groups-rings-fields

[viscid pewter]

solution spoilers: ||i think you have to take like (1, 0) and show it can only be generated by (1, 0) or like (-1, 0), and then consider (0, 1) so that's at least two required in the presentation||

[pine patio]

oh that makes sense

thanks

[pine patio]

is <(123),(456)> cyclic?

[mild laurel]

no

[pine patio]

how do i show it

[mild laurel]

Show that all the elements generated by those permutations have order 3

[next obsidian]

Shouldn't that group be Z_3 x Z_3

I think

[pine patio]

how does that help

[mild laurel]

If all the elements have order 3, and you have more than 3 elements in your group, then your group can't be cyclic

[pine patio]

i was trying a proof by contradiction

suppose g is the generator, then g has order 6

[mild laurel]

Why does g have order 6?

[pine patio]

we're given that <(123), (456)> is in S_6

so we only have 6 elements to cycle

[mild laurel]

uh

Do you know what it means to be a generator for a group?

[pine patio]

yes, so if H=<S>, then every element of S is a generator

right?

[mild laurel]

No. For example, in Z/4Z, 2 is not a generator, but 1 and 3 are generators

[pine patio]

what does / mean

[mild laurel]

quotient, those are the integers mod 4

[pine patio]

we havent done that yet

[mild laurel]

maybe an easier example is in Z, where 1 and -1 are generators but 2 is not a generator

[pine patio]

so what does it mean for an element to be a generator?

H=<g> if H is cyclic

so if H is not cyclic and H=<g,h> then what are g and h of H?

[mild laurel]

I feel like you might want to review your notes for these definitions

if H = <g,h>, neither of these necessarily have to be generators of H, although they could be

[pine patio]

so what does the third one mean?

[mild laurel]

Sorry, you're right. I was talking about being generators of a cyclic group

A group is cyclic if and only if its generated by one element

[pine patio]

ah that makes sense

[mild laurel]

Which is what I meant when I said that neither have to be cyclic generators of H, although they could be

[pine patio]

if H=<g,h> and H is cyclic, then isnt it true that atleast one of g and h is the cyclic generator?

so like <i, -1>

i^2=-1 so i is a generator and <i, -1> is cyclic

[next obsidian]

no

[mild laurel]

Consider that like Z = <2,3>, but neither 2 nor 3 are generators of Z

[next obsidian]

Consider something like the the integers mod 7

it's generated by 2,3

or that

[pine patio]

oh true

[pine patio]

so here in iii) they mean that the individual ones dont have to be generators, but they as a collection are considered generators

[next obsidian]

yes

[pine patio]

that statement seemed vague

[cyan marten]

One way to do it is to observe that a cyclic group has a unique element of a given order, but here there are two elements of order 3.

[mild laurel]

wait what

there are 2 elements of order 4 in Z/4Z

Z/9Z has 2 elements of order 3

[cyan marten]

Yeah

I was wrong

What is not wrong is that every cyclic group has a unique subgroup of every possible order.

In G = <(123), (456)> there are two (distinct) subgroups of order 3, so again it's not cyclic.

[next obsidian]

I mean compute the size

You have > 3 elements

but everything has order 3 or 1

[cyan marten]

Is a flat A-algebra just an algebra over A that is also flat as an A-module?

[prisma ibex]

yes

[next obsidian]

yes

[pine patio]

how do i prove this?

[next obsidian]

Which parts? Like

Just do it I guess lol

The first part of (i) is a computation

[pine patio]

i)

[next obsidian]

The second part is using Lagrange, the first isomorphism theorem

[pine patio]

we havent done those yet

[next obsidian]

Then do it as a set theoretic quotient

Also I realized that HK won’t in general be a group so you can’t do it anyway

You can define a relation on H x K

Saying that two elements are related if they map to the same element via phi

Then you can form H x K /~ as equivalence classes

This is in bijection with HK

Then you can show that each equivalence class has |ker phi| elements

So that the size of H x K / ~ is |H||K| / |ker phi|

And you know what ker phi is by the first part

[pine patio]

the order of an element is the number of elements in its equivalence class right?

[next obsidian]

Uhh...

You haven’t defined the equivalence relation

[cyan marten]

There's a simpler way to formulate this, I think: Constructing a bijection between |H cap K| and the preimage of an element.

[next obsidian]

Also I can’t think of a relation for which that would be true

[pine patio]

it's for the very first part

[cyan marten]

If we have g = hk then (hr^-1, rk) also has image g, where r is in H cap K. This way of associating preimages with H cap K gives a bijection.

[pine patio]

for every element a in H cap K, hk = (ka^-1)(ak)

oh i see

it's for each g in the statement that's putting me off

what do they mean by |phi^-1(g)|

the order of (h,k) right?

[pine patio]

how is it a bijection?

i can see that we have atleast |H cap K| elements in the equivalence class of hk

but how do i know its exactly |K cap K|

[cyan marten]

No! It's the cardinality.

[pine patio]

oh so |phi^-1(g)| just counts the preimages of g?

[latent anvil]

yup

the number of elements whose image is g

[pine patio]

now the question makes sense lol

[latent anvil]

Half the battle ;)

[pine patio]

so |phi^-1(g)| >= |H cap K|

can i somehow argue that |H cap K| <= |phi^-1(g)|?

then they'd be equal

[cyan marten]

As I said above, I think a simpler way is to construct a bijection between them. This has the advantage of still being true even when H cap K are infinite and we don't assume the axiom of choice

[pine patio]

but i dont see why it's a bijection

[cyan marten]

If (hr^-1, rk) = (hs^-1, sk) then r = s. On the other hand, if h'k' = hk,
then h' = hr^-1 and k' = rk for some r in H cap K.

[pine patio]

oh so if hk=h'k' then (h')^-1h=k'k^-1 \in H cap K

this is the r

the next part was easy

[pine patio]

can someone explain the notation in the wilson's theorem proof?

in the last line when they split the products

[golden pasture]

essentially pair each element up with heir inverse

say in mod 7:
1 <-> 1
-1 <-> -1
2 <-> 4
3 <-> 5

the product of each pair is 1

but {1} and {-1} are singletons

so you end up with -1

[scarlet estuary]

it honestly feels like a "hack"

it's IMO the first major example a student typically sees of field theory actually simplifying something

[pine patio]

ah makes sense

[cyan marten]

If dim A = 1, is Spec A just the set of prime ideals under the cofinite topology?

[golden pasture]

why would that be so?

consider A=Z

[cyan marten]

Hmmm..

V((n)) is the set of prime divisors of n, right?

Which is always finite

So it seems to me like closed = finite in this case

Well, with the exception V((0)) = Spec Z.

I guess I missed the fact that (0) is distinguished, unlike in a generic space with the cofinite topology

[golden pasture]

precisely

dim 1 has such minimal ideals

[cyan marten]

Yeah..

[pine patio]

how do i prove the second part?

im trying to show that K is closed

so if x is in H and y is in gH then y=gh for some h in H

then xy=xgh

how do i show that this is in K

oh it's the same G from before

nvm

how do i prove the next 2 bits?

i can imagine why this is true but idk how to word it

[pine patio]

how do i prove ii)?

[carmine fossil]

Let's say ($g_1,g_2$) is in K,let $n=\vert G_1 \vert, m= \vert G_2\vert.$
na+mb=1 for some integers a,b
Then $(g_1,g_2)^{(mb)}=({g_1}^{(mb)},e)=({g_1}^{(1-na)},e)=(g_1,e)$ is in K

[cloud walrus]

DrunkenDrake

[carmine fossil]

Since (g_1,e) is in K,that also implies (e,g_2) is in K

i.e., K=H_1 x H_2

[cyan marten]

If you prove 3, then 4 follows easily by induction. To prove it you may use the fact that H and gH are disjoint..

[pine patio]

prove that there is a counterexample?

or does it never hold if |G_1| and |G_2| are not coprime

[carmine fossil]

Consider Z_2xZ_4 and let K be the subgroup generated by (1,1)

H_1 will be {0} and H_2 will be {0,2}

But (1,1) is not in H_1 x H_2

[chilly ocean]

https://www.youtube.com/watch?v=c6FlpordfDk&list=PL8yHsr3EFj53Zxu3iRGMYL_89GDMvdkgt&index=5
Starting at 24:33, he starts talking about how to find all degree 6 irreducible polynomials over the finite field Z/2Z. I understand GF(p^n) is a subset of GF(p^m) iff n divides m and the venn diagram he drew of the finite fields GF(2), GF(2^2), GF(2^3), GF(2^6). However, I don't see why elements in GF(2^3)\GF(2) satisfy irreducible polynomials of degree 3, elements in GF(2^2)\GF(2) satisfy irreducible polynomials of degree 2, and GF(2^6)\(GF(2^2) U GF(2^3)) satisfy irreducible polynomials of degree 6

[mild laurel]

@chilly ocean GF(2^3) has dimension 3 over GF(2) so all elements in GF(2^3) must have minimal polynomials of degree 3 or leess

[warped bane]

$$ X^{2n}-2X^{n}*cos(n\theta)+1 $$

[cloud walrus]

Shura

[warped bane]

how to factor this in R[X]?

I get only solutions in C[X]

[chilly ocean]

why is that?

GF(2^6) has dimension 6 over GF(2) but not all elements of it have minimum polynomial of degree 6

[mild laurel]

Which is why I said "of degree 3 or less"

[chilly ocean]

Ah

[mild laurel]

If you take an element x in GF(2^6) for example

Then since it has dim 6 over GF(2), you know that 1, x, x^2, x^3, x^4, x^5, x^6 must be a dependent set

since there are 7 elements and the dimension is 6, just by linear algebra

[chilly ocean]

Right

So you get a polynomial such that x is a root of that

[mild laurel]

so you have some non-zero coefficients such that a_0 + a_1x + ... + a_6x^6 = 0

and that is your polynomial yes

The minimal polynomial must divide this polynomial

[chilly ocean]

Yeah

But do they need to be degree 3?

For the elements in GF(2^3) and not GF(2)

[mild laurel]

yeah

[chilly ocean]

And why is that?

[mild laurel]

well their minimal polynomials can't have degree 1, otherwise they'd be in GF(2)

also, the degree of the minimal polynomial must divide 3

so it must be 3

[chilly ocean]

wait why must the degree divide 3?

@mild laurel

[mild laurel]

Well if a has minimal polynomial of deg 2 over GF(2) for example, then GF(2)(a) must be degree 2 over GF(2)

But you can't have a degree 2 extension in between a degree 3 extension

Due to how dimensions multiply in towers of field extensions

[chilly ocean]

Ah

That is smart

So using this argument you can find the number of irreducible polynomial of any degree

[mild laurel]

Well, there are much easier arguments than using all this galois theory to see that the number of irreducible polynomials of degree n over GF(q) is

[chilly ocean]

Ah ok

Well I do know the main theorem

About how the subfields of a galois extension corresponds to the subgroups

[warped bane]

$$(X^{n}-e^{i\theta}) = \prod_{k=0}^{n-1} (X-e^{i\frac{\theta+2k\pi}{n}})$$

shall anyone tell me why is that true?

i know that $$X^{n}= e^{n\theta} ==> X = e^{i\frac{\theta+2k\Pi}{n}}$$

[cloud walrus]

Shura

[warped bane]

$$where k \in [IO,n-1I]$$

[cloud walrus]

Shura

Shura

[warped bane]

somebody?

[mild laurel]

idk what you're asking

the things on the right hand side are the roots of the polynomial on the left hand side? So it factors like that?

[warped bane]

oh right

sorry

[sour plume]

(it appears they in fact did understand what you were asking)

[ripe basalt]

Anyone know how to prove that if a map $\mathbb{E}: M_{n}(\mathbb{C}) \to M_{m}(\mathbb{C})$ which is positive and $\mathbb{E}$ maps to an abelian subalgebra of $M_{m}$ then $\mathbb{E}$ is completely positive?

[cloud walrus]

SU(n)

[ripe basalt]

I think if we find a nice form of E, then when we have the tensor we can use commutativity to cancel stuff out and result in the basic positive map. I'm just not sure which sort of representation i am going for

[ripe basalt]

Wait I think it's obvious

nvm

it's not obvious

[ripe basalt]

I was going to say we can apply stinespring's thm

[cloud walrus]

mirzathecutiepie

mirzathecutiepie

[uncut girder]

[lethal cipher]

@crude void ask in here

You have a real shot in here

[crude void]

@lethal cipher kk thx

So boiz here is my question

I need help coming up with a specific example of a group G and subgroups H and K s.t.

[latent anvil]

Do you know about direct products of groups?

[crude void]

@latent anvil ?? Not sure

[latent anvil]

do you understand what G × H means?

[crude void]

yes

[latent anvil]

then yes

Do you want me to give a full example or just a hint?

[crude void]

I don't know, I've been working on this for over 2 hours

[latent anvil]

well my hint would be to think about products

[crude void]

hmmm

I think I know where this is going

So is the choice of (Zmod10)^2 for the group is good?

wait maybe Z20 x Z50

[latent anvil]

Second is better

I'm not sure if the first works

[crude void]

yeah

[latent anvil]

it does but it's not obvious I think

[crude void]

So H can be like {m(1,0) | m is an integer}, and K = {m(0,1) | m is an integer}

[latent anvil]

yup!

[crude void]

AHHHH

[latent anvil]

Or even simpler, H = Z20×0 and K = 0×Z50

[crude void]

Thx for the hint, I wasn't thinking about products

[latent anvil]

Np

Subgroups which intersect trivially are closely related to products of groups

[lethal cipher]

The fact that we couldn't even process that 😆

[latent anvil]

If H, K are normal in G then HK (the set of products of h in H and k in K) is a subgroup and is isomorphic to H × K

[lethal cipher]

Over an hour and that wasn't even on our radar.

[latent anvil]

This is only if they're normal and intersect trivially though

[crude void]

Ok, thank you very much. Dackid, thanks for the brainstorm session too. Glad we got this resolved

[lethal cipher]

What does isomorphism mean with respect to groups?

[scarlet estuary]

this is one of the big lemmas on normal subgroups btw

its very useful

[scarlet estuary]

an isomorphism f between groups G and H is a bijective function G-> H such that f(ab) = f(a)f(b) for all a, b in G

note that the product ab is computed using the group operation from G

since theyre elements of G

wehreas the product f(a)f(b) is computed using the group operation from H

since theyre elements of H

[crude void]

^ we still haven't gotten to that yet but I think that would have been very useful

[lethal cipher]

Interesting, that is very different from the topology variant

[scarlet estuary]

its kind of the same thing

homomorpisms are "structure-preserving maps"

the structure of groups is their group operation

the structure of topological spaces is their continuous maps

the catch is that topological homeomorphisms explicitly require continuity to work in both ways [i.e. for the function and its inverse]

in groups, you get that "for free" from one direction

[lethal cipher]

Ahhh, so the key here is that f(ab) on H isn't some crazy new element, but it is directly related to a and b on G

[scarlet estuary]

yeah

the point is that

when we compute f(ab), we apply the group operation of G and then apply f

wen we compute f(a)f(b), we apply f and then the group operation of H

so if f(ab) = f(a)f(b), that means f "preserves" the "structure" of the groups under their respective group operations

[lethal cipher]

Speaking of, I had a question in my topology test (it's over, don't worry). Would it be possible to steal one of you in that channel momentarily? I really want to understand what I didn't catch

[golden pasture]

just ask

[rustic crown]

Is it true that every abelian group could be realized as a subgroup of (Q/Z)^S for some large enough set S?

[latent anvil]

I think so, yeah

so take a presentation of your group

[rustic crown]

I can see that $\text{Hom}{\mathbb{Z}}(A, \mathbb{Q}/\mathbb{Z})$ can be realized as a subgroup of $(\mathbb{Q}/\mathbb{Z})^A$ by applying $\text{Hom}{\mathbb{Z}}(-, \mathbb{Q}/\mathbb{Z})$

[cloud walrus]

det

[latent anvil]

this makes it Z^n/K for some submodule K and some cardinal n

yeah?

[rustic crown]

this Z^n is the direct sum or direct product?

[latent anvil]

direct sum

[rustic crown]

okie

[latent anvil]

so we can embed in Q^n/K

which I thought for some reason should be injective

why did I think that

[rustic crown]

cause its divisible

[latent anvil]

no I mean, why is Z^n/K -> Q^n/K injective?

I think you can just look at elements

yeah

the difference is in K

or it's not

doesn't matter about the bigger group

[rustic crown]

yea the kernel of Z^n --> Q^n/K is just K

[latent anvil]

right

so now I am thinking

K is free

submodules of free modules are free over a PID

[rustic crown]

(i only know the finitely generated case of this)

[latent anvil]

we should(?) be able to extend the elements to a scalar multiples of a basis for Z^n

(sorry, inf generated case is harder)

wait

no

the statement is immediately false, sorry

take any nontorsion group

or wait

is (Q/Z)^S a product or a sum?

I assumed sum

[rustic crown]

product

[latent anvil]

ahhh

so you can have nontorsion

tricky

anyways

here's what I'm thinking

so you should be able to pick a basis for Z^n such that K has a basis given by scalar multiples of those things

I would need to think harder to see if this is true

but I think it is

[rustic crown]

can we do zorn on the finitely generated proof?

[latent anvil]

hmm, maybe. Iirc it's pretty to show submodules of free modules are free and we're asking for even more

anyways

it seems plausible to me

which is good enough for me 😛

let's roll with it

now what does Q^n/K look like?

it looks like a direct sum of some number of Q's and some number of Q/nZ's

yeah?

because of how quotients of direct sums work

[rustic crown]

makes sense assuming that

[latent anvil]

well it suffices to embed each summand into a product like you want

because direct sums embed into direct products

do you agree with this?

[rustic crown]

yep

[latent anvil]

the factors Q/nZ are easy

so the whole problem reduces down to embedding Q into such a thing

ah wait I think i have a much simpler proof

do you know that any abelian group A embeds into Hom(Hom(A, Q/Z), Q/Z)?

[rustic crown]

nope

[latent anvil]

ah

well

that would simplify things

[rustic crown]

yea

[latent anvil]

sorry you said something related but I didn't notice

well uh

try to prove this

sorry

[rustic crown]

oh this is very much like the double dual situation

[latent anvil]

yeah!!!!

[rustic crown]

if I call A* = Hom(A, Q/Z)... then for any a, we have a map A* --> Q/Z (evaluation at a) which is just an element of A**

[latent anvil]

haha yeah

and you just need to show a |-> ev(a) is injective

[rustic crown]

i don't see an easy construction... if a is a non-zero element of A, then consider all maps B --> Q/Z where B contains a and doesn't let it die. this is non-empty as we can define <a> --> Q/Z easily. Not to hard to see zorn is applicable and so we have a maximal element M --> Q/Z. need to show M = A. so if b in A\M... then probably can extend to M+<b>...

[latent anvil]

sorry, give me 1 sec

sure so you have the right idea

there's a map <a> -> Q/Z nonzero on a

what do you know about Q/Z?

what properties does it have?

[rustic crown]

yea i think this works . consider the ideal {n such that n*b in M}. If this is zero, then i can send b to anywhere we want. else this ideal is dZ. So we already know where db goes, and want to see where b should go

we can divide by d as its Q/Z

[latent anvil]

I think you may be proving Baer's criterion

Which says divisible => injective

[rustic crown]

yea looks very similar

oh, the book i'm reading defined divisible only for principal ideals... so an R-mod D is divisible if r*D = D for every element of r of R

[latent anvil]

I've only ever heard it defined for PIDs

In which case it's the same

(Baer's criterion is more general though)

[rustic crown]

yea

Thanks!

(i'm done so you may ask >.< sowwy)

[tight otter]

is it sufficient to say that because $20^\circ$ is not constructible (via angle trisection) then neither is $40^\circ$, thus, $\cos(40^\circ)$ is not constructible, completing the problem?

[cloud walrus]

panoramatopia

[tight otter]

I feel like this is the correct logic but also under this logic that would imply that $20^\circ\cdot18=360^\circ$ is not constructible which does not make sense

[cloud walrus]

panoramatopia

[leaden finch]

can someone help me with real analysis

[tight otter]

wrong channel my guy

#advanced-analysis

[rustic crown]

in this case it isn't a problem, because A is constructible if and only if 2A is constructible.

but in general you need to look at the degree [Q(e^iA):Q]

[tight otter]

oooh yes that makes sense

thank you!

[rustic crown]

[cyan marten]

How does Z embed in a product of copies of Q/Z?

Oh nevermind

[cyan marten]

Since every abelian group can be embedded in a divisible group (say the injective hull), you can reduce to this case. Then you can use the structure theorem of divisible groups..

[rustic crown]

Dunno about the structure theorem...

but EShamrock's proof was pretty neat. embed A in Hom(Hom(A, Q/Z), Q/Z) which I think is called the Pontryagin double dual. So A can be embedded in (Q/Z)^Hom(A, Q/Z)

[cyan marten]

Well, essentially, it says Tor(G) can be embedded in the direct sum of copies of Q/Z.

[cyan marten]

Magic

[rustic crown]

oh, so handling the free divisible part... which must be Q^(n) needs the direct product

[cyan marten]

Yes.

[rustic crown]

[cyan marten]

Ohh.. Now I understand why this is the double dual lol

[latent anvil]

This was essentially my original proof

The thing with Q^n/K is an injective group which you embed into

and we have an explicit decomposition of it as a direct sum of Q/nZ's and Q's

So you just need to embed those

(also just shamrock is fine)

[rustic crown]

oh okie

[latent anvil]

And Hom(G, Q/Z) is the pontrayagin dual for finite G but in the infinite/topological case you want to map into the circle S^1 = R/Z

Q/Z is the finite order elements of this

(i don't really know this stuff, something about Fourier analysis on locally compact topological abelian groups)

[rustic crown]

Sounds very fancy

[latent anvil]

Yeah haha I've wanted to learn about it for years

But I haven't even gotten around to properly learning the usual Fourier transform

[proud bear]

shamrock, thought you went to sleep

[latent anvil]

Well, I saw it in an intro analysis course but didn't understand

dead shamrock lies sleeping in the sunken city of rleyh

[vestal snow]

Let $k$ be an algebraically closed field of characteristic $p$. There's a famous theorem by Boseck which says that if $E/k(x)$ is a function field extension of degree $p^n$, then all totally ramified points are Weierstrass EXCEPT when $E=k(x,y)$ where $y^p-y=\frac{ax+b}{(x-\alpha_1)(x-\alpha_2)}$. Can someone explain why the theorem fails in this case?

[cloud walrus]

Have a Banana, Bitch

[odd abyss]

Let E be a vector space on K, p and q projectors of E such that p ◦ q = 0. We consider r = p + q - q ◦ p.

1. Show that r is a projector of E.
2. Show that Ker (r) = Ker (p) ∩ Ker (q).
3. Show that Im (r) = Im (p) + Im (q). What is Im (p) ∩ Im (q)?
   Someone can help me with that pleas ?

[wraith obsidian]

I have no idea how to be helpful here since you just pasted an assignment without further context, like where specifically you're stuck on

[quaint ivy]

also #linear-algebra

most of these should follow by definition, maybe evaluating r^2 turns out to be r

[viscid pewter]

high out of my mind with sleep dep but i'm here to tell you that the commutator subgroup is a little bitch

at least i got there

[next obsidian]

commutator subgroup is based tho

[viscid pewter]

promise not to laugh

[next obsidian]

does not make promises he cannot uphold

[viscid pewter]

forgot how to prove it was normal and got stuck on it for ages

maybe 2 hours

g-1[x, y]g = g-1 x-1 y-1 x y g = (g-1 x-1 g)(g-1 y-1 g)(g-1 x g)(g-1 y g) = [(g-1 x g), (g-1, y, g)]

and then g-1[a, b][c, d]g = (g-1[a, b]g)(g-1[c, d]g) trivially so it all works out

[mild laurel]

the fact that the set of commutators is not always a subgroup is pretty neat

[next obsidian]

okay

so

I've done this many times

actually

Do you want to know the secret, based way to do this tho

Let x be in the commutator subgroup

Then look at

g^-1xg

Write out

g^-1xg = xx^-1g^-1xg = x[x^-1,g^-1]

then this is a product of two commutators so it's in the group

When someone showed me this I wanted to k-word myself since I've proven this twice and both times I did it the way you did by doing a shitty shitty computation

[viscid pewter]

yes

i remembered that that way existed but i couldn't remember how to do it

fuck

[next obsidian]

😭

[midnight solar]

all these permutations are cycles right

[next obsidian]

ewwwww

matrix notation for this

[magic owl]

honestly i like this notation for permutations

i always forget which way

the other one goes

[next obsidian]

but the only one that's a cycle there is the third one

the way the other one goes is merely a matter of convention

haha

[magic owl]

i kno

but i always forget the relevant convention

lol

[next obsidian]

I am team left to right

[midnight solar]

the first picture is my question and 2nd is my work

[next obsidian]

yeah and

[midnight solar]

1st is blue
2nd is purple
3rd is green
4th is yellow

[next obsidian]

only the third one is a cycle

[midnight solar]

the question is asking which is not a cycle

[magic owl]

3 of them

[midnight solar]

i can only chose one

[magic owl]

question is wrong

assuming your work is written correctly

i didnt check

[viscid pewter]

first one is a cycle

[magic owl]

it is not

oh wait sorry

[viscid pewter]

also 4th

[magic owl]

i see your point

[viscid pewter]

yeah, 3 of them are cycles

[magic owl]

yes okay these are all cycles after composition

my bad

thank u kaisheng

so the 2nd one isn't

[midnight solar]

ok why isnt the 2nd one

[magic owl]

as you can see

[viscid pewter]

it's two cycles

[magic owl]

if you remove the single-element parantheticals

you are left with one cycle

for all but one

[next obsidian]

oh lmfao

[magic owl]

i missed it 2

[next obsidian]

I feel really dumb tho hahaha

[magic owl]

well luckily im not sober

ill chalk it up to that

[next obsidian]

I am tho

chock

[viscid pewter]

lmao did i catch something you fancy folk didn't

[magic owl]

chock

r fancy brains cannot see the obvious

we are too enlightened

[viscid pewter]

ok time to take that to the bank and go to sleep

[next obsidian]

yeah

I'm too smart for easy things

scrubquote

like the guy who complains he's too good at fighting games

which is why he loses to scrubs

[magic owl]

oh i think i know who ur talking about

[next obsidian]

this isn't just 1 person

there's a lot of these people hahaha

[magic owl]

oh im thinking of a specific person

[midnight solar]

ok maybe im still not understanding how does it being (123) and (56) not make it a cycle

[magic owl]

who became famous

[next obsidian]

I bet you're thinking of Lowtier God tho

[magic owl]

maybe

[next obsidian]

A cycle

like

by definition

[magic owl]

each thing in parantheses

is a cycle

[next obsidian]

Idk how to explain this, what's your definition of cycle?

this is just like saying words tho haha

[magic owl]

u can tell its a cycle by the way it is

[next obsidian]

Amazing

[midnight solar]

but why do single ones get the pass

[magic owl]

they represent the identity permutation

so you can ignore them

(if you didn't then being a cycle would not be well defined)

like

we agree (123) is a cycle

but (123)=(4)(123)

so if the left hand side is a cycle

the right hand side needs to be for this notion to make any sense

because they represent the same element of S_4

I don't think so

[midnight solar]

no i got it

[magic owl]

see

[midnight solar]

thanks

[magic owl]

np

[midnight solar]

ok another question does the operation matter when you are talking about isomorphism. Say i have two groups G and H. Here is my definition; G and H are isomorphic if you have a bijective homomorphism. But what if G and H have differenet operations does that change anything?

i was thinking how would you check group sturcture to show they are isomorphic

[magic owl]

The theorem you want to think about is

If G and isomorphic

you can take two elements in G

and send them to H

multiply them in H

bring them back to G

and get the right answer

so in some sense

they have the same multiplication

(if you have an isomorphism)

Well

sorry

you need to be able to do this in both directions

otherwise you just get monomorphism

[midnight solar]

ok ok got ya

thanks

[magic owl]

mirza this is just the homomorphism condition

the condition im refering to is different

$f^{-1}(f(x)f(y))=xy$

[cloud walrus]

max

[magic owl]

ah i see I thought you were converting what i said into symbols

[midnight solar]

trying to think of a way of proving it without picking an example

to prove that something is not isomorphic can you just show a case where it isnt

[next obsidian]

whaaaaaa

we can't call Z and R having teh same operation

You can obtain Z

as a restrictino of R's but...

I mean I guess I see what you mean, but I don't think that's relevant

sure

but there's a much deeper issue with Z and R haha

Anyway BigWall

until you obtain invariants it's hard to distinguish things as being isomorphic

the like coarsest way is just if the sets are different sizes

sincea bijection would imply same size

yeah

Like, it can be hard to show two things are not isomorphic

You could check if one of them is abelian, and the other isn't

compute the size of the center

look at subgroups

etc.

[midnight solar]

thats what i was thinking cardinality right

[next obsidian]

yeah

R is really big

😔

I wanted to know all the numbers...

😭

[midnight solar]

what was the answer you wanted me to answer to that question

No i think i got it now

[glacial bloom]

any tips on how to solve this? I don't see how order plays into the 2nd homomorphism theorem

[mild laurel]

just consider both sides of the isomorphism in the 2nd homomorphism theorem

since everything is finite, both sides must have the same order

[glacial bloom]

got it thanks

[scenic sage]

{1,2,3} and {a,b,c} are isomorphic but like i can desin function not matching this

but this does not show that they are not isomorphic

[scarlet estuary]

the typical way to prove two things arent isomorphic is to prevent an isomorphism invariant that they differ on

such as cardinality, maximal order, whatever

"isomorphism invariant" here is jargon for "thing that isnt changed by an isomorphism"

so for example, isomorphisms (since they are bijections) shouldnt change the cardinality (size) of your structures

so if your two things have different cardinalities, they must not be isomorphic

thats a fairly easy example, typically they'll be a bit tougher than that

but theres no general technique; you'll have to give us more specifics if you want more specific help

[this isnt the ONLY way to prove two things arent isomorphic, fwiw; another common approach is to assume they are and then apply some other theorem interchanging the structures to derive a contradiction]

[if that sounds vague, it's intentional; again, these really have to be tailored to the situation]

[but for example, if you want to show something is not isomorphic to a certain quotient group, you could try looking at the isomorphism theorems for groups and seeing if you get a contradiction if you replace the quotient group with your other group]

[i'd look for differing invariants first, though.]

[midnight solar]

My response was badly worded. I was trying to ask if cardinality would be enough to justify not an isomorphism. I understand now. They explained it well last night and @scarlet estuary comment helps as well.

[oblique leaf]

Why is it that if ord(a,b)=mn, then ord(a)|m and ord(b)|n?

[carmine fossil]

That's called Lagrange theorem

Order of an element of a group divides the order of that group

You are looking at "a" as a member of Z_m

[oblique leaf]

Doesn’t “a” have to form it’s own subgroup under Z_m for lagrange to apply?

[carmine fossil]

Consider <a>

Number of elements in this subgroup=|a|

[oblique leaf]

Oooh got it! Tysm

[chilly ocean]

Do quadratic residues fall into AA?

[uncut girder]

so you have the vector space of binary cubic* forms

and you define a GL(2) action on it

linear substitution of variables

[delicate bloom]

wait why is this a binary quadratic form

this looks like a homogeneous cubic polynomial

[uncut girder]

i meant cubic sry

[delicate bloom]

oh ok cool continue

[uncut girder]

so a binary cubic form can also be represented as a 4-tuple

then the action of a matrix

will be as follows

[delicate bloom]

should I be reading this as a 2x2 block matrix of 2x2 sub matrices

also the (g.x)' makes me think this is the derivative of something not g.x itself or, well ok

[uncut girder]

nah its actually g.x

[delicate bloom]

I guess I can wait til the end to ask questions or we'll stop too often to get anywhere exciting

[chilly ocean]

Can I ask a question when you guys are done?

I'll wait, it's cool

[uncut girder]

okay so you have a discriminant function for binary cubic forms

and then you can verify

i mean that's pretty much all I wanted to say for now

I verified that in Sage

The the orbits of GL(2) in the vectorspace of binary cubic forms actually parameterize cubic number fields

so that's cool

but i have to understand that stuff better

I'm writing some notes on this stuff if you wanna check it out after im done

@chilly ocean go ahead

[delicate bloom]

what makes this binary cubic form interesting in particular

the orbits parametrizing cubic number fields seems cool though

[uncut girder]

its exactly the parameterizing cubic number fields that's interesting

if you can understand the orbits, you can understand various counting questions of cubic number fields

[chilly ocean]

,tex I'm taking a course on cryptography and we covered quadratic residues and non-residues. I know that $\mathbb{QR_N}$ is the set of quadratic residues, and that $\mathbb{QNR_N}$ is the set of quadratic non-residues. I also know that $x^2 \equiv q (\text{mod} p$, and therefore the set $\mathbb{QR_5} = {1, 4}$, since $1^2 \equiv 1 (\text{mod } 5)$ and $2^2 \equiv 4 (\text{mod } 5)$, but what's the calculation for the set of quadratic non-residues? I know that that set is the set of numbers that aren't in $\mathbb{QR_N}$ but how is it calculated?

[cloud walrus]

Dark Angel
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[uncut girder]

such as how many cubic number fields do there exist with discriminant bounded by N, asymptotically as N -> infty?

[chilly ocean]

Oops

[uncut girder]

you can actually give an asymptotic formula for this count

using this parameterization

[delicate bloom]

ooh neat

[uncut girder]

this is not the only space which parameterizes numbe fields

for example, you can consider binary quadratic forms under a GL(2) action

this parameterizes not only quadratic number fields, but also ideal classes of quadratic number fields

Gauss used this to prove the class number formula for quadratic number fields, i think

[delicate bloom]

interesting, does this generalize to quartic and quintic etc number fields?

[uncut girder]

yes

basically yes

quintic is the highest we've gone i believe

Manjul Bhargava is famous for considering the quintic case

there's still a lot we dont know

[delicate bloom]

interesting, I would have figured this would have all been worked out

cool

[chilly ocean]

,tex I'm taking a course on cryptography and we covered quadratic residues and non-residues. I know that $\mathbb{QR_N}$ is the set of quadratic residues, and that $\mathbb{QNR_N}$ is the set of quadratic non-residues. I also know that $x^2 \equiv q (\text{mod } p$, and therefore the set $\mathbb{QR_{5}} = {1, 4}$, since $1^2 \equiv 1 (\text{mod } 5)$ and $2^2 \equiv 4 (\text{mod } 5)$, but what's the calculation for the set of quadratic non-residues? I know that that set is the set of numbers that aren't in $\mathbb{QR_N}$ but how is it calculated?

[delicate bloom]

I guess it becomes pretty laborious to try to do any of this work without a computer

[chilly ocean]

Sorry for the interruption, there's the question I had when you guys are finished

[uncut girder]

no worries

[delicate bloom]

well computing one necessarily means you've computed the other

since it's the complement of the other set

so it's a waste of time to "compute" them both

that make sense? @chilly ocean

[cloud walrus]

Dark Angel

[chilly ocean]

Well, kind of. I understand it, but I thought there would be a formula for the set of quadratic non-residues like there is for the set of quadratic residues

I don't know why LaTeX thinks a 5 is that weird symbol it seems to have replaced it woth though

[delicate bloom]

what do you mean formula

it's more of an algorithm

square the numbers from 1 to (p-1)/2 mod p, that's the set of quadratic residues

remove those from the set of 1 to p-1 and that's the set of quadratic nonresidues

[chilly ocean]

,tex Like, you know how the set of quadratic residues is the set of solutions to the equation $x^2 \equiv q (\text{mod } p)$? Like that but for quadratic non-residues

[cloud walrus]

Dark Angel

[delicate bloom]

just write $x^2 \not \equiv q \mod p$

[cloud walrus]

Merosity

[delicate bloom]

I guess a worse way to get them would be to take the quadratic residues, pick a single quadratic nonresidue, then multiply it by every quadratic residue and that is a bijection onto the entire set of quadratic nonresidues

there's nothing holy about having a formula, I wouldn't worry about it

[chilly ocean]

I was trying to compute the two sets and I came up with this:
,tex
\begin{displaymath}
1^2 \equiv 1 (\text{mod } 5) \
2^2 \equiv 4 (\text{mod } 5)
\end{displaymath}

[cloud walrus]

Dark Angel

[chilly ocean]

meaning that 1 and 4 are quadratic residues, and that what isn't there, i.e {2, 3} are QNR. But for that, I had these inequalities, and I can't see where the 2 and 3 are coming from like I can with QR
,tex
\begin{displaymath}
3^2 \not \equiv 9 (\text{mod } 5) \equiv 4 \
4^2 \not \equiv 16 (\text{mod } 5) \equiv 1
\end{displaymath}

[cloud walrus]

Dark Angel

[chilly ocean]

What's the bot got against line breaks?

[delicate bloom]

quadratic residue just means it's a square

quadratic non residue means it's not, so there is no way to write it as x^2=q mod p

the N in QNR means it's not the formula x^2=q mod p

[chilly ocean]

Oh, so because 1 and 4 are square numbers, they're in QR, and because 2 and 3 aren't squares, since sqrt(2) and sqrt(3) are irrational, they're in QNR?

[delicate bloom]

sort of

[chilly ocean]

And you only focus on the range 1-4 since it's modulo 5

[delicate bloom]

sqrt(2) and sqrt(3) don't really exist here

sqrt(6) could be considered to exist though, cause we're working with equivalence classes

[chilly ocean]

What does exist mean?

[delicate bloom]

well 6 = 1 mod 5

so could just say sqrt(6) = sqrt(1) = 1 mod 5

[chilly ocean]

Oh, I see

My brain went into philosophy mode and I was wondering if any numbers exist

[delicate bloom]

but it might not be a really good idea to try to do that, or at least there has to be a lot more discussion

like for instance should sqrt(4)=2 mod 3 or should sqrt(4)=sqrt(1)=1 mod 3?

so it's more sane to just think in terms of what squares rather than square roots

[chilly ocean]

So, if I want QNR, it's easier to just compute QR and look at what isn't in that set?

[delicate bloom]

yeah exactly

and you only have to check up to (p-1)/2

[chilly ocean]

Rather than trying to compute QNR first

[delicate bloom]

since after that you are getting the negatives

like mod 5 you do 1^2, 2^2 but then 3^2 = (5-2)^2 = (-2)^2 = 2^2 which you've already done

similarly 4^2 = (5-1)^2 = (-1)^2 = 1^2

[chilly ocean]

So it's pretty much pointless to have a formula for QNR?

[delicate bloom]

yeah

although there's quadratic reciprocity

[chilly ocean]

I suppose you could write {x | x \in Z_N}\{QR_N} for that though, since you're computing the set Z_N and removing all the quadratic residues

[delicate bloom]

well you don't have to 'compute' the set Z_N

but yeah

[chilly ocean]

Thanks for explaining that. I never really considered whether there might be an actual formula for QNR before, since it's just whatever isn't in QR

[delicate bloom]

look into Legendre symbols and quadratic reciprocity if you're interested in this more, it will give you a way to determine more about these

but it's not as efficient if you're just determining all of them like that

[chilly ocean]

I believe we covered Legendre symbols in my course so I'll have to look back at the material.

[wraith obsidian]

I think the word „formula“ is too vague to be helpful here (and adding „actual“ doesn't make things more precise). It's not clear whether you mean a declarative description of some sort or an algorithmic way of computing things.

[chilly ocean]

We had to use Legendre symbols for the Goldwasser-Micali encryption system

[delicate bloom]

here check out: https://brilliant.org/wiki/legendre-symbol/

they give an example of computing if 74 a quadratic residue mod 131

[chilly ocean]

I mean more like an algorithm, like how to determine if a number is in QR, we just solve the equation x^2 \equiv q (mod p)

[delicate bloom]

this would be more precise if you're wanting to determine a specific number

[chilly ocean]

I'll check that out, thanks

[delicate bloom]

rather than the entire set

[chilly ocean]

I think my programmer brain is taking over lol. I just applied the algorithm to compute a specific number multiple times to get the whole set

It was really inefficient

[delicate bloom]

also this is more #elementary-number-theory than #groups-rings-fields for future reference

[chilly ocean]

Ah, OK, will keep that in mind

Sorry about that

[delicate bloom]

idc lol

[clever rampart]

Does representation theory (well, mostly the basic stuff for finite groups) go here?

[carmine fossil]

Yes

[clever rampart]

https://cdn.discordapp.com/attachments/488104216678760469/820344932606017536/309013702982500352.png

Before this part, there was an exercise on when a conjugacy class of G splits in H (which I was able to do), but I am not sure how to use that for this, whether I should try computing inner products of characters with Frobenius reciprocity or something.

[mild laurel]

Yeah it should be frobenius reciprocity

[vital anchor]

would this be the place to ask about taking quotients of Z-modules? it's for homology computations

ooh nvm i got it

[thorn delta]

what does it mean for a cyclic R-module to have order r, where r is in R?

[uncut girder]

It means it's probably R/<r>

[thorn delta]

ah yea, found our definition. its basically that

[restive star]

Alright I have a question that seems super simple, but I can't quite solve it

Let K be an algebraic extension of F. Let σ : K → K be an injective homomorphism of
fields so that σ(a) = a for all a ∈ F. Prove that σ is surjective

So I took some element y in K that isn't in F

clearly there will be a polynomial with it as a root

but then after that I'm stuck

[latent anvil]

@restive star do you see how to handle the finite case?

Where [K : F] is finite I mean

[restive star]

hmmm

Im assuming using properties of a vector space?

[latent anvil]

yup

[restive star]

then yeah I think I get that part

but I don't see how it generalizes to the general case

[latent anvil]

I'm not actually sure this helps, it's just the first thing that jumps out to me

Since any algebraic extension is sort of built out of finite ones

[latent anvil]

So let's follow this chain of thought

You have y in F

Which you want to show is in the image

It satisfies some f in F[x]

yeah?

[restive star]

yep

so then I thought to plug it into such an f

and then take sigma of both sides

[latent anvil]

that's a great idea!

[restive star]

unfortunately it doesn't seem to work unless I'm missing something

[latent anvil]

It doesn't work exactly, but it tells us something really useful

If y has minimal polynomial f then σ(y) also has minimal polynomial f

right?

[restive star]

oooh yeah

[latent anvil]

I think you can combine this with our discussion of the finite case above to get a proof

(but let me know if you need more help)

[restive star]

Yeah I'm not really seeing where to look for the final jump

[latent anvil]

Sure

[latent anvil]

Can you think of a way to rephrase this in terms of the image/preimage of an extension of F under σ?

hello Brofibration

[sturdy marsh]

hello EShamrock -> BShamrock

[latent anvil]

What's the rofibration group anyway?

[latent anvil]

Sorry Randle, maybe this is too much of a jump

[restive star]

yeah sorry I'm really thinking haha

[sturdy marsh]

sigma takes roots to roots, f has finitely many roots

[latent anvil]

Oh, that's a more direct way too it

[restive star]

ooooooooooh

[latent anvil]

Although you want "roots in K"

[sturdy marsh]

yes

[restive star]

so then the finite case pretty much applies?

[latent anvil]

I was thinking of taking the extension of F by all the roots

But brofib is saying something even simpler

If R is the set of roots of f in K then σ restricts to a function R -> R, yeah ?

[restive star]

yep

[latent anvil]

Well, this is an injection between finite sets of the same size

[restive star]

which is therefore surjective!

right?

[latent anvil]

Yeah!

and then we win

[restive star]

you're really good at teaching haha

thanks so much!

[latent anvil]

ty haha I have a lot of practice working as a TA

[sturdy marsh]

is TAing fun?

[latent anvil]

Yes!

I love it so much

One of the worst things about the pandemic for me is that TAing online is just awful

I'm going to try it again next quarter for the first time in a year

I really love explaining things to students and seeing it click

It's also really helped my public speaking skills

I got comments when doing my practice talk on dold kan that I sounded like I'd really practiced my lines and spoke with confidence

[sturdy marsh]

noice

[latent anvil]

I had not practiced at all

[sturdy marsh]

they don't let undergrads TA over here I think

which is a good thing ig?

more jobs for grad students

[latent anvil]

¯\_(ツ)_/¯

There's only two TA positions for undergrads in math at uw

For honors calc and honors analysis

But cs has a crazy huge TA program

Like a majority of the hours worked by course staff in uw cse are undergrads

Or something crazy

(cs has insane funding so this isn't coming at the cost of grad student ta positions or anything)

[chilly ocean]

how tensor product???

[uncut girder]

What

Do you have a question based ttera

[chilly ocean]

yes

how tensor product

[uncut girder]

Tensoring changes your base ring/field

[chilly ocean]

does tensoring change my life

[uncut girder]

Its gives you more perspectives

[magic owl]

tensoring doesn't necessarily do that tho

[bleak abyss]

You just take bilinear product and divide by two

[magic owl]

like if M,N are R modules M\otimes N is still an R module

[uncut girder]

Right so I was referring to extension of scalars

[magic owl]

sure yeah

but i would not call that the "point" of tensoring

[uncut girder]

Yeah it's just one of many

[next obsidian]

Oh yeah well

Actually M (x) N is an abelian group

you need them to be bimodules

[restive star]

Alright I have another question haha

Prove that $F_{p^n} = F_p(α)$ for some element α

[cloud walrus]

randleS

[restive star]

I cant seem to find such an element for a general field

[sturdy marsh]

the group of units has p^n -1 elements

so every element satisfies x^p^n - x

[restive star]

whereas in the original field, every element satisfies x^p-x, right?

[sturdy marsh]

yes 0 satisfies that equation too, so every element satisfies it

oh wait I misread

elements of F_p satisfy x^p - x, yes

but that may not be very helpful

anyway, try to figure out what happens when you adjoin a root of x^p^n - x to F_p

[restive star]

alright I'll try that. Thanks!

[sturdy marsh]

hmm perhaps what I had in mind isnt the best way to do it

[restive star]

yeah I'm working on it and I cant quite see how to figure it out

any other tips?

[oblique river]

@chilly ocean simply tensor the modules together

and that's the tensor product

if you have two things just ask yourself "what would these look like if they were one thing?" and that's the tensor product

it's like those websites where you upload two pictures of people and it tells you what your baby would look like

that's the tensor product

[sturdy marsh]

If G is a finite abelian group of order n such that there are at most d elements of order d for each d dividing n, then G is cyclic

[stark sigil]

How are we defining $\bF_{p^n}$ in the course? e.g. $\bF_p[x]/p(x)$ for some irreducible degree $n$ polynomial?

[cloud walrus]

Icy001

[restive star]

field of order p and P^n respectively

sorry I shouldve specified

[stark sigil]

That's not quite a definition

unless you prove theorems about existence and uniqueness of such fields

[restive star]

ah well typically we do use Fp[x]/p(x)

[stark sigil]

so you'll just have to prove that x generates the field

[restive star]

I just realized that we already proved in class that Fpn is the splitting field of Fp

so I can literally just use that to get that Fp(r) is Fpn for any root r of x^p^n -x and just be done lol

[sturdy marsh]

it doesnt work for any root

for example, 0 is a root

[oblique river]

@latent anvil @chilly ocean smh why

smsmh

[restive star]

oh shit yeah I'm a moron haha

could I say it works for any nonzero root though?

[sturdy marsh]

no

1 is a root

and it doesnt work for 1

but again, like icy said, show that X generates it

if X did not generate it, it would sit inside an extension of lower degree

and this is a problem

[restive star]

wdym by x? Im sorry I can tell Im asking stupid questions. I'm really bad with this stuff.

cause wouldn't that just be some root of x^p^n - x?

resulting in the same issues with it potentially being 0 or 1

[sturdy marsh]

no worries, F_p^n = F_p[X]/(f) for some irred poly of degree n right?

[restive star]

yep

[sturdy marsh]

by X, I meant the equivalence class X + (f)

[restive star]

ooooh ok

[sturdy marsh]

in F_p[X]/(f)

[restive star]

thx

[latent anvil]

why am I being whyd

Oh right

[sturdy marsh]

smh

[latent anvil]

I thought it was a very good explanation

[oblique river]

why is why

[latent anvil]

But also very worthy

It can be both

[oblique river]

ok i'll take that

😌

[latent anvil]

I think of tensor products as sort of transferring the structure of one thing onto another I guess

Change of scalars, tensoring with A/I, tensoring with a localisation, tensoring an algebra with a polynomial ring

[oblique river]

I feel like understanding tensor products was a real mathematical reckoning for me because of that

like, on the one hand

[latent anvil]

lots of examples where you can encode some operation as tensoring with the ultimate example of that operation

[oblique river]

tensoring is a symmetric operation

[latent anvil]

right

[oblique river]

but on the other hand it's not

[latent anvil]

That's a really good observation haha

[oblique river]

like in those examples, with base change or like tensoring A with k[x] or something -- those two things aren't playing the same roles

X \otimes_K L

[latent anvil]

the tensor and the tensee

[oblique river]

hahahaha exactly!

hahahaha

[chilly ocean]

tensor product with a larger ring can act as "extending" the ring of scalars (e.g. complexification), and it has the property that bilinear maps on the cartesian product correspond to linear maps on the tensor product

this is about all i know

other than the definition

[latent anvil]

petthewolf

[chilly ocean]

i can't, nitro ran out

[latent anvil]

damn

[oblique river]

you gotta find another simp

[latent anvil]

Anyways yeah tterra a tensor is a section of the tensor bundle

smh

You should know this

[chilly ocean]

what other important things should i know about the tensor product (of modules)?

genuine question

[latent anvil]

There's a bunch of rules

[chilly ocean]

this time its not a shitpost

[sturdy marsh]

tensor hom

adjunction

[latent anvil]

M (×) A/I ≈ M/IM

[chilly ocean]

[sturdy marsh]

tensoring with a field kills torsion

[latent anvil]

M (×) (N (+) L) = (M (×) N) (+) (M (×) L)

[oblique river]

yeah so actually shamrock's description of "transporting one structure onto another" is really good i think

[sturdy marsh]

torsion tensor divisible is 0

[latent anvil]

more generally tensors commute with arbitrary direct sums

[oblique river]

things like his example of M x A/I = M/IM

[latent anvil]

Concretely: Tensoring with Q over Z kills torsion

[oblique river]

transferring the structure of A/I onto M is how I think about it

[latent anvil]

whoops lol

I just did a 3 hour review session with a classmate for my analysis final

[oblique river]

for me, when I'm tensoring things, it's usually something like, tensoring one field extension with another

[latent anvil]

I'm a little woozy

[chilly ocean]

[oblique river]

i've been drinking liquor for 4 hours

so i'm also a little woozy

[latent anvil]

hahaha

I was going to do either that or my French exam

(exclusive or)

[oblique river]

french exam on a saturday? 🤢

[chilly ocean]

wee wee

[latent anvil]

I get 48 hours and can choose 2 to do it in

So it's not so bad

but having it on the weekend is annoying

[chilly ocean]

why is everyone taking french

[latent anvil]

je ne sais pas

[sturdy marsh]

to read ega/sga/fga

[chilly ocean]

AGoomer chat

[latent anvil]

je ne connais pas ? idk

[oblique river]

i took french because i lived in texas and everyone spoke spanish and i wanted to be different

also because im gay lmao

[latent anvil]

lol

[chilly ocean]

ah is it time for me to take it as well

[latent anvil]

I took Japanese in high school and everyone was a weeb and it was so fucking embarrassing

[oblique river]

oui

tu peut parler avec sham et moi

[latent anvil]

oui !

[oblique river]

voulez-vouz parler avec moi, ce soir

(that's a reference btw)

(for those of you uncultureds)

[chilly ocean]

[dusty peak]

https://ega.fppf.site/

[chilly ocean]

i took two semesters of russian

what can i read?

[latent anvil]

Russian

Is this a trick question?

[chilly ocean]

no

[latent anvil]

Idoit

[chilly ocean]

online killed my interest in learning russian

[carmine fossil]

Perelman

[sturdy marsh]

privyet

[chilly ocean]

privyet

kak dela?

[sturdy marsh]

cyka blyat

[chilly ocean]

yuo are dso funy

[sturdy marsh]

is that russian?

[chilly ocean]

i am not convinced language courses are very fun in an online format

[latent anvil]

I have evidence to support this view

My prof isn't teaching next quarter so I need to adjust to a new french prof

Scary

[chilly ocean]

just whip out your 🥖

[latent anvil]

[stark sigil]

Another interesting tensor product: $B\otimes_A B$ for a ring map $A\to B$

[cloud walrus]

Icy001

[latent anvil]

yes!

Trying not to scare tterra with ag

[sturdy marsh]

also restricting sheaves to subschemes

[latent anvil]

Lol

[stark sigil]

I spent today working out Galois descent and comonadic descent and I had to deal with objects such as $V_\bR\otimes_\bR(\bC\otimes_\bR\bC)$

[cloud walrus]

Icy001

[sturdy marsh]

this is a special case of extension of scalars

but induced reps

if H < G is a subgroup

and if M is an H module

then Z[G] (x)_H M is a G-module