groups-rings-fields

[next obsidian]

I’m not sure actually

I think Fall is Max right

Maybe Sándor does winter

[latent anvil]

Okay anyways

Is the map I defined above an open immersion? Y/n

You have ten seconds

[next obsidian]

Fuck you

Idk

I think so

The map on sheaves is injective

And the topological map is

[latent anvil]

Wait really?

[next obsidian]

So I want to say you use first isomorphism for topological spaces

[latent anvil]

Wow

That's pretty cool

[next obsidian]

Look at II.2.18 b)

Sheaf map injective iff ring map is

For maps of affine schemes

[latent anvil]

Oh yeah I realized that

It was the underlying map on topological spaces open I was surprised by

[next obsidian]

Wait what

[latent anvil]

Don't you need that to be an open immersion?

[next obsidian]

Idk I just guessed

But honestly I feel like morally this should be

Like why the hell wouldn’t a localization embed as an open immersion

[latent anvil]

That's my question lol

Well, take a stalk

[next obsidian]

Don’t we need the map to be open?

[latent anvil]

Or localize at a prime or w/e

?

That's a localization

[next obsidian]

Is open,

[latent anvil]

what

[next obsidian]

Oh yeah

Sure I se

What you mean

Idk, I don’t see how this helps

[latent anvil]

The open bit?

You want to describe this space

Well it comes with a kind of embedding into a nice space

[next obsidian]

I mean viewing it as a sub scheme

[latent anvil]

Let's try to understand that

[next obsidian]

Isn’t this going to be the same thing basically as it being primes of k[s,t] which avoid that set?

[latent anvil]

this could theoretically let me draw a picture tho

Also I'm not sure what you mean

[next obsidian]

I mean primes of this are primes of k[s,t] localized at that set

[latent anvil]

Yes

[next obsidian]

Of d(x)r(t)

So primes of the localization are primes which avoid said set

[latent anvil]

I thought you gave up on that though

As too hard

[next obsidian]

I said I was honestly fine with that

[latent anvil]

But we can still maybe say things about the geometry of this scheme

Oh

[next obsidian]

I don’t see how I can draw a picture of Spec k[s,t] even

And I feel like the geometry of that embedding is exactly the same info as that of that ring theoretic description

[latent anvil]

I mean, I draw a plane

And pretend k = R but it's algebraically closed

[next obsidian]

Sure

Okay let’s take a step back

Primes in k[s,t] are either 0, height 1, or maximal

Right?

Err...

[latent anvil]

yeah, because dim 2

Is that wrong?

Idk

[next obsidian]

Is maximal necessarily height 2

[latent anvil]

Don't trust me

[next obsidian]

I know height 2 is maximal

[latent anvil]

You mean could maximal be height 1?

[next obsidian]

Yeah

[latent anvil]

Hmm I say no

The quotient is a field

[next obsidian]

Yeah

[latent anvil]

So dim 0

[next obsidian]

Oh yeah

Nice

Okay

[latent anvil]

ht + dim of quotient = 2

[next obsidian]

Are all height 1 ideals principal?

[latent anvil]

I think so

Sounds hard

Oh yes

They are

I proved this at some point

It's like a codimension 1 hypersurface is defined by a single equation

You want to show there's an irreducible element

Right okay

Let p <= k[x, y] be prime and height 1

Take any nonzero element

Factor it

One of those factors has to be in p

so p contains an irreducible f

Thus (f) <= p

But height 1, so (f) = p

@next obsidian yeah?

[next obsidian]

Ah yeah

[latent anvil]

This works in a ufd

Height 1 prime is principal

[next obsidian]

Okay so remember the Jacobson ring thing

[latent anvil]

Hi

*no

[next obsidian]

Well

f.g. Algebras over a field are Jacobson

This means a lot of things

But I ended up on the stacks page for them

[latent anvil]

Yeah I mean that's true

[next obsidian]

So k[s] and k[s,t] are both Jacobson

[shy bluff]

Wait quick question

[latent anvil]

yeah liria?

[shy bluff]

Are all non units of a ring contained in its ideals?

[next obsidian]

yeah

I mean every element is contained in an ideal, namely (1)

but if a is not a unit then (a) is proper

[shy bluff]

oke just wanted to double check

Thank

[stone fulcrum]

Every element is contained an a maximal ideal

[next obsidian]

So Brendan, if R is Jacobson and you have a map R -> S which is of finite type then S Jacobson means the map Spec S -> Spec R sends closed points to closed points

[stone fulcrum]

Iff choice

[next obsidian]

This means maximal ideals pull back to maximal ideals right?

[latent anvil]

Weird okay

[next obsidian]

So your height 2 ideals pull back to maximal ideals of k[s]

so they contain some non-zero polynomial P(s) right?

[latent anvil]

Hang on

You don't need this I think

err okay maybe you do in the non algebraically closed case hmm

Okay nvm I agree now

[next obsidian]

But anyway you need to contain a non-zero polynomial p(s) right>

[latent anvil]

But yeah sure

[next obsidian]

So the maximal ideal becomes everything in the localization?

[latent anvil]

I've definitely done this more explicitly

[next obsidian]

Since you have a unit

[latent anvil]

Hmm

one sec

You're saying take a maximal ideal of k[s, t]

[next obsidian]

yup

[latent anvil]

That pulls back to something nonzero in k[s]

[next obsidian]

maximal

ideal

[latent anvil]

why do we need maximality?

[next obsidian]

uh

I mean sure

[latent anvil]

So there's some nonzero element in k[s] intersect m

[next obsidian]

yeah

I guess that's all you need

[latent anvil]

That's a unit in S^-1 m

Right?

[next obsidian]

Yeah

[latent anvil]

Okay cool

[next obsidian]

So you only need to look at height 1 ideals

[latent anvil]

And that's easy

[next obsidian]

which you said are all principal

[latent anvil]

They're principal

[next obsidian]

and you know what those are

specifically which ones stay prime

[latent anvil]

So it's just, do they factor like f(s) g(t)

Right

[next obsidian]

yup

and finally you got (0) just chillin

[latent anvil]

Okay so the image in A^2 contains no closed points

[next obsidian]

Uhhh

[latent anvil]

It contains some plane curves

Whack

[next obsidian]

yeah i guess so yeah

[latent anvil]

Oh wait sorry they can't factor like f(t) g(s) and be prime

[next obsidian]

Wait so do you think you can show that a maximal ideal of k[s,t]

[latent anvil]

Okay so that's a complete description

[next obsidian]

contains some p(s)

without the Jacobson bs

[latent anvil]

I think I have done it in the past lol

Not sure I could do it now

I'm rusty

[next obsidian]

Actually I think I can do it easy

sort of

so I don't need p(s)

I need p(s) or p(t) right?

[latent anvil]

Right

[next obsidian]

as long as you can show there's at least one thing of f(s)g(t)

primeness implies it

[latent anvil]

right

[next obsidian]

now why is that true

[latent anvil]

If not

All those are nonzero in the quotient

Seems weird

[next obsidian]

yeah

Idk, I at least have a proof now

rather, a complete description

This problem was whack

and honestly after doing II.1 and II.2

I was like, okay the comm alg I know should be good for a while

[latent anvil]

Yeah this is so cool

[next obsidian]

but man II.3

[latent anvil]

lollll

[next obsidian]

just really fucks you for not knowing it

What Sandor said that you'd get boned not knowing it really makes sense now

[latent anvil]

Something for me to look forward to 🥴

[next obsidian]

haha

I mean once you do II.1 and II.2 you at least have a sense of the basics IMO

[latent anvil]

Maybe I will do that in my summer break

[next obsidian]

I mean you can't do shit really

but like

[latent anvil]

After the actual break

[next obsidian]

Yeah

[latent anvil]

Twitter voted for it

[next obsidian]

I think I'm gonna start working through Matsumura again

so I'll be back to Matsumura posting cool stuff

[latent anvil]

I wish I had been working with you all along

But I didn't have time and wouldn't have wanted to

[next obsidian]

I'm only on section 4!

[latent anvil]

Haha I'm not gonna catch up

[next obsidian]

Fair haha

So cool I can write another proof, contingent on some comm alg thing I'll learn at some point lol

[latent anvil]

I think you can do it explicitly

Still

[next obsidian]

Yeah, I guess

I'm just tired and want to actually feel like I'm making progress

I know I'll properly learn the stuff in the future anyway so ¯_(ツ)_/¯

[latent anvil]

No for sure

This seems like it was a good day

[next obsidian]

Yeah, I got up at like 10

which is like 5 hours earlier than it has been

so i've been getting stuff done

Oh also Brendan

I think I have an easier way to see what the tensor product is

So like you can write this as S^{-1}k[s] (x) T^{-1}k[t] where S = k[s] \ {0} and likewise for T right?

This turns into ST^{-1}(k[s] (x) k[t]) = ST^{-1}k[s,t] right?

then ST is exactly that set we described

I think you have to be careful about what exactly you're tensoring over but I think you can do that bimodule thing where you can like do associativity of tensor products over different rings when the thing is a bimodule

[latent anvil]

Hmm plausible ig

[kindred mist]

for some reason google is not being very helpful here

having trouble thinking of an example of a group G where TorG is not a subgroup of G

i.e. where there exists g, h in TorG where gh does not have finite order

I think we have to look in nonabelian (infinite) cases, just not sure which, e.g. direct products and S_X for example don't seem to be helping a lot here

[oblique river]

how explicit does it need to be? consider the group <a,b | a^2, b^2>

a and b are finite order, but ab has infinite order, kind of by construction

if you want something explicit, try playing around with invertible matrices

[kindred mist]

do you mean a^2 = b^2? I'm not sure I get the presentation you stated

explicit or not explicit is fine

[latent anvil]

a^2 = b^2 = 1 in that notation

[kindred mist]

ty

[next obsidian]

So the residue field of (s - a) in k[s] is just k right?

This is for all a, zero or non-zero?

Yeah I'm pretty sure this is what it is

[delicate bloom]

yeah, a can be 0

[next obsidian]

Yeah but the computation is correct either way right?

I mean "computation"

[delicate bloom]

I don't know what you mean with or without quotes

[next obsidian]

I just mean that the residue field of (s - a) is indeed just k

[delicate bloom]

yes

[next obsidian]

really it's k[s]/(s - a)

but that's k

although it matters since I'm tensoring that it be of that form for my purposes

cool thanks

[delicate bloom]

I think the a=0 case is the most obvious personally

[latent anvil]

same

[shy bluff]

Question: For this, K[x]/(x^n) would just be all polynomials up to degree n right? then its maximal ideal is just (x)?

[delicate bloom]

sort of

when you say polynomials up to degree n, is that including degree n?

[shy bluff]

Not including degree n

Bceause x^n is just 0 in the quotient

[delicate bloom]

then you're good

and yeah, maximal ideal is (x)

[shy bluff]

Ok that's what I thought

[delicate bloom]

👍 cool

[shy bluff]

I think I'm starting to get some of this stuff

[delicate bloom]

you might like p-adic analysis before too long

[shy bluff]

I don't even know what those words means

[next obsidian]

You can think of it as looking at what happens when you quotient out by powers of an ideal

successively

[shy bluff]

Yea

[next obsidian]

You get a topology and stuf

[shy bluff]

Oh

[next obsidian]

and you can get cool results

[shy bluff]

Ithought you were talking about the question I posted

[next obsidian]

This is an example

[shy bluff]

lol

Ah

[next obsidian]

Sort of

if you take p = (x)

then you're sort of looking at that

I only vaguely know the idea of it though lol

[next obsidian]

It's well known that if $M_\mathfrak{m} = 0$ for all maximal ideals $\mathfrak{m}$ that $M = 0$, if $M$ is finite, when combined with Nakayama's lemma this implies that if $M\otimes_A \kappa(\mathfrak{m}) = 0$ for all $\mathfrak{m}$ that $M = 0$. One can actually weaken the conditions here

[cloud walrus]

Mathemagician:

[next obsidian]

If $f\colon A\to B$ is a homomorphism of rings, and $M$ is a finite $B$-module, then if $M\otimes_A \kappa(\mathfrak{p}) = 0$ for all $\mathfrak{p}$ prime ideals of $A$, then $M = 0$

[cloud walrus]

Mathemagician:

[next obsidian]

one can weaken the assumption very slightly so that you only need to know this is true for all $\mathfrak{p}$ of $A$ such that $\mathfrak{p}$ is the inverse image of a maximal ideal of $B$

[cloud walrus]

Mathemagician:

[latent anvil]

Weird

[next obsidian]

Let’s see if I remember the proof. Suppose that M isn’t 0, then for a maximal ideal P of B you know that M_P/PM_P ≠ 0 by Nakayama. Let p be the contraction of P in A, and let S = A \ p. Note that M_p = M_S when localizing wrt A, and this is further equal to M_f(S) when localizing wrt B. Let T = B \ P then note that f(S) < T, this tells you that (M_f(S))_T = M_P.

Note that pM_P is a subset of PM_P so that M_P/pM_P is non-zero.

From here you get that 0 ≠ M_P/pM_P = (M_f(S)/pM_f(S))_T = (M_p/pM_p)_T = (M_p (x)_A k(p))_T so that in particular M_p (x)_A k(p) must be non-zero

The first step is because there must be a maximal P such that M_P isn’t 0 then by Nakayama M_P/PM_P is non-zero

Further whack thing. Let M be a finite A-module and U_r the set of primes such that M_p is generated by r elements. This is open in Zariski. If M is also of finite presentation then U_F which is the set of primes such that M_p is free is open.

[vestal snow]

I think that there is a typo in this proof. $p_2$ should be equal to $(B_{q_1}p_2)^c \cap A$

[cloud walrus]

Have a Banana:

[fading wagon]

#help-8 on Cayley-Bacharach Theorem

[steady axle]

I think that there is a typo in this proof. $p_2$ should be equal to $(B_{q_1}p_2)^c \cap A$
@vestal snow i think its correct. $(B_{q_1}p_2)^c$ means $(B_{q_1}p_2) \cap A$ so it doesnt matter

[cloud walrus]

bert:

[vestal snow]

Hmm I'll have to look at it again

I skipped it for now

I have two questions about this proof:

1. Where did we use that x^-1 is not in B? (figured this out)
2. How do we know that x is algebraic over k?

For 2), I can show that x must satisfy some polynomial equation over k, but I couldn't show that it is not the zero polynomial

Here's my progress so far.

Since $k' = k[\bar{x}], \bar{x}=\bar{b_0}+\cdots +\bar{b_n}\bar{x^n}$ where $b_i \in B$. Assume that the polynomial obtained by subtracting $\bar{x}$ from both sides is the zero polynomial

This implies that $b_0, b_2,\cdots , b_n \in m'$ and $b_1-1\in m'$

[cloud walrus]

Have a Banana:

[vestal snow]

Since 1 is not in m', b_1 is not in m'

I don't know how to proceed

[charred pewter]

I’m not sure where to go from here^

Like how could I prove that A^-1 is in the form of H

[somber rivet]

Well, look at how each entry of $A$ is defined, rather than looking at it in terms of $x$ and $y$. We have the following:

\

$a_{11} = x$

\

$a_{12} = y$

\

$a_{21} = -y = -a_{12}$

\

$a_{22} = x+4y = a_{11}+4a_{12}$

\

Now, look at $A^{-1}$, which you have found correctly. Do each of the entries obey the general rules above? If they do, then the given matrix belongs to $H$.

[cloud walrus]

Abhijeet Vats:

[somber rivet]

@charred pewter

[charred pewter]

Sorry I was trying to decipher that cause TexIt wasn’t working lol

[somber rivet]

Yea lmao Idk why it didn't render it

but hopefully, it's more understandable now

[charred pewter]

Oh okay, so those equations for each component should confirm that A inverse is an element of H

[somber rivet]

Yeap. a_{11} and a_{12} are fixed in this instance

a_{21} and a_{22}, then, depend on them

[charred pewter]

Got it

Thank you!

[somber rivet]

You're welcome.

[next obsidian]

@vestal snow about your original question, you assumed that A is a subset of B so you don’t need to contract the ideal. For the second, can I see more context?

[vestal snow]

Sure thing

@next obsidian

[next obsidian]

Oh

I think I may have gotten it?

Do you know Zariski’s lemma?

It says that for field extensions, K \ k that K being finitely generated as an algebra implies that it’s finitely generated as a vector space

[vestal snow]

I do know it, but that comes after this theorem

[next obsidian]

Oh

Well... I was gonna say if you use that...

Then k’ = k[x] so it’s finite over k

[vestal snow]

I'm sure that you can prove the lemma without using this theorem, but the book must be using a different argument

[next obsidian]

Otherwise I’m sort of at a loss as to how you get the result otherwise

I guess hmm

[vestal snow]

@oblique river If you have time, can you look at this?

[vestal snow]

@next obsidian I figured it out

[next obsidian]

Oh nice, what did you do?

[vestal snow]

It's much simpler than I had thought

If x is not algebraic, k[x]=k' is ismorphic to the polynomial ring of one variable over k

and that is not a field

[next obsidian]

Yup

Oh lmfaooo

Yeah, that’ll do it lol

Funny enough another problem I have sort of hinged on remembering that if an extension isn’t algebraic that one of the transcendental elements forms an isomorphic copy of a polynomial ring

Which is basically what you pointed out here 🙃

[vestal snow]

It seems like the proofs get more and more subtle as one delves deeper into algebra

Why is it possible to define f_1 that way?

Intuitively, you wouldn't get any "contradictions" defining the map that way.

because the map agrees with the polynomial relation

[charred pewter]

How do I get the elements of an alternating group of order n

I read a forum that said that if n is odd, An is generated by (123)(1234...n) and that if it's even, it would be (12)(1234...n)

Is that accurate?

[next obsidian]

@vestal snow when you adjoint u^{-1} what you're really doing is localizing at the set {1,u,u^2,...} in which case the definition of that map is exactly the map induced by the universal property of localization

@charred pewter I have no clue if that claim is true, it seems fake to me, it doesn't seem like that has a high enough order, if say, n = 5 then the order of that element is 15 but A_5 has 60 elements. In general it's just "the set of even permutations" which like

¯_(ツ)_/¯

[vestal snow]

Ahh that makes sense

[charred pewter]

So basically for A_12 for example, I need to get all the even permutations from S_12

[vestal snow]

In integral domains, all localizations are embedded in the field of fractions right?

which is why adjoining u^-1 is consistent with localizing at the set {1,u,u^2,...}

[next obsidian]

Yeah Aur that's the definition

Yup Banana

I mean all you need is that u isn't nilpotent

then localizing at {1,u,u^2,...} will actually be an embedding

but adjoining u^-1 when u is nilpotent doesn't even make sense because if you just take (1/u)^n for a high enough n you're dividing by 0

[next obsidian]

When you start a proof by saying "by quotienting out by J we can assume WLOG that J = 0" to simplify a proof and you feel like a baller 😎

[charred pewter]

When referring to A_4, a forum poster said that to identify the subgroups of order 3 of A_4, it "suffices to find the permutations in A_4 of order 3"

Ok so does this mean that I literally stick every element in A_4 of order 3 into a subgroup and call it H?

Or am I reading this wrong

How can I identify a subgroup of order 3

[stone fulcrum]

@charred pewter
Any subgroup of order 3 would be Z/Z3, because that's the only group of order 3. Ergo, all subgroups of order 3 are actually generated by an element of order 3.

This logic works for any subgroup of prime order

[charred pewter]

Ok so if I take an element such as (123) in A_4, how can I make that generate the subgroup

[stone fulcrum]

So yeah, what permutations have order 3?

[charred pewter]

Well, I have the list of permutations with order 3, I just don't know which ones belong in a subgroup with each other

[stone fulcrum]

(123)
(123)(123)
(123)(123)(123) = e
Is a subgroup of order 3

[charred pewter]

(123) is a subgroup by itself?

[stone fulcrum]

No. It wouldn't be closed, (123)(123) is not in that.

[charred pewter]

Wait so I need a different element?

[stone fulcrum]

For what?

[charred pewter]

Nevermind, I got confused there

I'm confused, what would the elements of H be?

[stone fulcrum]

What's H? Haha

[charred pewter]

H is a subgroup of order 3 of A_4

[stone fulcrum]

Such a group isn't unique, there's many ways I can make it. One such way I did above, generating a group with (123)

H = {(123), (123)(123), (123)(123)(123)}

[charred pewter]

Yeah, I changed my wording to point out that it isn't unique, sorry about that

[stone fulcrum]

See why that H works?

Or, see what it means to generate a group using an element?

[charred pewter]

If I have an element sigma, I take sigma^2 and sigma^3 to find a group of order 3?

[stone fulcrum]

Only if sigma³ = identity yeah

Or if sigma is an element of order 3

[charred pewter]

Which (123) is

[stone fulcrum]

Then σ, σ², σ³ has to be a group

[charred pewter]

So H={(123),(123)(123),(123)(123)(123)} is a subgroup of A_4

[stone fulcrum]

Now, if I take my H and simplify it:
H = {(123), (321), e}
It looks a lot cooler

[charred pewter]

True, that looks better

[stone fulcrum]

And the symmetry pops real nice

[charred pewter]

That's a pretty cool way to generate a subgroup

[stone fulcrum]

Yeah that's a subgroup of order 3, of A4

Sadly you can only get Z/nZ subgroups this way

But every element belongs to such a subgroup which is cool

[charred pewter]

So only subgroups in which the element in question gives you the identity at the nth power?

[stone fulcrum]

Order of the element = order of the subgroup it generates

Which is not a trivial thing to say, since the order of the element is the lowest n such that a^n = e

[charred pewter]

Oh, okay

And so if I have G=A_4 and H={(123),(321),e}, how do I go about finding the 4 cosets

I know H is already a coset itself

So I find elements not in H but in G and multiply by H to find distinct cosets?

[stone fulcrum]

Pick an element not in H.
Multiply that into every element in H

That's a coset that contains that element

[charred pewter]

I see

Thanks, Kaynex

[stone fulcrum]

Np! Feel free to ask if you have anything else!

[charred pewter]

If I have a group G whose elements are in the form a+b(sqrt(-2)) where a,b are rational numbers, how do I find the inverse for any element

Idk how to do it in this instance

[solemn rain]

so lets call this group G

let x be in G

x= a+bsqrt(-2) for a,b in Q

you want to find y in G such that x*y=1

where *denotes the group operation

@charred pewter

[charred pewter]

Yes

[solemn rain]

so an element y

would look

a'+b'sqrt(-2) for a',b' in Q

so just solve this basically

(a+bsqrt(-2))(a'+b'sqrt(-2))=1

for y

[charred pewter]

Oh, I forgot to mention that G is a group with respect to multiplication

my bad

[solemn rain]

yea

no no it's okay

Q[sqrt(-2)] is a field

it's a group both with multiplication ( without 0 ) and with addition

[sturdy marsh]

Hint: Look at (a+b sqrt(-2))(a-b sqrt(-2))

[charred pewter]

Wait so the inverse is the conjugate?

[sturdy marsh]

not exactly

[solemn rain]

what do you think

[sturdy marsh]

you might have to modify it a bit

just write it out and see

[solemn rain]

xy=1 --> y=1/x

[charred pewter]

Wait, so if I just say 1/x is the inverse, it actually does make sense because a and b are rational numbers

lmao..

Is it really that easy

[sturdy marsh]

no, did you try writing it out?

[charred pewter]

Well, I tried writing it out the way mo2men said it first, I was going to try it both ways

hold on

[solemn rain]

you have to show that 1/x

you have to show how 1/x looks like

interms of y

so that if i give you an element you get me its inverse

[charred pewter]

I could show that 1/(a+bsqrt(-2)) = (1/a)+(1/b)sqrt(-2)=q+rsqrt(-2) for some q,r rational numbers?

and therefore I have an inverse in the same form as any element of G?

[sturdy marsh]

try multiplying out the number with its conjugate

you should get a rational number

what does that give you?

[stone fulcrum]

That's the right thinking! But very wrong algebra

1/(a + b√[-2]) = ...?

Solve that and you're done

[charred pewter]

ok yeah that was a very dumb thing for me to say

LOL

[stone fulcrum]

Conjugate is the right brain direction. Include it somewhere

[charred pewter]

Don't I just get a^2+2b^2?

[sturdy marsh]

yup

Can that be zero for a or b nonzero?

[stone fulcrum]

There should be a √[-2] in there

[sturdy marsh]

I think Aur is talking about the norm

[stone fulcrum]

Oh did I miss that? Haha mb

[charred pewter]

I'm talking about multiplying a+bsqrt(-2) and its conjugate

that's what I got

[sturdy marsh]

Yup, so you have x xconjugate = some rational number

that rational number is nonzero if a or b is nonzero

so it's invertible

[charred pewter]

Oh so by multiplying by its conjugate and getting that result, it proves x has an inverse because a or b are nonzero?

[sturdy marsh]

Divide by that rational on both sides

you get an explicit formula for the inverse

[charred pewter]

oh

so it should be 1/(a^2+2b^2)?

[sturdy marsh]

the inverse?

nope

you have (X)(X conjugate)/rational number = 1

[stone fulcrum]

You want to start with
1/(a + b√-2)
Multiply top and bottom by the conjugate

Know what I mean by that? Haha

[charred pewter]

rationalizing the denominator?

[stone fulcrum]

Yaya

[charred pewter]

I was doing so much mental gymnastics rn lol

I'm so bad, I can't believe it's that easy

[stone fulcrum]

It's easy to do, but the importance of what you did is worth thinking about. Every element of the form a + b√(-2) has an inverse of the same form

"We should make a group out of this"

[next obsidian]

If G is a finite group, and G has a unique maximal subgroup, then G is cyclic. One can prove this. Does anyone know if this is still true in the infinite case?

[fickle brook]

i assume you mean unique maximal proper subgroup?

do you have a proof of the finite case handy?

it might be possible to locate where exactly finiteness is used and use that to construct a counterexample

[next obsidian]

Yeah it is

Also I know exactly where it’s used, by maximal I mean proper the same way a maximal ideal is proper

Anyways, let M be the maximal subgroup. Let g be in G\M, then consider the subgroup <g>. If this is proper it must be a subset of a maximal subgroup, but this cannot be M as g is not in M. You conclude that <g> = G

The fact that you conclude that <g> lies in a maximal subgroup is where you need finiteness

I can think of groups without any maximal subgroups (Q is an example) but can’t think of any with 1 unique maximal subgroup, but such that not every element lies in a maximal subgroup

Maybe you can take a product involving Q? 🤔

Like say, Q x Z/4Z?

I will be the first to admit that I don’t know Jack about monoids, but maybe that’s what’s needed to form quotients?

Or conversely, that’s what is the kernel in a category theoretic sense

I think that’s probably the actual reason, but someone else has to confirm or deny

Don’t worry so I can give you a rundown of what it would be

Basically a kernel is associated to a map, so just let like G and H be stuff, think algebraic objects for now and f: G -> H

Then the kernel is an object with a morphism, call it K and say g: K -> G

Such that if Z is any other object with a map h: Z-> G such that hf = 0

Then h factors through g, i.e. you get a map h’: Z-> K such that gh’ = h

Yeah

So let’s try and see why the kernel in groups is actually a kernel right?

Oh sorry one little thing

fg must be 0

So the kernel should “map to 0”

Yup

[cloud walrus]

summoid:

[next obsidian]

The map should be the inclusion of K into G I think

I’m just struggling off the top of my head to see why a map would factor through the kernel, but I think I’m just totally missing something lol

I’m gonna think about it for a bit haha

Oh this is so dumb lmao

So literally

If h: Z -> G

Is such that fh = 0

Then the image of h is a subset of kerf

Right?

By definition of ker f

So you can just consider h as a map into the kernel instead

Oh and I forgot one other crucial fact lol, sorry

The map induced into the kernel must be unique

So there’s only one map from Z -> K such that the triangle of maps commutes

So in the case of groups the map is unique because for that triangle to commute, the induced map has to agree with the original

Because you compose with inclusion

Which does nothing

Uhhhhhh

I think it’s cancellation for injective

let i:K -> G be the inclusion

This is clearly injective

And we are saying that if ih = ih’ then h = h’

So it’s left cancellation for injectives

Also in general this property that if ih = ih’ then h = h’ is called a Monomorphism

It generalized injectivity

Yup exactly

So yeah I bet that’s why kernels for monoids are defined differently

I mean

If I understood what you wrote correctly

Isn’t the kernel in group the equivalence class of stuff that maps to 0?

What do you mean?

From what I gathered you have the equivalence relation that x ~ y it fx = fy right?

And the kernel for monoids is the quotient by that relation?

Oh what

Oh then it won’t be

Not at all

So like

A group is a monoid right?

But the kernel categorically is something which cares about the category you’re in

So if you take the group kernel of a group homomorphism

Versus the monoid kernel of a group homomorphism (considering it a monoid homomorpbism)

The two are different

Because for the group kernel you only worry about groups Z mapping into G

But for the monoid one you care about all monoids mapping into G

And yeah

So you won’t be able to realize the monoid kernel as the group kernel

Kernels are unique up to unique isomorphism

So like, if it did somehow play the role of kernel

You’d need K to be isomorphic to that weird subset of M x M

And like... yeah no way

Sort of

It’s not quite a diagonal

Because you care about them being the same after they’re mapped into the target

But yeah

Yup

So like if a group Hom is injective

Then the kernel is literally just the identity

But then your monoid kernel is the diagonal

Which aren’t iso

A bit

Enough to get by

If you want to learn a bit

I cannot comment on that since I don’t know what an action network is

But a looooot of things are categories

Maybe post this in #category-theory

I’m gonna go to bed anyway haha

If you want to learn some category theory

I recommend Riehl’s category theory in context

It’s available legally, for free on the web

Sure, it’ll be there whenever haha

[next obsidian]

A while ago I posted something in this, and couldn't figure out for the life of me how to solve it. The problem was as follows, let $B\subseteq A$ be integral domains, $A$ a finitely generated $B$-algebra such that $A$ has finitely many primes lying over $0$. Show that Frac$(A)$ is a finite extension of Frac$(B)$. I finally have, what I believe to be, a proof.

We know that $A = B[a_1,\dots,a_n]$, and it follows immediately that Frac$(A) = \text{Frac}(B)(a_1,\dots,a_n)$. We show that this is an algebraic extension, implying that the $a_i$ actually generate Frac$(A)$ as an algebra, but then by Zariski's lemma Frac$(A)$ is a finite extension of Frac$(B)$ as desired.

Suppose that some $a_i$ is transcendental over Frac$(B)$, then Frac$(B)[a_1,\dots,a_n] = C$ has transcendence degree $d > 0$ over Frac$(B)$. By Noether Normalization, there exist elements $\gamma_1,\dots,\gamma_d$ algebraically independent over Frac$(B)$ such that $C$ is finite over Frac$(B)[\gamma_1,\dots,\gamma_d]$, so in particular it is integral over this subring. As polynomial rings over a field have infinitely many primes, this subring has infinitely many primes, and then by the lying over property (going up) it follows that $C$ has infinitely many primes. We can view $C$ as the localization of $B[a_1,\dots,a_n]$ at the set $B\setminus{0}$, so in particular there are infinitely many primes of $B[a_1,\dots,a_n] = A$ which intersect $B$ only at $0$, which is to say there are infinitely many primes which lie over $0$. This contradicts our hypothesis, so that we conclude all $a_i$ are algebraic, and as argued before Frac$(A)$ is a finite extension of Frac$(B)$.

[cloud walrus]

Mathemagician:

[river fern]

"From Lagrange's theorem, there are at most d solutions to x^d = 1 mod p"
Why does this follow from Lagrange's thm?
Why can't there be more than d sols?

[stone fulcrum]

What is d?

@river fern

In relation to the group structure

[river fern]

In the context of what I read, d divides p - 1

Someone told me why this is true

(Actually works for any d I think)

Apparently there are two lagrange theorems

One of them basically says that there are at most d roots to a polynomial of degree d in a field

Which trivially implies the desired result

[stone fulcrum]

That's not a ton of info. d mineswell just be p-1 for your question. Then, everything is a solution @river fern

[river fern]

No, I was trying to see if this works for any d (it does due to lagrange)

[stone fulcrum]

No it doesn't rofl

[olive mirage]

I mean, it's true for any d because that group is cyclic

and cyclic groups have a unique subgroup of every order

[river fern]

Yeah it's cyclic if d divides p - 1

[olive mirage]

well, the point is, U(p) is cyclic

so every subgroup is cyclic

[stone fulcrum]

"it's"

[river fern]

Oh ok

[olive mirage]

this is circular, though, becuase the proof that U(p) is cyclic uses the fact that this equation has at most d solutions 😛

[river fern]

Yeah

Lol the thing I was reading was the proof that a primitive root exists

Which implies that U(p) is cyclic lol

[olive mirage]

yeah, to my eye, the easiest way to prove that involves combining the structure theorem of finitely generated abelien groups with very basic facts about polynomials over fields

(e.g. a degree d polynomial has at most d roots)

[bleak abyss]

I think there's a lower tech proof actually

If G is a finite group such that there are fewer than phi(d) elements of order dividing d, then G is cyclic

That's just because, let's say G has order n, well G is the union of the set of elements of order d for d dividing n. But n = sum_{d|n} |{elements of order d}| ≤ sum_{d|n} phi(d) = n

So yeah you have an element of order n

And then you apply this result to a finite subgroup of F^x

@olive mirage

[river fern]

^Yeah that's the proof I was reading

[bleak abyss]

Can this also prove that units in Z/p^n are cyclic? I think that's a thing

Problem is you can't use the fieldness

[olive mirage]

I think all of the proofs of eveyrthing around finite fields are some of the coolest bits of math

[kindred rivet]

Wait dami

Do u mean (Z/p)^n

Or Z/(p^n)

[bleak abyss]

Latter

[river fern]

U(Z/p^n) isn't a finite field though

Since (1) + (p - 1) = p, which is not an element in the set for n > 1

[latent anvil]

no, but it's contained in a finite field and the multiplication coincides

[river fern]

ok

[latent anvil]

Solutions to x^d = 1 are the same in the group of units and in the field Z/pZ

[river fern]

What about Z / p^2 Z?

Does it still hold?

[latent anvil]

hmm maybe not

[river fern]

I feel like no

[bleak abyss]

Wait which finite field is it contained in?

[latent anvil]

I don't see why it would off the top of my head

Fp?

Oh lol

Misread

I just didn't read the n

Lmao

[river fern]

ok haha

[latent anvil]

Ignore everything I've said

lol

[bleak abyss]

Lmao I've never been gaslit that hard before

[latent anvil]

look I know a good deal of algebra and speak confidently

This excuses me from having to read

[bleak abyss]

Lmao

[latent anvil]

anyways units in Z/p^nZ are cyclic if p is odd

But I don't remember why

And I think it can fail for p=2

[river fern]

i think i read the proof a long time ago and forgot

reading one now

[latent anvil]

I'll just claim it's obvious from the classification theorem

:^)

[river fern]

I think it's true for 1, 2, 4, p^n, and 2 * p^n

[latent anvil]

Ah yeah (Z/8Z)^* is klein four

[river fern]

I'll just claim it's obvious from the classification theorem
@latent anvil is this the thing for abelian groups?

[latent anvil]

yes but I was kidding

[bleak abyss]

I'll at some point look this up

[latent anvil]

I think it came up recently in your stuff

We talked about it

Some kind of sum over roots of units

[bleak abyss]

Yeah a Gauss sum computation

I wonder if you can manifest them as a subgroup of Q_p

(them = group of units of Z/p^nZ)

[latent anvil]

Isn't there something about the totient function at those values

p, p^k, 2, 2p^k

Like is φ(n) = n - 1 exactly then?

No that's wrong

[bleak abyss]

Should be phi(p^n) = p^n - p^{n-1}

But yeah once you know the result for p^k then 2p^k becomes CRT

[latent anvil]

Yeah

[delicate bloom]

group of units of Z/p^nZ isn't a subgroup of Q_p

[bleak abyss]

F

[latent anvil]

@bleak abyss does it help if it's a subgroup of C?

:^)

[bleak abyss]

Lol

[river fern]

There's a proof here but it's kinda a hard read imo https://people.maths.bris.ac.uk/~mazag/nt/lecture8.pdf

[delicate bloom]

it's a subgroup of C_p @latent anvil no need to hop ship

[bleak abyss]

Oh, the moment you can stick it in as a subgroup of a field you're done

The above argument holds for finite subgroups of the multiplicative group of a field

[smoky cypress]

What's the automorphism group of the symmetric group

[oblique river]

it's S_n unless n = 2 or 6

[smoky cypress]

Oh

Odd

But interesting

[oblique river]

if n = 2 it's trivial and if n = 6 it's a group of order 1440

which is 2*|S_6|

[smoky cypress]

The automorphism group of C_n is (Z/nZ)* right?

[oblique river]

yes

[latent anvil]

@bleak abyss i was being dumb

the p^kth roots of unity is Z/p^k Z, not the group of units

blerggh

[delicate bloom]

eh?

[bleak abyss]

Lmao sham

[smoky cypress]

Is $x^{p^{nd}}-x\in\bF_{p^n}[x]$ the product of all irreducible polynomials of degree dividing $d$?

[cloud walrus]

Whoever:

[smoky cypress]

https://math.stackexchange.com/questions/106721/product-of-all-monic-irreducibles-with-degree-dividing-n-in-mathbbf-q#:~:text=In the finite field of,be Xqn−X.

[latent anvil]

Dami I'm so out of practice with algebra

It's been like 2 months

qq

tfw I won't have time to take an algebra class until next winter, maybe even next spring :(

[next obsidian]

Why are you talking to Dami

About that lol

In #groups-rings-fields

[bleak abyss]

It's because of my "lmao sham" comment

[next obsidian]

Ah

[vestal snow]

Do you need vector calc or differential equations at any point in algebra?

More specifically, can I make a case to my faculty advisor to let me substitute those two required courses for others?

[scenic sage]

as far as i am, it is not needed

i mean they prolly have connections but you can learn algebra without them

[vestal snow]

Great!

[next obsidian]

I don’t think you can make a case to substitute those two, maybe they’d let you substitute it for analysis or something but I don’t think a program will let you get a degree if you don’t know what a gradient is

[vestal snow]

I studied vector calc on my own in high school and I'm planning on taking real analysis I and II

I talk to him tomorrow so we'll see how it goes

[vestal snow]

Can splitting rings be constructed for arbitrary rings?

Let R be a (commutative) ring and let f(x) be a polynomial in R[x]. Consider the new ring A = R[x]/(f(x)). Then A has a root of f'(t), call it a (Here f' represents the passage of the coefficients of f into the new ring A).

Then we can look at the image of f'(t) under the natural homomorphism from A[t] to A[t]/(t-a). We can check by looking at the coefficients that the image of f'(t) in this is 0 (A slicker way would be to say that t=a in this ring, but I feel like it isn't as precise). Therefore, f'(t) = (t-a)h(t). Now repeat the same argument with h(t) and the ring A

There are unnecessary details (like defining f'(t) instead of just viewing R[t] as a subset of A[t]), but I wanted to be as explicit as possible

[pure crest]

i think you are implicitly using the Kronecker's theorem, so you need R to be a field

other than that your arguement makes sense

[vestal snow]

What do you mean?

Can you point to where exactly in my proof I might be using that?

I would consider the proof to be a very elementary construction similar to that of splitting fields

Ahh I can see what might have confused you

I meant a splitting ring, not a splitting field (I edited my original message)

[pure crest]

ohh right, that makes sense now

its a good proof

i was just learning a similar proof for fields

i thought you were doing the same for fields

[vestal snow]

It's almost the same, except the resultant ring is not a field

at least not necessarily

[elder valley]

@vestal snow I think you may need that the leading coefficient of f is a unit to ensure h has coefficients in A

[pure crest]

https://math.stackexchange.com/questions/1314208/perfect-field-of-characteristic-p

i am kinda confused as to where they use the fact that g(x) is not separable

[pure crest]

ohh is see

why does $ g′(x)≡0 $, imply that g is a polynomial in $x^p$?

[cloud walrus]

boat:

[pure crest]

@open torrent

[solemn rain]

i thinnk it has to do with the fact that

if the derivative of a polynomial has phi as a root then phi is a multiple root of the polynomial @pure crest

[vestal snow]

@elder valley yeah you're right. I forgot to add that in the proof

[next obsidian]

Here's something insane. Let $A \subseteq B$ both be integral domains, and $B$ finitely generated over $A$ such that Frac$(B)$ is a finite extension of Frac$(A)$. Then there exists $\alpha \in A$ such that $B_\alpha$ is finite over $A_\alpha$.

[cloud walrus]

Mathemagician:

[next obsidian]

Proof: We know that $B = A[b_1,\dots,b_n]$ as it is finitely generated over $A$, furthermore Frac$(B) = \text{Frac}(A)[b_1,\dots,b_n]$, from here we see that all $b_i$ are algebraic over Frac$(A)$, so that there exists polynomials
$$p_i(x) = x^{n_i} + a_{i(n_i - 1)}x^{n_i - 1} + \dots + a_{i1}x + a_{i0}$$
such that $p_i(b_i) = 0$

[cloud walrus]

Mathemagician:

[next obsidian]

Each coefficient $a_k $ is of the form $\frac{\beta_k}{\alpha_k}$, so by setting $\alpha = \prod \alpha_k$ we can replace all $a_k = \frac{\beta_k}{\alpha_k}$ with $\frac{\beta_k\alpha/\alpha_k}{\alpha}$, and we still have that $p_i(b_i) = 0$, but now all the coefficients used are all over a common denominator.

[cloud walrus]

Mathemagician:

[next obsidian]

By localizing $A$ at $\alpha$, we see that $p_i$ exists as a polynomial over $A_\alpha$, so that we still have that $p_i(b_i) = 0$ for all $i$, but then that implies that $B_\alpha = A_\alpha[b_1,\dots,b_n]$ is integral, and finitely generated over $A_\alpha$ so that it is also finite

[cloud walrus]

Mathemagician:

[elder valley]

@vestal snow I think there's still an issue even if f is monic, because h should have coefficients in A

Take R=integers and f=x^3-2 for example. Taking out a root leaves a quadratic that doesn't have integers coefficients

[vestal snow]

The coefficients of h must be in A because f'(t) is in the ideal generated by t-a in A[t]

@elder valley

[elder valley]

Oh I got confused with the notation

[elder valley]

Anyone know of any Hartshorne reading groups?

[delicate oriole]

I’m so glad I don’t have to take this

[elder valley]

algebra you mean?

[pure crest]

why is the dimension of $\mathbb{Q}(\sqrt{2}, \sqrt[3]{2}, ... ,\sqrt[n]{2})$ over Q finite?

[cloud walrus]

boat:

[latent anvil]

@pure crest you can find this by adjoining sqrt(2) to get a field K1, then adjoining cbrt(2) to get a field K2, and so on, up to Kn

Does that make sense?

It then suffices to show that the dimension of K_(i+1) over K_i is finite for each i

Do you see why this is sufficient? I can explain if you need

[cloud walrus]

boat:

boat:

[pure crest]

and so on right

i am just wondering if there is a more elegant way of doing that

[delicate bloom]

what's not elegant about that?

[oblique river]

Specifically for this problem, you could show that the field you care about is a subfield of $\bQ(\sqrt[lcm(2,3,\ldots,n)]{2})$

[cloud walrus]

Buncho Bananas:

[oblique river]

i.e. the "lcm(2,3,...,n)"-th root of 2

which clearly has finite degree

but honestly I don't think that's worth the effort here

[pure crest]

why does the LCM be the only root that matters?

[oblique river]

just prove that if you adjoin any number of algebraic elements to Q, the degree is finite

i.e. if $\bQ(a1), \bQ(a2), \ldots, \bQ(an)$ are all finite degree over $\bQ$, then so is $\bQ(a1, \ldots, an)$

[cloud walrus]

Buncho Bananas:

[pure crest]

right

i think that would be better

[oblique river]

in some sense the fact that they're all roots of 2 doesn't really matter at all and it kind of misses the point

"the point" being that these are just a finite collection of algebraic elements, and thus always generate a finite-degree extension of Q

[pure crest]

how would you prove it in the general sense?

i think it would only be true of a1 is algebraic over Q

[oblique river]

I mean, all of them need to be algebraic over Q

"a1 is algebraic over Q" is equivalent to "Q(a1) is finite degree over Q"

[pure crest]

right

i think i understand now

thank you very much

i really appreciate the beauty in algebra over calculus

[oblique river]

me too :)

[knotty mason]

why is there an inner and outer semidirect product?

can you always form C_2 semidirect product G?

[solemn rain]

wdym

whats C_2

[knotty mason]

the cyclic group of order 2

[solemn rain]

okay so

whats your question

[next obsidian]

The inner semidirect product is formed specifically using conjugation

while an outer semidirect product is just via any map into the automorphism gropu

The thing about the inner semidirect product is that you also have an inner direct product and an outer direct product, the outer just being G x H for arbitrary G,H

But in a group G if you have H_1 and H_2 both normal with trivial intersection you can form the inner direct product H_1 x H_2 which is expressed set-theoretically as the set H_1H_2 as a subset of G

The inner semidirect product is what you get where you loosen that up a bit and only require one of the subgroups to be normal

[knotty mason]

why is conjugation always good enough?

thanks! I see the connection there

[next obsidian]

Because that’s an automorphism

Specifically conjugation by h is an inner automorphism, since that is the set of automorphisms of the form of conjugation by some element

Another connection to the word “inner”

[knotty mason]

ahh!

[shy bluff]

what exactly is Z[alpha]?

Or did my prof typo and mean Z[x]?

[oblique river]

the Z[x] should be Z[alpha]

[shy bluff]

Wait but for part b

Is Z[alpha] like Z[x] but instead of x you have powers of alpha?

[oblique river]

I'm saying that the definition in the statement of the problem is for Z[alpha]

Z[x] is the ring of polynomials with integer coefficients

and Z[alpha] = {a + b*alpha : a,b in Z}

[shy bluff]

Oh ok

Thank you!

[oblique river]

👍

[elder valley]

Is Z[alpha] like Z[x] but instead of x you have powers of alpha?
@shy bluff this is also true. it's just that alpha^2 is an integer, so larger powers of alpha reduce down and you're left with only linear polynomials in alpha, which is why the definition in the problem is only a+b*alpha

[shy bluff]

Yea

okie thats' what I thought

[kindred mist]

Maybe this is a silly question but let A \in M_n(F), for a field F, am I correct that detA is defined the same way regardless of what field the coefficients are in? And that the invertible matrix theorem holds regardless of what field the coefficients are in?

[oblique river]

yes

[kindred mist]

ty

oh one other thing

furthermore that all the following usual definitions (for matrix determinant) from M_n(R) are suitable for any field right?

cofactor expansion, Leibnitz formula and the formula that uses Levi-Chivita symbol

[kindred mist]

<@&286206848099549185> ^^^ sorry for the ping

(if I can at least get verification about cofactor expansion then that would be really helpful, thanks)

[river fern]

Are there any well-known results about the size of center of a group of size p^n?

I think for p^3, it's >= p

[shy bluff]

So given that I did a and b, how do I go about doing c here? All that I really know about prime ideals is that if it's prime and ab is in it, then either a or b is in it?

[elder valley]

@shy bluff there's a theorem you probably saw about quotienting a ring by a prime ideal

[shy bluff]

Uhhh what does said theorem look like?

[elder valley]

let I be an ideal of a commutative ring R with unity. then I is a prime ideal if and only if R/I is an integral domain

[elder valley]

furthermore that all the following usual definitions (for matrix determinant) from M_n(R) are suitable for any field right?
@kindred mist the only things i can think of that won't hold are stuff that use the ordering of R. so things like positive definite doesn't make sense in other fields

[kindred mist]

yeah

ty Auvera

[shy bluff]

oh ok

So I just need to prove that that's an integral domain

[elder valley]

yeah, but that's pretty immediate

given (a) and (b)

[somber rivet]

Yeap, what’s the problem?

[charred pewter]

So pretty much all I know is that an isomorphism is a homomorphism that's also a bijection and that to be a homomorphism you need to prove that f(x*y)=f(x)+f(y) for some x,y in G

[somber rivet]

So you need to show 3 things over here, really:

1. f is a homomorphism
2. f is injective
3. f is surjective

Work with the definitions to do this

[charred pewter]

How do I do this with matrices and complex numbers, I don't understand the relationship here

[delicate bloom]

what matrix maps to 1? what matrix maps to i?

[somber rivet]

The definition of f is given to you. In this case, f takes a_11 and turns it into the real part of a+ bi. Then, it takes a_12 and turns it into the imaginary part of a+bi

So, you can think of it like that if you wish

of course, a_11 and a_12 refer to the entries in the matrix

[charred pewter]

So I can prove 1,2, and 3 using the complex numbers instead of the matrices?

[somber rivet]

Well, you’ll need to work with matrices too. Don’t think of the matrices as being an entirely different thing. GL_2 is just one group being mapped to another group. So, do almost exactly what you would normally do if you were proving the bijectivity of a given function

[charred pewter]

So I have to prove that for all x,y, f(x)=f(y) => x=y to prove that it's an injection

[somber rivet]

For all x,y in G, f(x) = f(y) —> x = y

That’s just the regular definition of injectivity

[charred pewter]

I still can't imagine how I'm going to be able to show this with matrices. I take two matrices A and B, and I say that f(A)=f(B) which means that the complex numbers they map to should be equal?

[somber rivet]

Yes and so, what can you say about the real parts and imaginary parts?

Like, let $f(A) = a_1+b_1 i$ and $f(B) = a_2 + b_2 i$. If $f(A) = f(B)$, then:

$$a_1+ b_1 i = a_2 + b_2 i$$

So what can you say further?

[cloud walrus]

Abhijeet Vats:

[somber rivet]

You should also explicitly write out what A and B are

[charred pewter]

Well then given the group we're working with we'd have A_11 = a1, A_12=b1, A_21=-A_12=-b1 and A_22=A_11=a1

And then the same scenario with B's components

I said it wrong sorry

You can see that since a1+b1i=a2+b2i, then a1=a2 and b1=b2 and so A=B

..i think

[somber rivet]

That's exactly right

Have a bit more confidence

(Would've looked nicer in latex but that's a minor complaint)

[charred pewter]

Yeah, I have no idea how to use latex

I wish I did so my typing in here could look better

[somber rivet]

That proves that f is injective. Now, you need to show that f is surjective and that it is a group homomorphism

[charred pewter]

The thing is I don't exactly know how to prove surjectivity.

[somber rivet]

Use the definition

Let $z \in \bC \setminus {0}$. Can you find a matrix $A \in G$ such that $f(A) = z$?

[cloud walrus]

Abhijeet Vats:

[somber rivet]

Constructing A using the definition of f shouldn't be hard at all

[charred pewter]

Well, yes. You just find the real and imaginary components of z and plug them into A, right?

[somber rivet]

Yeap. So, let $z = a+bi$. Then, we define the matrix $A$ to be:

$$A = \begin{pmatrix}a&b\-b&a \end{pmatrix}$$

[cloud walrus]

Abhijeet Vats:

[somber rivet]

Now, to ensure that $A \in G$, we just need to make sure that its determinant isn't 0. But it's determinant is just $a^2+b^2$ and since $z \neq 0$, it follows that one of $a$ or $b$ is nonzero. Hence, $a^2+b^2 \neq 0$.

[cloud walrus]

Abhijeet Vats:

[somber rivet]

Does that make sense?

[charred pewter]

Can’t we use the given information that a^2+b^2 is nonzero?

The professor includes it in the definition of G

[somber rivet]

Yeap, that's exactly what I've used.

See, what i've done is to just specify a 2 x 2 matrix

That's all I've done. And by the definition of f, that gets transformed into a+bi. But how do I know if my chosen matrix actually belongs to G? To show that concretely, I'd have to show that the determinant is nonzero

[charred pewter]

And so that proves surjectivity?

[somber rivet]

Yeap

[charred pewter]

And lastly, we need to prove f(x*y)=f(x)+f(y), right?

[somber rivet]

Yeap. In this case, the operation on $G$ is standard matrix multiplication. The operation on $\bC \setminus {0}$ is, presumably, multiplication since it wouldn't make sense to call it a group if the operation being considered was addition instead.

[cloud walrus]

Abhijeet Vats:

[charred pewter]

So we can do that by multiplying the previous matrices A and B and then using the components of the resulting matrix to see if they're equal to a1+b1i +a2+b2i, right?

[somber rivet]

You're multiplying the complex numbers

[charred pewter]

Oh

Wait, are you referring to the RHS?

[somber rivet]

Yes, I am

We're considering $\bC \setminus {0}$ as a group under multiplication since it seems reasonable to assume that that's the intended operation.

[cloud walrus]

Abhijeet Vats:

[charred pewter]

Oh, right

So, multiply matrices A and B, use their components to find the resulting complex number, then see if it's equal to (a1+b1i)(a2+b2i)

And that should end up proving it's a homomorphism, thus proving that f is an isomorphism, correct?

[somber rivet]

Indeed

[charred pewter]

Awesome

Thank you so much for helping me and explaining these concepts to me

[somber rivet]

You should review your basics thoroughly

You're tripping up over very basic things that aren't related to group theory. That needs to be fixed if you are to go any further without struggling on the basic stuff.

[charred pewter]

I'll work on that, it has me worried too

[somber rivet]

Another thing also is with regards to starting the problem too

It would be one thing if you couldn't complete one step because you weren't paying attention to the algebra or the specific details given in the problem

It's a whole other thing to not be able to even start to prove basic things like injectivity and surjectivity. Please do look over these and work on some simpler problems to get a good grasp of them before you do more of this.

Otherwise, it's going to be very painful moving forward

Anyways, you're welcome

[shy bluff]

@elder valley what do you mean "given (a) and (b)"?

[elder valley]

you can use (a) and (b) to get that the quotient is an integral domain pretty easily

[next obsidian]

Are there any well-known results about the size of center of a group of size p^n?
@river fern any p-group has non-trivial center so it has to be at least size p

[smoky cypress]

Is there a way to construct a new field from two fields?

Like product of two rings

The product of two fields is not a field because we have elements like (1,0) which is a 0 divisor

[oblique river]

not in a categorical way

[smoky cypress]

What do you in by categorical?

[oblique river]

like, category theory

the category of fields is really bad

[smoky cypress]

I know a slight amount of category theory, and I’m not sure what you mean by you can’t combine two fields in a categorical way

[oblique river]

like, there's a general notion for what it means to be a "product" of two objects in any category

but not every category has products

and the category of fields doen't

basically this just says that there isn't a nice "product" operation on fields

[smoky cypress]

What are the properties of a nice product in a category?

[oblique river]

sorry you'll have to ping me tomorrow, i'm really tired and wont be able to give a real explanation now

[smoky cypress]

Ok sure

[next obsidian]

A product is something with this property

Let X,Y be objects, and Z any object with morphisms f:Z-> X and g:Z-> Y

The product of X and Y is something, I’ll denote it X x Y with maps pi_X: X x Y -> X and pi_Y: X x Y -> Y such that there is a unique morphism h: Z -> X x Y such that f and g factor through h

I.e. f = pi_X•h

And g = pi_Y•h

This object is super general since it’s over ANY Z with any maps to X and Y

So a category need not have them, like for fields for example

In set it’s literally the set X x Y

And in abelian groups it’s G x H

[river fern]

Looking the last example of this https://brilliant.org/wiki/burnsides-lemma/#:~:text=Forgot password%3F-,Burnside's Lemma,to be counted as distinct. (Finding the number of ways to put n indistinguishable items into m indistinguishable boxes), does this imply that you can apply burnside's lemma to efficiently find the number of nonneg solutions to x1 + x2 + ... + xm = n where the xi's are nondecreasing?

(Just want to make sure)

Hmm actually it might be slow if m gets large (since Sm will be really large)

I wonder if there is an efficient algo to find the number of nonneg solutions to that equation if m is large

[next obsidian]

There is no closed form for that number btw

It’s the same thing as finding number of conjugacy classes in S_n for example

And it’s not well understood from what I know

[raven nexus]

What's your favorite 'bang-for-buck' short exact sequences which betray some deep relationship between the objects in the sequence that is non-obvious, or describe an interesting relationship that is obvious, but is of important consequence?

[river fern]

Any hints:

The direction trivial => normal is where I am stuck
I've been able to show that Hg is a subset of gH for any g in G, but I'm not able to show gH is a subset of Hg

[plucky flicker]

Hey guys! I'm new here, and I have a question in algebraic geometry, I hope I'm in the right room. So, recently I started to do some algebraic geometry, and I need some explanation about the coordinate ring. I mean, what's the main idea of it, what motivates to define such an object?

[oblique river]

@river fern it turns out that just shwoing Hg subset gH for all g in G is enough

[river fern]

what if H is infinite?

[oblique river]

if you want to also show that gH is a subset of Hg for some g: first, by what you've already shown, Hg^(-1) is a subset of g^(-1)H

now multiply on the left and right by g

[river fern]

ah ok thanks!

[oblique river]

:)

yeah it's a clever little trick which can save you in times like this

@plucky flicker this channel is fine for algebraic geometry, but people also sometimes post AG stuff in the "topology and geometry" channel.

the coordinate ring of a variety is like the ring of functions on that variety

the general idea is that you should be able to go between "a space X" and "the ring of functions on X"

the first is the "geometry" and the second is the "algebra"

in algebraic-geometry-world, our "functions" aren't arbitrary functions but polynomials

[plucky flicker]

i got it, and why is the coordinate ring exactly the quotient ring?

[oblique river]

the quotient ring of what?

can you provide any more context?

there are lots of quotient rings in the world haha

[chilly canyon]

k[V] is defined to be k[X]/I(V/k)

[oblique river]

(yes I do know that... I was talking to boti because I wanted them to practice being more precise when they talk about math)

[chilly canyon]

(I'm following along, I also had the same question !)

(oh sorry !)

[oblique river]

(ah okay! sorry I didn't realize)

so the idea is that, well all of our functions are polynomials, so we're going to start with k[X]

but, some polynomials might end up giving the same function on V

for example, let's suppose V is the circle x^2 + y^2 = 1

or, x^2 + y^2 - 1

so I = (x^2 + y^2 - 1)

here are two different polynomials which give functions on V: f(x,y) = 3x and g(x,y) = 3x + y(x^2 + y^2 - 1)

what can you tell me about the functions f(x,y) and g(x,y) on V?

f(x,y) and g(x,y) are certainly different polynomials, but if you just consider them as functions on V, what can you say?

[plucky flicker]

they are same?

i mean on V

[oblique river]

yes -- but why?

[plucky flicker]

because the second part of the g polynomial, i mean y(x^2+y^2-1) vanishes on V, leaving g as 3x

[oblique river]

yeah, exactly

the polynomials in I are exactly the polynomials which vanish on V

(by definition!)

which means that if you take some function f(x,y) on V, if you add any element of I to that function, it's going to be the same function!

because you've just added 0 to it

so this is why we quotient by I

when you quotient by I, we get that f(x,y) + I and g(x,y) + I are now in the same coset