delta]

That's fair

[shy bluff]

Wait question, why is cd in (ab)? We have that if it's an ideal then we have that cdab is in(ab) isnt' it? Not just cd?

[shy bluff]

Wait how can we have a, b in K[x] but not in (ab)? by definition, we have to have that a(ab) in (ab) right? As (ab) is an ideal?

and a is an element of K[x]?

like in order for it to be an ideal, it must have that for all r in R, i in I, ri, ir in I right?

And here we have that our ideal is (ab), and our ring K[x], so we have to have that for all k in K[x] that kab in (ab)

so we have to have aab in (ab) no?

[plucky bolt]

let S be an abelian subgroup of the pauli group on n qubits such that S does not contain -1. let G be a set of generators for S. let g be an element of G. show that there exists an element p of the pauli group such that p anticommutes with g but commutes with every other element of G

whys this true/is it obvious?

they just included this fact casually as part of a proof but i dont see why it's true immediately

[elder valley]

Wait how can we have a, b in K[x] but not in (ab)? by definition, we have to have that a(ab) in (ab) right? As (ab) is an ideal?
@shy bluff yes that's correct. aab and bab are in (ab), but these aren't necessarily equal to a and b

[shy bluff]

Oh right because a and b are a + (ab) and b + (ab)?

[elder valley]

Those are their images in the quotient yeah

I'm not sure what you're trying to prove, I didn't follow the whole conversation

[shy bluff]

And by "show that a + (ab) and b + (ab) are zero divisors", what does that mean?

[solemn rain]

(a+(ab))c for some nonzero c is 0

thats what it means for a+(ab) to be a zero divisor

[elder valley]

It means they're nonzero in the quotient and their product is 0

[shy bluff]

Show that if K[x]/(ab) is a field, exactly one of a or b is a unit

This was the neither is a unit case

Ok so and that's just ac + (ab) right?

Oh if c is just b, then it just goes to ab + (ab) which is 0 in this?

[elder valley]

Yeah, but it's b+(ab), not b

[shy bluff]

Err yea

[solemn rain]

try multiplying

(a+(ab))(b+(ab))

[shy bluff]

And so if neither a nor b is a unit, we see that (a + (ab))(b + (ab)) = ab + (ab), which is just equal to 0 in this quotient ring?

[elder valley]

That's true regardless if they're units or not

[solemn rain]

maybe not?

a and b not being units

gurantees that ab wnt be 1

right?

[shy bluff]

Oh

Hrm so I want to find c, d not equal to a or b such that (a + (ab))c = (b + (ab))d = 0?

Or rather, show that such a c and d exist?

[elder valley]

No, you just want to show the quotient has zero divisors, which you've almost done above

[shy bluff]

Isn't that showing that the quotient has zero divisors?

[elder valley]

That would do it but you don't need them different from a and b

[shy bluff]

Wait I'm confused, you said that that's true regardless of if they're units or not

Right?

[elder valley]

Yes

Meaning you haven't used that condition yet

[shy bluff]

The condition of neither are units?

[elder valley]

Yeah

[shy bluff]

If neither of units, than we know that ab will never be identity right?

[elder valley]

Sure

[shy bluff]

And there can't be c such that ac = 1 or bc = 1

[elder valley]

Yeah

[shy bluff]

Is this headde in the right direction lol

[solemn rain]

if ab =1 ab+(ab) is not 0

( if 1 is in (ab),(ab) is the whole ring )

so i think u used

[elder valley]

Hmm not really. You need to show zero divisors. You got half of it up there by showing their product is 0

[solemn rain]

the condition?

[shy bluff]

@solemn rain that's for the both are units case

[elder valley]

ab+(ab) is always 0 because ab is in (ab)

[shy bluff]

Yea

What if I multiply this by some c and we get that neither a nor b nor c are 0 but their product is 0?

[elder valley]

You don't need to. a+(ab) and b+(ab) will be your two zero divisors, you just need to show it

[solemn rain]

oh okay

lmao mb

[shy bluff]

To show that the product is 0 I want to show that the product is equal to some multiple of ab right?

[solemn rain]

brain bad

[elder valley]

Pause for a sec. If I is an ideal of R and x,y are in R, what's the condition that x+I is equal to y+I?

[shy bluff]

If they're sent to the same equivalence classes?

[elder valley]

And how do you know whether the equivalence class will be the same?

[shy bluff]

That is a uh good question

And I am not sure

[elder valley]

x+I=y+I if and only if x-y is in I

[shy bluff]

Ok

[elder valley]

That's why ab+(ab) = 0+(ab), because ab-0 is in (ab)

[shy bluff]

Ok but what does it mean to show that a + (ab) is a zero divisor?

It means to show that there exists c such that (a + (ab))c = 0 right

[elder valley]

It means it's nonzero and that you can multiply it with another nonzero element to get zero

[shy bluff]

Ok, so we want to show that there exists non zero c in K[x]/(ab) such that (a + (ab))c = 0

[elder valley]

You've shown that (a+(ab))(b+(ab))=0+(ab), so if you show that a+(ab) and b+(ab) are nonzero you're done

Then they will be zero divisors

[shy bluff]

Oh wait what

Oh so that first bit already proved that they

Oh

I just needed to show that a, b are nonzero

[elder valley]

You need that a+(ab) is nonzero not that a is nonzero. Same for b

[shy bluff]

Well a + (ab) nonzero means that a not in (ab)?

[elder valley]

Yeah

[shy bluff]

Then I want to show that ai not in (ab) for some i in (ab)?

[elder valley]

I would do a contradiction. Suppose a is in (ab)

[shy bluff]

If a is in (ab) then we have that for all k in K[x], ak, ka in (ab)

[elder valley]

That's not right. What does the set (ab) look like

[shy bluff]

I am not sure? We have {abr for all k in K[x] } I think?

Err

Abk

[elder valley]

Yeah, so a must be one of those elements if it's in there

[shy bluff]

(ab) = {abk for all k in K[x]}

Suppose a is in (ab), then we have that aab, aba in (ab)?

[elder valley]

No, it's that a=kab for some k

[shy bluff]

Wait why

Oh right

Ok I see

[elder valley]

Because things in (ab) have the form kab and a is one of them

Or abk, doesn't matter

[shy bluff]

So then a = kab for some k in K[x], but this is a contradiction as we'd need k = b^-1 but b is not a unit so it doesn't have an inverse?

[elder valley]

You can give a better argument. Rearrange the equation

[shy bluff]

a = kab <=> a - kab = 0, but a is not zero?

[elder valley]

Factor

[shy bluff]

Oh I can factor that?

a(1 - kb) = 0, which implies that either a is 0 or that k in an inverse of b, which is a contradiction?

[elder valley]

Yep. You assumed b wasn't a unit so that case is a clear contradiction. Why is a=0 bad though

[shy bluff]

Well we have that a is nonzero in the first place don't we

[elder valley]

Was that assumed in the problem?

[shy bluff]

Ah, that's not specified

Hrm a = 0 is bad because then (ab) is just 0?

[elder valley]

Yes, why is that bad

[shy bluff]

Then we're quotienting by 0

But idk how that works for rings lol

[elder valley]

I've gotten slightly lost in our proof method. We assumed a and b weren't units and trying to show the quotient isn't a field

[shy bluff]

Yes

[elder valley]

To do that we show a+(ab) and b+(ab) are zero divisors of each other

[shy bluff]

Sure

That's what we were doing earlier yea?

[elder valley]

So if a=0, we get that (ab)=(0), so that means K[x]/(ab) is isomorphic to K[x] because quotienting by 0 does nothing

Yeah I was just trying to keep track of where we were

[shy bluff]

Oh wait

Quotienting by 0 does nothing?

Well I guess that makes sense

But also wourd bc dividing by 0?

[elder valley]

No it's quotient ring, not actual division

[shy bluff]

Yea

Yea ok because the quotient is basicaly everything thaat gets sent to 0

And if your quotinet is just 0 then only 0 is sent to 0

Ok

[elder valley]

Kernel you mean*

[shy bluff]

Pardon?

You mean the quotient function?

[elder valley]

Well you can think of it as collapsing the ideal to 0, but that's the same as the kernel of the projection map K[x]-->K[x]/(ab)

[shy bluff]

Yea the kernel of the projection map is just 0

If ab = 0

[elder valley]

Yeah so we get the quotient is just K[x]

Which is never a field

[shy bluff]

Oh right it has to be K[x]^x that's a field

[elder valley]

What

This proof was very messy

I wouldn't have done contrapositive in the first place

[shy bluff]

Oh uh K[x] isn't a field bc x doesn't have an inverse right

[elder valley]

Right

[shy bluff]

Oke

Thank you for the help, sorry for being obtuse :x
I'm having a hard time understanding the material, to the point where idk what I can even ask unforhunately

[elder valley]

you're getting better with quotients i think 🙂

[shy bluff]

I kind of get what it is, that it's just the ring with everything sent to equivalence classes and that everything that's in your ideal is basically a thing to send to 0

Like I don't know what I can and can't do in them though

Liike factoring

That was a surprise to me

[elder valley]

yeah basically. the important things to remember in getting started in my opinion are:

1. elements look like r+I for any r in R
2. operations work as r+I + s+I = r+s+I, and (r+I)(s+I) = rs + I
3. r+I=s+I if and only if r-s in I

you can factor in any ring, and we factored in K[x], not in the quotient

[shy bluff]

Err yea but like "You can factor in a ring?????" Was my thought when you said factor

Which makes sense as it's commutative

[elder valley]

oh

that's one of the defining properties of a ring

x*(y+z) = xy + xz for all x,y,z in R

[shy bluff]

Oh right

But like factor out from the middle

As we had a - kab

But I guess same thing

[elder valley]

oh you need commutativity for that

[shy bluff]

Yea

That bit is commutativity

[shy bluff]

how does this work? so if it's an ideal it's closed under subtraction, so I see that the first one, the sum, is an ideal

but what about the other two?O

[timid hull]

Obviously first and second holds good and the 3rd doesn't hold

[stone fulcrum]

Check the conditions!

Usually "additive subgroup" is pretty easy to break

[timid hull]

Just multiply f(x) by 2 and conclude that 3 doesn't hold

[shy bluff]

what do you mean by multiply it by 2?

[smoky cypress]

Ideals are closed under multiplication, and 2 is an element of Z[x]

[oak perch]

One of the many stipulations for an ideal is its elements are closed under addition.

Notice: f(0) = odd means f(0) = 3 and g(0) = 5 are in the third set.

Notice also: Their sum f(0) + g(0) = 8 is not in the set.

I'm assuming the operations are the "normal" sum of polynomials and the "normal" product of polynomials.

[whole basalt]

Just to make sure I understand the intuition here

Say M is an R-module. Then Ann(M) is the set of all ring coefficients that when multiplied by M give zero?

So for example if you look at Z/6Z as a Z-module, then Ann(Z/6Z) = 6Z?

[oblique river]

Ann(M) is a subset of R, yes

[smoky cypress]

I found this beautiful stackexchange answer

https://math.stackexchange.com/questions/501275/is-the-ideal-generated-by-an-irreducible-polynomial-prime/816406

[solemn rain]

probably sums up like

50% of basic ring theory lmfao

[latent anvil]

If I were an integral domain I would simply be a PID

[next obsidian]

if I were an integral domain I would simply be a field

[stone fulcrum]

If I were an integral domain my multiplication wouldn't be associative

[next obsidian]

🤢

Imagine posing as a ring without associative multiplication

[smoky cypress]

Are non associative operations even interesting?

Ah I guess cross product

[woven delta]

Exponentiation

[glad juniper]

Cross product isn't even the right operation. The wedge product is associative.
Exponentiation isn't a real operation, it's just using exp and products and your favorite branch of log.
No one cares about the octonions.
Therefore, non-associative operations aren't interesting

Also, a-(b-c)=(a-b)-c, right?

I guess subtraction and division also aren't real operations since they're just addition and multiplication with inverses

[woven delta]

What do you mean "not an operation"?

[smoky cypress]

Also, a-(b-c)=(a-b)-c, right?
@glad juniper

An operation is just a function S^n->S, and a binary operation is when n=2

[glad juniper]

That's why I said "aren't real operations" rather than "aren't operations". I mean morally, the things worth elevating to the status of "operation" that aren't easily definable in terms of other important stuff, tend to be associative. Other things we call "operations" technically satisfy Whoever's broad definition, but I'm arguing, tongue half in cheek, that they aren't deserving of the name

[woven delta]

Lol "morally"

I don't agree with that statement at all

Also lie algebras have operations which aren't associative

The cross product being an example

[glad juniper]

To be clear to anyone reading, I was using "morally" in the weird mathematician sense (see http://eugeniacheng.com/wp-content/uploads/2017/02/cheng-morality.pdf) not the usual English sense.

In the study of Lie algebras, the fact that they call it a "bracket" and write it with brackets instead of some operation symbol like × or ★ seems like evidence it's not like a usual "operation".

The cross product is like, an accident because we happen to live in 3D. If we lived in 4D no one would name that as an operation

[woven delta]

I know, I hate the use of the word morally

[glad juniper]

fair

[woven delta]

And it gets called an operation