groups-rings-fields

[scarlet estuary]

the isomorphism, in this case, is very obvious

just multiply the elements of the first group by 10

[old hollow]

Thank you, yeah I didn't understand what was meant by "structure preserving" but

Ok

[scarlet estuary]

its easy to check that this is a homomorphism

f(a+b) = 10*(a+b) = 10*a + 10*b = f(a) + f(b)

and its also clearly bijective (the inverse is just dividing by 10)

hence its an isomorphism

[old hollow]

Ok wow thank you

[scarlet estuary]

the isomorphism isnt always that simple

as previously mentioned, the isomorphism from (R, +) to (R>0, *) is exp

[old hollow]

Is there a simple example of a homomorphism that isn't isomorphic

[scarlet estuary]

sure, map everything to 0.

(that is, the identity)

[old hollow]

Well, I guess like a "good"example like an actual function I guess

[scenic sage]

f(x) = 0

[old hollow]

I know mapping to 0 is a function but

[scarlet estuary]

there's still plenty of examples of those

for example

take the group of integers under +

and the group of integers mod 5 under +

then you can construct a homomorphism from the first group to the second

(just take every integer mod 5)

[scenic sage]

you can in general take any two groups with not equal order

[scarlet estuary]

but this homomorphism certainly isnt invertible

after all, would 3 map back to -2, 3, 8? something else?

[scenic sage]

then there is no possible isomorphism

[scarlet estuary]

yeah groups are isomorphic iff everything about them is the same

you can always construct homomorphisms betwene groups

usually multiple

but they're only isomorphic in special circumstances

[old hollow]

Ah, yeah that's interesting that there's infinitely many numbers 3 would map back to

[scarlet estuary]

yeah, the map in that example is surjective but not injective

hence its not invertible

[old hollow]

Surjective means every codomain value has 1 or less elements that map to it from domain?

[scarlet estuary]

1 or more

[old hollow]

Oh shoot ok

[scarlet estuary]

injective would be 1 or less

[old hollow]

Injective and surjective is bijective then

[scarlet estuary]

yeah essentially

well not "essentially"

literally

thats literally what bijective means

[old hollow]

Lol

[scarlet estuary]

its where "bi" comes from

"two"

and this is equivalent to being invertible

(the proof of which is probably a good exercise if you're unfamiliar with it)

[old hollow]

No, I'm not familiar

So the exercise would be to prove bijective mappings are invertible?

[scarlet estuary]

and invertible functions are bijective

it's an iff statement

[old hollow]

Oh ok

[scarlet estuary]

so for one direction, you can assume an arbitrary function f is both injective and surjective, and want to prove that it has an inverse

for the other direction, you can assume that an arbitrary function f has an inverse f^-1, and want to prove that f is both injective and surjective

its just definition-pushing

[old hollow]

Ah

[thorn delta]

lol hello @old hollow

[old hollow]

Oh hi lol

[dawn kiln]

Yo what's up @old hollow

[shy bluff]

how does part b here work? what does the leading coefficient of f being a unit have to do with anything?

Like I get that R[x]/(f) is isomorphic to polynomials of degreee less than or equal to n

But what does the unit bit have to do with anything?

[oblique river]

what you wrote isn't true

if the leading coefficient isn't a unit

that's the whole point of the problem

when the leading coefficient is a unit, what you wrote is true

[shy bluff]

hrm

A unit just means that it's invertible with respect to multiplication right

[oblique river]

yes

that's the definition

also to be clear, it's talking about the leading coefficient being a unit in the ring R

[shy bluff]

yes

And the other coefficients we dont' care about right

[oblique river]

correct

the only thing that matters is the leading coefficient

[shy bluff]

Wait so leading coefficinte, it's just like... if it takes on the from of a_nx^n + ... a_1x + a_0 it just wants a_n in R to be invertible wrt multiplication?

[oblique river]

yes

[shy bluff]

hrm

I dont' thnik that I see what the unit bit has to do with this as this is supposed tob e isomorphic as a group right?

[oblique river]

why don't you try to write down a proof

or just look at an example

Z[x] / (x-1) is isomorphic to Z

but Z[x] / (2x - 1) isn't

or maybe even better, Z[x] / x is isomorphic to Z but Z[x] / 2x isn't

can you see why?

[shy bluff]

I think that I have a problem with understanding quotients then

Well I think I understand but I don't know how to formalize

For Z[x]/x it's because 1 is a unit in R, so call it sinverse a or such, and then we can write aZ[x], but 2 isn't a unit, so any element of Z[x] divided by 2 doesn't mean anything?

If that makes sense?

[oblique river]

I think I understand what you're saying

and yes that sounds correct

see if you can see where that would come up in the proof

[shy bluff]

well to show that R[x] / (f) is congruent to R[x]^(n) we'd want a quotient function whose kernel is (f) right?

err sorry isomorphic

[oblique river]

we're only asking for isomorphic as a group

but I guess yeah if you just think of R[x]/(f) as a group

then sure you can think of it that way

[shy bluff]

wait wdym? don't you have isomorphism theorems for both rings and groups?

[oblique river]

yeah

I'm just saying make sure to not confuse them

[shy bluff]

ah ok

[oblique river]

also I gotta go -- sry to step out on you but hope you can make some progress

[shy bluff]

oke

[chilly ocean]

@shy bluff hey I am a junior and senior this school year and graduate HS this year and wanted someone to talk to if u want accept my friend request

I wanted to major in medical research btw

[shy bluff]

Ok...?

[molten pendant]

can a product of non-finitely generated ideals be finitely generated

[latent anvil]

That sounds plausible

Oh I think I have an example maybe

Let A = k[x1, x2, x3,...]

Let J = (x1, x2, x3,...)

Then J is a non finitely generated ideal of A/J² but J*J = 0

@molten pendant

[molten pendant]

ah very nice

thanks

[shy bluff]

Wait question, is 2x not in (x)?

[molten pendant]

it is

if 2 is in the ring at least

[shy bluff]

Wait another question

Hrm

No nevermind I try this again first

Yea ok @oblique river I got nothing, I tried to write it out using the exmalpe of Z[x]/x and Z[x]/(2x) to compare them

but I'm not gettintg anywhere

Actually wait I can define f to be a function that takes a_nx^n + ... + a_0 and sums all the odd componets and discrads the evene ones?

[shy bluff]

I think that its kernel is (2x)

Like I think that this $$\phi(a_nx^n + \cdots a_1x + a_0) = \sum_{i = 0}^{n/2}a_{2i + 1}$$

[cloud walrus]

Liria ^(;,;)^:

[shy bluff]

Is a function that goes from Z[x] to Z and has kernel (2x)?

... no it needs to sum up all the terms that have odd coefficients, not the ones that are odd-placed

[oblique river]

Think of the proof for why Z[x]/x is isomprhic to Z

for any polynomial, you discard all of the higher power terms

because they're all multiples of x

but that same argument doesn't work with Z[x] / 2x

the polynomial x is nonzero in tehre

as is x^2

and x^3

and x^4

and those are all distinct

[latent anvil]

Let A be a ring which isn't a field

if I say "A-representation of G", would you assume I mean a linear action of G on an A-module or a homomorphism of G into GL(n, A)

[shy bluff]

Hrmmmm how do you write that thouch @oblique river

I can see that it's true but idk how to prove it?

@oblique river is it because if you have that the leading coefficient is a unit, we have that x^n is in the ideal generated by f, and so we can go and quotientmit out?

[oblique river]

@shy bluff not exactly, x isn't in the ideal generated by x - 1 for example

@latent anvil those are nearly the same thing but the first is more general. that said I'd probably assume the second.

[shy bluff]

Hrmmm

[latent anvil]

We're talking about the second type of thing in my reu paper (specifically representations of the braid group) and I was feeling a little ick about it

I guess it's probably fine, it's what the papers/books I've been looking at say

[runic palm]

is this correct guys?

[latent anvil]

@runic palm post this in #prealg-and-algebra

[runic palm]

ah yeah sorry

[shy bluff]

@oblique river do you have any further hints

[oblique river]

one way to prove the statement is to use the division algorithm for polynomials

which only works if the thing yuou're dividing by has a leading coefficient as aunit

[next obsidian]

@shy bluff If the first element is a unit you can assume WLOG that the polynomial is monic by just multiplying by the inverse of the leading coefficient and if (f) = (g) if and only if f and g differ by a unit

Now assume for simplicity's sake that f is irreducible, and monic, and degree n, then do you think you can show R[x]/(f) is isomorphic to R^n?

If you can do that then you'll actually be done

(I think)

[shy bluff]

What does it mean for a polynomial to be monic?

[next obsidian]

leading coefficent is 1

This is maybe a situation you've seen before like Z[x]/(x^2 + 1)

[shy bluff]

ah

hrm

[next obsidian]

If you can do that then the Chinese Remainder Theorem should give you the general result

[shy bluff]

Well I just checked in our notes, and we have this which I guess is what I'm trying to prove

We have'nt learnedthe chinese remaindetheroem yet at this point

[next obsidian]

Oh oof

[shy bluff]

Or at least I don't think that we can use it at this point

[next obsidian]

I mean you can always just prove it

But if you did have it you can just break f up into a product of irreducibles

[shy bluff]

So is this basically just proving the division algorithm? Like show that there's a function phi that is the quotient of (f) and that this is equal to just doing the division algorithm when f has leading coefficient being a unit?

[elder valley]

I would say use first isomorphism theorem with that division algorithm. No need for irreducibility or monic

[shy bluff]

Yea I just need to find the function that works with for the first isomorphism theore m

And the function that works for that is the division algorithm right?

[elder valley]

Depends what you mean by that. Division algorithm isn't a function itself

[shy bluff]

hrm

Well how does the quotient function work then

[elder valley]

Well it would map from R[x] to R[x]^(n), meaning the input would be any polynomial and the output needs to be a polynomial of degree less than n

[shy bluff]

hrm

Yea then that's not just the quotient function right

[elder valley]

You mean the polynomial q in the theorem?

[shy bluff]

Isn't the theorem just this?

[elder valley]

Yeah. Wasn't sure what you mean by quotient function. If you're using that then you need to define a phi

[shy bluff]

Yea, then I need to define a phi that sends polynomials in R[x] to R[x]^(n)

I would say "just truncate it"

but that doesn't seem right

[elder valley]

That would give you a map, but it won't have the right kernel because the higher order terms don't matter in that case

[shy bluff]

yea

I need the kernel to be f

Err to be (f)

[elder valley]

Yeah. So suppose you want to define phi(g), which should be some polynomial of degree <n. And since the kernel should be (f) you should use f somehow to define phi(g)

[shy bluff]

phi should send the polynomial g to q(x)d(x) + (f) right?

Err sorry + some element in (f)

[elder valley]

What are q and d?

[shy bluff]

um dunno that's the division algorithm though right

[elder valley]

Division algorithm takes 2 polynomials as input. The theorem you linked up there started with f and g, and it states the existence of q and r

[shy bluff]

well we'd always have the same f as being one of the inputs to the division algorithm in our phi

[elder valley]

So to be precise you should say what two polynomials you're applying the theorem to, and what the two polynomials you get from the theorem

[shy bluff]

hrm

[elder valley]

Yeah that's right

[shy bluff]

Let phi(g) be the quotient? (is that the right word?) of the polynomial g by the polyonmial f for any g in R[x]?

[elder valley]

Quotient is the q, remainder is the r

[shy bluff]

hrm

but is that idea right?

[elder valley]

Close. You have f to begin with, and you're trying to define phi at some g. So use those as the inputs

You're looking for a polynomial of degree less than f. The theorem hints at that

[shy bluff]

phi(g) = q, q is a polynomial such that deg(qf) = deg(g), and such that qf + r = g for some r in (f)?

[elder valley]

Well deg(q) will just be deg(g)-deg(f), or q=0. But that won't guarantee that it's degree is less then f's

Read the theorem again. It tells you something about the degree of the output

[shy bluff]

deg(r) < deg(g)?

Is there not a theorem that says that deg(qf) = deg(q) + deg(f)?

[elder valley]

The f and g are swapped for our problem

We're dividing g by f

[shy bluff]

ok so deg(r) < deg(f)

hrm

[elder valley]

Yeah

[shy bluff]

g = qf + r, where deg(r) < deg(f)

oh wait no phi should send phi(g) to r, not q right?

[elder valley]

Is there not a theorem that says that deg(qf) = deg(q) + deg(f)?
@shy bluff yes that's true if both polynomials are nonzero

[shy bluff]

The same way how for like Z/nZ you go and take the remainder

Not the quotient

[elder valley]

oh wait no phi should send phi(g) to r, not q right?
@shy bluff yes that's it

[shy bluff]

oh

Well then that was disappointing on my end (as in I am disappointed that I didn't realize that sooner)

[elder valley]

Yeah it's exactly the same idea. Sometimes people call this modding by f, just like modulo a number

[shy bluff]

yea

I swapped the words

quotient and remainder were swapped in my head :x

[elder valley]

Ah

This division algorithm is used all the time to prove tons of stuff with polynomials

But yeah. Given some g, there are q and r from division algorithm such that g=qf+r with either r=0 or deg(r)<deg(f). Then define phi(g)=r

[shy bluff]

Yea

I had uh difficulty with the word quotient/remainder a lot in elementary... it's showing up again now lol

[elder valley]

By the way, your R[x]^(n) was defined as <= n but it should be <n for part b to be true

Because division algorithm gives strict inequality

[shy bluff]

oh

Yea prodbadly a typo, the prof makes a lot of typos lol

[charred pewter]

How do you add to a cyclic subgroup

Like if G= Z/24Z and H=<3> how do you compute 1+<3>

[old hollow]

Wdym

{0,3,6,9,12,15,18,21} is the subgroup I think

[charred pewter]

Isn’t the generator 3^n for all integers n?

[old hollow]

Well 3 generates only those elements, it doesn't generate the entire group

[charred pewter]

Wdym

What power of 3 gives 6

I’m confused on how that subgroup was generated, it looks more like 3n

[old hollow]

Oh sorry

Ok so you meant G = (Z/24Z)*, the multiplicative group mod 24

Yes it would be 3^n, but <3> doesn't generate the entire group

[charred pewter]

When would it stop then

That’s one of the things that’s confusing to me

[old hollow]

Ok so

<3> = {3^n | n in Z }

And 3^n only generates the elements 3 and 9, no matter what n is

Try computing 3^n mod 24 and plug in any value of n, it only gives you 3 or 9

Never mind I'm dumb

Ignore all of that

[charred pewter]

Well I mean I checked it with n = 1,2,3,4 and you were right

Or were you saying ignore how you said the elements are generated

[old hollow]

Well one issue

Actually don't ignore it

Idk why but I put 3^156 mod 24 and it gave me 8

But in wolfram alpha it gave 9

[charred pewter]

that’s pretty weird

But basically to generate the elements for H =<3> in G= Z/aZ, it’s 3^n (mod a)?

[old hollow]

So yeah <3> generates {3,9}

Yeah change it to a

[charred pewter]

Nice

Ok I wish my professor said this stuff in his vids

[old hollow]

Well there is an important thing to notice

When you say Z/nZ you mean the additive or multiplicative group?

When you do 3^n it means it's the multiplicative, since you're multiplying 3 by itself a bunch of times

[charred pewter]

The paper doesn’t say, it literally just says Z/24Z

It doesn’t have an asterisk

Or any other symbol

So I’m going to assume multiplicative

[old hollow]

Yeah

[charred pewter]

He makes typos sometimes so I checked my email to see if he addressed it

And he didn’t

So I’m just assuming multiplicative

[old hollow]

people usually denote it (Z/nZ)^x or (Z/nZ)* or Z_n^*

but yeah you can assume the group operation is multiplication

[charred pewter]

Anyway, now that we have that the cyclic subgroup <3>= {3,9}, how do we perform 1+<3>? Add 1 to each element?

Also, thanks for clarifying that

[old hollow]

well I think so

I'm not familiar with what 1 + <3> means but I assume yeah, you add 1 to each element

[charred pewter]

It’s odd because he doesn’t even address it in the videos

But I’m gonna go with adding to each element, thanks for helping me

[charred pewter]

Ok I found this in the textbook

So I think we were wrong to assume it was the multiplicative group

And we should have assumed it was the additive group

But I don’t understand how 0+H=3+H

And so on

Ok so new question: How does one generate the elements of the additive subgroup H=<3> in G=Z/24Z

Is it just 3n (mod 24)?

[stone fulcrum]

Something like Z/24Z is always additive, yes

H = <3> is the subgroup generated by 3. That is,
H = {3, 3+3, 3+3+3, 3+3+3+3...}

Or, using multiplicative notation
H = {3¹, 3², 3³, 3⁴...}

But remember I'm using 3² to mean 3 + 3 here

[charred pewter]

Okay, yeah that confused me for a second

I'd rather write H={3(1), 3(2),...,3n}

Does that work?

Or is that not good notation?

[stone fulcrum]

It works too haha

[charred pewter]

Okay, so the elements of <3> are {0,3,6,9,12,15,18,21}?

And to perform 1+<3> I just add 1 to every element?

My professor is asking us to compute 1+<3> and 7+<3> and then asking if they're the same

[stone fulcrum]

Exactly

@charred pewter

[charred pewter]

Okay, thank you for helping me :)

[charred pewter]

What's the order of the additive group of integers mod n

n-1 right?

[scarlet estuary]

not quite...

remember that 0 is in this group

[charred pewter]

oh so it's just n?

[old hollow]

Yes

Keep in mind for the multiplicative group this isn't the case

Cause the multiplicative group only has numbers coprime to n

[latent anvil]

lol why did you post this here

does the ged involve group theory?

anyways <@&268886789983436800> someone is asking for test answers

oop they deleted

[scarlet estuary]

they got banned

[chilly ocean]

Damn feels bad

[latent anvil]

i mean

Break the server rules, get banned

[rose axle]

they were banned 👍

[chilly ocean]

it's not a feels bad if someone blatantly breaks the rules

[charred pewter]

if I have G=S_42 and H=A_42, then the order of S_42=42! and the order of A_42=42!/2, and the number of cosets is |G/H|=2

So if one of the cosets is A_42 itself, how do I find the other coset?

[stone fulcrum]

Anything that's not in A_42 rofl

Since there's only 2, you've got "in" or "not in"

You can also pick some element you know that isn't in A_42, and multiply it by an element in A_42. This will generate an element in the other coset

[charred pewter]

oh lmao

[vestal snow]

If B is integral over A then B/b is integral over A/b^c

This doesn't make sense to me

A/b^c is not a subset of B/b

It's given that A is a subring of B

[limber haven]

Can anyone answer this , I am not getting. Option 2 is incorrect.

Other ,from 1 ,3 ,4 ,. Which is correct

[vestal snow]

Seems like 4)

Because of Lagrange

[woven delta]

Yeah

[limber haven]

Okkk

[woven delta]

All elements are of order 1, 2, 4, or 8. There's always one element of order 1, there are no elements of order 8, and only 1 of order 2. Therefore the other 6 elements must be of order 4

It's easy to see that such a group can't be abelian. This is because if it was abelian it would be forced to be Z/4 x Z/2, (Z/2)^3, or Z/8. All of these options are impossible

This group must in fact be the quaternion group

[hot lake]

if there were no possible group you would have to say that all of 1 2 3 and 4 would be right

[limber haven]

Ohkkkk.got it.

Only identity element have order 1 in any group. No other element can have order 1 , right?

[hot lake]

yeah

[stone fulcrum]

Can you prove it?

[limber haven]

And suppose group have order n. So all its elements have order divisible by n.

[stone fulcrum]

Yes. That's Lagrange

[next obsidian]

A/b^c is not a subset of B/b
@vestal snow you can view it as a subring of B/b. Just send a + b^c to a + b, and you’ll find that this is injective

If you have a,a’ in A such that a - a’ in b, then actually a - a’ in b^c so the cosets were equal in A/b^c to begin with

You can either manually show it’s well-defined, or write this as the map induced by A -> B -> B/b induced by the universal property of the quotient A/b^c

And suppose group have order n. So all its elements have order divisible by n.
@limber haven this is wrong, they have order dividing n

[limber haven]

Oh..yeah...I meaned that , but confused while writing. Thanks for correcting

[vestal snow]

@next obsidian Thanks

[upbeat burrow]

Group theory “show that a power of a cycle need not be a cycle”.

[next obsidian]

Just come up with some contrived example

[upbeat burrow]

If I find a power of a cycle results in disjoin permutations, is that correct?

[next obsidian]

yeah

I feel like almost all of them won't be

[upbeat burrow]

Ah, okay. Just wanted to make sure that’s what it’s asking.

[next obsidian]

(1234)^2 = (13)(24) I think

[upbeat burrow]

(1234)^2

Yeah haha that’s what I have

[next obsidian]

Maybe in general you can make a statement about like divisibility

since that cycle has length 4 and you square it

[upbeat burrow]

So you’re saying if the powers aren’t coprime to the cycle length then that doesn’t hold?

[next obsidian]

Maybe?

I mean if the power is coprime to the length then the order is still going to be the length

I just don't know if that necessarily means it's still a cycle

[upbeat burrow]

Oh I see what you’re saying

[zinc depot]

How would I use this graph to compute k * k = -1

[knotty mason]

what are the arrows

I guess k = red,green since that takes you from 1 to k

so start at k and follow red,green and see where you end up

[zinc depot]

Red arrows represent multiplication by i, green arrows by j.

start at k, now you need another k, so which arrow do you choose?

[scarlet estuary]

to get from 1 to k, you follow red, then green

so to multiply k by k, start at k

then follow the red arrow, then the green arrow

[zinc depot]

in the reverse case, so like i * -k = j, how do you do that? start at i (which is 1 red arrow from 1 to i) then how do you reverse k, to get j? k is red, green, so from i i'd assume you do green, then red and get the j that way? i.e.

[scarlet estuary]

so first let's try and figure out

what -k corresponds to

to get to -k from 1, we take a green arrow, then a red arrow

so -k is green, then red

so now let's start at i and apply -k

start at i, follow the green arrow, then follow the red arrow

and you end up at j

[zinc depot]

okay, awesome. that makes sense, thanks!

[scarlet estuary]

note that, if you came up with something different for -k, it'd still work

like we can go from 1 to -k by taking green -> red yes, and this is the fastest way

but we could also start from 1 and take red -> green -> red -> red

and indeed, if we start from i and take red -> green -> red -> red, we still end up at j

as long as it's a valid path from 1 to -k, you can use it to figure out multiplication by -k

[zinc depot]

okay, good point that makes sense

[next obsidian]

Is this some way to represent the quaternions or something?

with two generators

[zinc depot]

yes

[next obsidian]

I guess red is i and green is j

weird I've never seen something like this

[scarlet estuary]

its just a cayley graph

[next obsidian]

Yeah I've never seen one lol

If $B$ is an $A$-algebra what is $B\otimes_A A(x)$?

[cloud walrus]

Mathemagician:

[next obsidian]

If we were looking at $A[x]$ the answer would be $B[x]$, but I don't think the answer in this case is $B(x)$

[cloud walrus]

Mathemagician:

[next obsidian]

I mean I guess you localize $B$ at the image of $A\setminus{0}$ and then append an indeterminate

[cloud walrus]

Mathemagician:

[shrewd halo]

Just finished up a problem and it's pretty cute so I thought I'd share: show that the only finite-dimensional real vector spaces that are also fields are R and C. It's the kind of problem with an aha moment, very satisfying.

[next obsidian]

Can't you do this with Galois theory or something?

[shrewd halo]

No need

[next obsidian]

I was just thinking about the Galois theory proof that C is algebraically closed

And I feel like you could jerryrig that into a proof of that

Although it seems overkill

[latent anvil]

I think I can

it's a finite extension of R

so it's algebraic over R

So it fits between R and C

so it's either R or C

[shrewd halo]

Ah. If you use Galois for your proof of the FTA then sure, but the simplest proof I could come up with goes through that

[next obsidian]

I mean that's a terrible way to prove FTA

but it's the first thing that came to mind

[shrewd halo]

Well, right instinct then

bc the intuitive leap for me was to move over to algebra and not think linear algebra

[latent anvil]

hang on

you're basically proving a finite extension of R is either R or C

So this is literally what I said

Just the fact that C is algebraically closed

And [C : R] = 2

Or that [C : R] is prime lol

[shrewd halo]

yeah, there's nothing more to it than that

[latent anvil]

This counted as galois theory in my head

For some reason

So I was searching for a different proof

[shrewd halo]

ah gotcha

[zinc depot]

simple question (just learned about quaternions), but this graph basically shows that you can generate every element in the quaternion group with just elements i,j -- so I can say these two elements are the generators, yeah?

[next obsidian]

I mean you can't say "the"

in good faith

since they aren't the only generators, but the two together do generate the group

[bleak abyss]

the

[zinc depot]

can you say they 'are' generators?

[next obsidian]

yeah

[zinc depot]

okay, i see

[bleak abyss]

"a generating set"

[next obsidian]

Okay, wtf is $k(s)\otimes_k k(t)$

[cloud walrus]

Mathemagician:

[next obsidian]

So I thought you want to localize $k(s)$ and then adjoin t

[cloud walrus]

Mathemagician:

[next obsidian]

but $k(s)$ is already a field so localizing does nothing

[cloud walrus]

Mathemagician:

[latent anvil]

hmm no

hang on

[next obsidian]

But there's no way that it's just $k(s)[t]$

[cloud walrus]

Mathemagician:

[latent anvil]

You're not localizing k[t] over k to get k(t)

[next obsidian]

True

I was just realizing that

Because I'm supposed to describe Spec $k(s) \otimes_k k(t)$

[cloud walrus]

Mathemagician:

[latent anvil]

So uh

Morally

[next obsidian]

And this seems... hard

[latent anvil]

I want it to be k(s, t)

right?

[next obsidian]

same

but like, it isn't

[latent anvil]

Oof

[next obsidian]

because then Spec is 1-point

[latent anvil]

So?

[next obsidian]

This is for the problem that the point set of product is not product set

there's no way it would be 1-point if he's asking you to describe it

metaproof

[latent anvil]

oh lmao

Yeah for sure

Hmm

This seems hard

[next obsidian]

Yeah, it's fubbed

[latent anvil]

But also it seems easy

And like I'm dumb

[next obsidian]

I'm starting to think you actually can't say anything really nice

[latent anvil]

what if you just write down two primes

Done

[next obsidian]

Like maybe $A\otimes_B B(x)$ isn't anything great

[cloud walrus]

Mathemagician:

[next obsidian]

So this isn't to like

show the product set isn't the set

he wants you to "describe it"

so I think you can classify the primes somehow

but I don't even know what this set looks like lol

[latent anvil]

This feels like it should exist though

Like nicely

Ugh

[next obsidian]

Yeah

[latent anvil]

okay so universal property memes

[next obsidian]

I agree, but maybe it just doesn't and rational functions are evil

uhhh

[latent anvil]

Easy to define k algebra maps out

Right?

[next obsidian]

I forget the rational function universal property

of?

maps out of what?

[latent anvil]

Use field of fractions+polynomial ring universal properties

Maps out of the tensor

It's the pushout in k alg

[next obsidian]

uh yeah i guess

yeah

[latent anvil]

So what's a map out of it?

[next obsidian]

can you map to k by forgetting the variables

like reduce mod t

for k(t) and k(s)

[latent anvil]

what

[next obsidian]

Like send a polynomial to the constant

evaluate at t = 0

is what I mean

except maybe not

since like, fractions

[latent anvil]

I'm not sure what you're saying

Here's what I'm saying

This is a coproduct

[next obsidian]

I'm trying to come up with maps from k(s) and k(t)

to the same target

[latent anvil]

Into?

[next obsidian]

to induce a map

into k

but I don't think that works

[latent anvil]

I'm saying into an arbitrary target

Oh you mean to get maximal ideals?

Hmm I guess

[next obsidian]

idk

[latent anvil]

But no seems unlikely

[next obsidian]

you said "what's a map"

So i tried to come up with an example lol

[latent anvil]

Lol I meant what's the functor Hom(k(t) (×) k(t), —)

[next obsidian]

oh

uhhhhhhh

Hom(k(t), Hom(k(t),-) right?

[latent anvil]

No I meant k algebra maps

hom(k(t) (×) k(s), A) = hom(k(t), A) × hom(k(s), A)

since (×) is the coproduct in k alg

[next obsidian]

sure

Wait so why are we doing this?

Is this enough to figure it out by Yoneda or whatever

[latent anvil]

I mean yoneda tells us this functor determines the tensor

I'm hoping it's easy to write down

[next obsidian]

Even though we're looking only at algebra homs?

[latent anvil]

wym?

[next obsidian]

I mean I was looking at general module homs right?

when I wrote my thingy

Oh nvm

I agree

[latent anvil]

so a map into A is a pair of maps from k(t) and k(s)

[next obsidian]

yeah

[latent anvil]

Which is the same as two injective maps from k[t] and k[s]

Which is the same as an element of A transcendental over k?

Sorry a pair of elements

Let K/k be some field

[next obsidian]

hmmm

I see actually

yeah

[latent anvil]

then Hom(k(s) (×) k(t), K) is the set of pairs of transcended elements over k

In K

And these are the same primes

Err not quite

[next obsidian]

wait but

[latent anvil]

Every prime comes from such a map

[next obsidian]

why are you only looking for field extensions of k

oh to classify the primes

sure

[latent anvil]

Yeah

[next obsidian]

Don't we need to look at integral domains then?

Or are all primes maximal you think?

[latent anvil]

no

integral domains embed into fields

[next obsidian]

ah sure

right right

we look at the image

might not be surj

[latent anvil]

Okay so now I understand the difference between this and k(s, t)

[next obsidian]

okay

pls explain because I'm like mabye 60% there

[latent anvil]

You can't map $\Q(s, t)$ to $\R$ by picking out $\pi$ and $\pi+1$

[cloud walrus]

shamrock:

[latent anvil]

Maps out of k(s, t) are pairs of algebraically independent elements

[next obsidian]

oh

[latent anvil]

but maps out of this tensor are pairs of elements which are separately transcendental

[next obsidian]

oooo

weird

[latent anvil]

But maybe have nontrivial relations between them

Yeah that's cool

[next obsidian]

So maps out of k(s,t)

[latent anvil]

But I think it really shows why they have to be different

[next obsidian]

pick out all uh

transcedental sets of size 2 or whatever

[latent anvil]

Yeah and just to be clear I'm thinking about mapping into a field

[next obsidian]

yeah

is it okay to limit yourself to that case?

oh wrt to prime ideals

is all we're talking about

so it is actually okay'

[latent anvil]

Okay so let's consider maps into k(r) maybe?

When do two such maps have the same kernel

[next obsidian]

why into k(r)?

[latent anvil]

¯\_(ツ)_/¯

Maybe I can justify this uhhh

No I can't

Just felt like it

Seems simpler

We know there are maps for sure

Lots of them

maybe we could go in completely the other direction

[next obsidian]

So I just saw this thing

I was flipping through A-M

[latent anvil]

we've described the functor of points of Spec k(s) ×_{Spec k} Spec k(t)

That's pretty based right

[next obsidian]

is this thing we've described f.g. over k?

[latent anvil]

Can we use that to show that our space can't have one point

Doubt

Yeah definitely not

It's infinite dimensional

over k

As a vector space

[next obsidian]

as an algebra

[latent anvil]

Remember dimension and tensor plays nice

oh sure

Still no I think

[next obsidian]

If you can it's a Jacobson ring

[latent anvil]

It should contain k(t) as a subring

Right?

[next obsidian]

which in particular says all primes are intersections of maximal ideals

if it's integral you can also get the same result but I think f.g. is more possible

[latent anvil]

So we have an algebra inclusion k -> k(s)

[next obsidian]

yeah

[latent anvil]

Tensor with k(t)

Which is free and thus flat

Oh sorry I'm dumb

[next obsidian]

is k(t) free?

[latent anvil]

Finitely generated doesn't imply subrings finitely generated

Yeah we're working over a field

[next obsidian]

what dim?

countable?

[latent anvil]

countably infinite?

yeah

[next obsidian]

huh okay

[latent anvil]

Err

No that sounds wrong actually

Assume k is infinite and has cardinality κ

[next obsidian]

yeah

[latent anvil]

All the elements 1/(t-a) are linearly independent

and that's a linearly independent subset of size κ

[next obsidian]

oh yeah

[latent anvil]

So it's probably dimension |k|

Weird

[next obsidian]

hmm okay

idk this is really weird

man fuck Spec of tensor products

[latent anvil]

anyways I strongly doubt it's finitely generated

You started with two really big things

you tensored them

[next obsidian]

yeah

[latent anvil]

man I like commutative algebra

I always like, say I dislike math

And then find out I'm wrong

bad vibes imo

[next obsidian]

This is hard 😦

[latent anvil]

Yeah

What are we doing again?

Trying to prove Spec has more than two points?

[next obsidian]

No

to describe it

[latent anvil]

Well I described its functor of points

That's good enough

Lol jk

[next obsidian]

I mean

only when mapping into fields

[latent anvil]

oh right

well you can fix what I said

[next obsidian]

you described the like

[latent anvil]

To work in general

[next obsidian]

Wait

That's describing Spec haha

probably those functors are iso or whatever

you described the prime ideals via the maps that produce them

[latent anvil]

wym?

Well I didn't describe Spec quite right

[next obsidian]

Like if you took the functor which looks at maps from that tensor but only into fields

You didn't?

[latent anvil]

You still need to understand when different maps have the same kernel

[next obsidian]

oh

[latent anvil]

Alex if I ever get into topoi you have to kill me

no mercy

[next obsidian]

Doesn't that have to do with the relatiosn between the two trascedental elements you picked

[latent anvil]

Put me down like a dog

Yeah the kernel is all the relations

[stone fulcrum]

Oh god are we doing the Spec(Z) question again?

[next obsidian]

Man, I think at some point I'll end up learning topoi or something idfk. If AG keeps going the direction it's been going

No

not Spec(Z)

[latent anvil]

No kaynex

Spec Z is pretty tame

[stone fulcrum]

Okay thank God

[next obsidian]

This is Spec (k(s) (x)_k k(t))

much worse

[stone fulcrum]

Wow your mother is tame

[next obsidian]

Also describing Spec Z is easy

[latent anvil]

My mom is a nice lady

[stone fulcrum]

Yes I'm sorry she didn't deserve that

[next obsidian]

I don't think tame and nice are mutually exclusive

[latent anvil]

Okay so what are the primes

Cmon Alex

You got this

Figure it out

[stone fulcrum]

2,3,5,7...

[latent anvil]

hm

no

[stone fulcrum]

Jk sorry everybody

[latent anvil]

You forgot 0, -2, -3, -5, -7,...

[next obsidian]

(0),(2),(3),(5)

[latent anvil]

Lol

[next obsidian]

(57)

[latent anvil]

Okay so let p be a prime

Somehow in the fraction field of the quotient

We have some elements

Right?

Like uhhh

[next obsidian]

Wait what

what quotient

[latent anvil]

Maybe the images of s (×) 1 and 1 (×) t

By the prime p

[next obsidian]

okay

I'm not certain why you jumped to the fraction field

[latent anvil]

idk I like fields

[next obsidian]

okay

[latent anvil]

We can stick with just the quotient

Maybe it's a field even?

[next obsidian]

Uhhhh

That'd be news to me

if all primes are maximal

[latent anvil]

Yeah me too

but like

why not

Yknow

[next obsidian]

I guess ¯_(ツ)_/¯

I just don't have any reason to suspect it's true atm

[latent anvil]

I'm just trying to understand these primes

[next obsidian]

Is it an integral domain?

[latent anvil]

the quotient?

[next obsidian]

No

the ring itself

[latent anvil]

Oh you mean the tensor

[next obsidian]

yeah

[latent anvil]

I think so actually

[next obsidian]

then for sure quotient by 0 isn't a field

[latent anvil]

Because tensoring over a field is nice

[next obsidian]

so maybe height 1 primes

[latent anvil]

Oh yeah lol

[next obsidian]

Is it Noetherian?

Because if so, and it's dim = 1

[latent anvil]

X

[next obsidian]

then we know the quotients are dim 0

which is close to being a field

[latent anvil]

Seems weird for this to be noetherian

[next obsidian]

same tbh

[latent anvil]

I mean anything about this is weird

[next obsidian]

yeah

[latent anvil]

It's just a weird ring

[next obsidian]

I hate it

[latent anvil]

But okay is it an integral domain

who tf knows

:(

[next obsidian]

yeah idfk

this is so garbage

lol

I'm gonna see if $A (x)_k k(t)$ has some nice rule

[cloud walrus]

Mathemagician:

[latent anvil]

X

Again

I mean you can describe maps out like I did

[next obsidian]

https://en.wikipedia.org/wiki/Tensor_product_of_fields

[latent anvil]

what about it?

[next obsidian]

This doesn’t seem tremendously helpful

Actually

[latent anvil]

yeah lol I read through and didn't see anything

The composite thing is only in the finite case

[next obsidian]

I found pdf

By Keith Conrad

[latent anvil]

Noice

[next obsidian]

Where he talks about this exact thing

https://kconrad.math.uconn.edu/blurbs/linmultialg/tensorprod2.pdf

[latent anvil]

Lmao

[next obsidian]

Example 7.9

[latent anvil]

Okay cool

This is sweet

So it is an integral domain

The elements b(x) d(y) are units right

Yes

[next obsidian]

Yeah so weird

Now understanding prime ideals there are still gonna be whack

Okay wait

Can you describe this as a localization if k[t,s]

Specifically you localize at all things of the form b(x)d(y)

[latent anvil]

Yeah right?

Like exactly that

I don't see why not

That's a multiplicatively closed set

[next obsidian]

Then the primes are primes of k[t,s] which avoid them right?

[latent anvil]

oooooh

Nice!

[next obsidian]

Is there a better way to describe those?

Are they generated by an element which can’t be factored that way and is irreducible,

k[t,s] is not a PID since it’s height 2

Err dim 2

And I don’t think it’s ideals are generated by at most 2 elements

I feel like we had that Final problem which showed some ideal can’t be generated by < 3 elements

But I don’t remember how many variables that was

[latent anvil]

You can find a complete classification of primes in that ring online I think

I've seen it before

And it's elementary

[next obsidian]

Really?

Huh

[latent anvil]

Now saying you should

*not

[next obsidian]

I remember there’s one for like Z

[latent anvil]

Yeah a polynomial ring over a PID

[next obsidian]

And it was fucky

[latent anvil]

Yeah it's the same thing

Iirc

[next obsidian]

Oh

I forget if Reid has something about this

Also just to be sure

If f(x,y) cannot be factored in that form

And is irreducible then nothing in the ideal generated by it can be right?

Where it gets weird is if you have two such things and look at (f(x,y),g(x,y))

[latent anvil]

Good q

Hmm

[next obsidian]

Can they cancel them out

[latent anvil]

Probs

[next obsidian]

Also Brendan I think like UFD stuff implies it’s not possible

Like if gf(x,y) factored into that wouldn’t f have to be a subset of that factorization

[latent anvil]

What's not possible?

[next obsidian]

If f doesn’t factor that way

[latent anvil]

oh I think I see what you mean

[next obsidian]

That something in its ideal does

[latent anvil]

Yeah I sed what you mean

Being of this form means your irreducible factors are in k[x] or k[y]

[next obsidian]

Yeah

[latent anvil]

If gf has that property, f does too

Because subsets

[next obsidian]

You want the other direction

[latent anvil]

I do?

[next obsidian]

Oh nvm

You’re right

[latent anvil]

Yee

[next obsidian]

But what happens when you introduce sums

I feel like things can get fucky

Like 1 - xy and 1 + xy?

[latent anvil]

Sounds likely

[next obsidian]

I think both are of that form

[latent anvil]

yeah that's a good point

they are

[next obsidian]

Actually that generates all of the ring

[latent anvil]

Oh lol

[next obsidian]

So trying to find a little less contrived of an example

Maybe like 1 + x + xy and 1 - xy?

[latent anvil]

Yeah that contains 2 + x

But it's proper

[next obsidian]

Yeah

[latent anvil]

oof

[next obsidian]

So yeah things can get weird

I mean okay

I think I’m happy with primes of k[t,s] which avoid that set

[latent anvil]

hmm

Okay so like

There's a scheme map into Spec k[s, t]

Maybe we try to think geometrically

that's a "plane"

we're looking at a weird subset of the plane

is it closed? Probably not

X

Is it open? Probably not still

Open is more likely though

But still seems wrong

[oblique river]

hey sorry I tried to read back to figure out what question you're talking about

but I still couldn't figure it out

can I interrupt and ask what the question/situatoin is?

[latent anvil]

Understand the fiber product of Spec k(s) and Spec k(t)

[next obsidian]

What is Spec k(s) \otimes_k k(t)

[latent anvil]

Sorry we've been going for a while lol

[oblique river]

oh I see

[latent anvil]

Ah fuck

I was looking at my phone

And walked into a stop sign

Fuck being yall

[next obsidian]

OOF

[latent anvil]

*tall

It was just the edge

I didn't see it

Okay well my headphones got knocked onto the ground but aren't damaged

[oblique river]

sorry are you just trying to understand the ring?

or like, what part of the structure do you want

[next obsidian]

The primes

[latent anvil]

I'm not even sure

[oblique river]

isn't it just a point?

[latent anvil]

I don't know the original question

Nope

[next obsidian]

Nope

[oblique river]

like, it's the pullback of a point over a point

[latent anvil]

yeah lol it sure is

[oblique river]

hmmmmm

so have you found examples of prime ideals?

[latent anvil]

We have determined that it's an integral domain which isn't a field

[next obsidian]

It’s the subring of k(s,t) where the denominators factor as the product of a d(s) and a r(t)

[latent anvil]

So there are at least 2 primes

[next obsidian]

Not as a sum

[latent anvil]

What's not as a sum?

[next obsidian]

The denominator

Isn’t a sum of products of polynomials in t and s

[latent anvil]

Oh I see

Yeah lol

[next obsidian]

Yeah so Sham you do have that scheme map

[latent anvil]

Right

[next obsidian]

But I don’t know what it does for you

[latent anvil]

But it's kind of shit yeah

I just want to draw a picture ig

It's an injection

But not a good one I think

[next obsidian]

How do you know it’s injective

Oh nvm lol

The map of rings is injective cuz ID

[latent anvil]

I mean I was thinking properties of localizations

Not sure what you mean by ID

We know the primes are really primes of k[s, t] which don't intersect some open subset

That means that pulling back along k[s, t] -> k(s) (×) k(t) is injective

[next obsidian]

I mean the map of rings is injective so the induced map on specs is

[oblique river]

sorry I'm still a little confused because I don't understand why fiber products of schemes don't agree with fiber products of topological spaces

[next obsidian]

Well I guess this is for the map of sheaves

This is the whole point Buncho lol

[latent anvil]

I mean buncho I have a dumb example

Over Z

[oblique river]

yeah I'm just looking for any example

to help me reset my intuition

[latent anvil]

Z/2Z and Z/3Z

Over Z

They're both fields/points

The fiber product is empty

[oblique river]

yeah but you can see that on the topological space level

[latent anvil]

Wym?

[oblique river]

because they map to different points of Spec Z

[latent anvil]

oh

Good points

[oblique river]

so the fiber product of them is definitely empty

[latent anvil]

*point

[next obsidian]

The example Hartshorne has you compute is A^1 x A^1

Over k

You get your copy of the product in the form of like (f(x),g(y)) right?

[latent anvil]

Wym?

[next obsidian]

Primes of k[x] are an irreducible f(x)

[latent anvil]

Sure

[next obsidian]

Likewise for k[y]

[latent anvil]

Yup

[next obsidian]

So you can see this in the primes of the form (f(x),g(y))

Right?

[latent anvil]

oh you're saying there's a map A^1 (fiber product over a point in top) A^1 -> A^2?

[next obsidian]

Yeah

[latent anvil]

And it's not surjective

[next obsidian]

Yeah

[latent anvil]

I'm not sure exactly what kind of example buncho wants though

[oblique river]

hmm yeah I think that example might help me think about this

basically like I just wanted an example where the naive fiber product of topological spaces was wrong

[latent anvil]

Yeah so in variety land products aren't topological products

And the same kind of issue shows up with schemes

[next obsidian]

But the underlying point set is the same

[latent anvil]

But things can get even worse

[next obsidian]

In schemes even the point set can be fucked

which is big rip

[latent anvil]

Oh fair yeah, I was thinking topologically

Forgetting and then taking the (topological space) product isn't the same as taking the (variety) product and forgetting

[next obsidian]

Oh I emailed Sándor btw lol

[latent anvil]

Ah nice

For what exactly?

[next obsidian]

About the fucking fraction field finite thing

Generically finite bs

[latent anvil]

Sure

Lmk what he says

I miss sandor :(

Won't have a class with him until spring

[next obsidian]

I mean me neither

[latent anvil]

aren't you doing AG stuff?

Is he only taking over in the spring?

[next obsidian]

Oh yeah

I forgot he was doing that too