groups-rings-fields

[celest brook]

professor I would prove this, but there isn't enough space on overleaf for me to fit it

[loud root]

im really lost here

would appreciqate some help

ive been staring at it for so long

the last line is what im confused by

[wraith obsidian]

it's like saying „2ℤ + 3 and 2ℤ + 1 are the same coset, because 1 = 3 - 2, and the thing we're subtracting is in the ideal“
Just here we have „(f) + x^n and (f) - (an-1x^n-1 yadda yadda + a0) are the same coset, because (yadda yadda) = x^n -f, and the thing we're subtracting is in the ideal“

in essence, Q consists of „polynomials up to differences in (multiples of) f“, and x^n and -(yadda yadda) are the same thing up to difference of f

[loud root]

@wraith obsidian i dont know why this has took me so long to get my head around - maybe because ive been reading algebra for 10 hours today - but I understand it much better now. it doesn't feel really natural yet as today is the first time im seeing polynomial rings and ideals like this, but i can get through it if i concentrate. thank you so much

[wraith obsidian]

You might want to take a break, I know the half hour rhythm everybody tells you about doesn't necessarily work with abstract stuff due to long warmup time – but otoh 10h consecutively definitely sounds like overload.

Also, concepts like these just take some time to sink in

[golden pasture]

10 hours sounds too long lol

i do at most like 2 hours

before me shut down

[loud root]

oh right, thanks, then i feel better about thinking at the speed of a snail rn xD

Yeah i should take breaks

[wraith obsidian]

I have some spare ones if you like

<br><br><br><br>

(…sorry about that)

[next obsidian]

Is A[[x]]/I = (A/I)[[x]] when I is an ideal of A?

I think I can show this by doing some really fucking stupid projective limit bs, but I would rather not lol

I can't show that the kernel of A[[x]] -> (A/I)[[x]] is the extension of I, because I would have to sum an infinite number of elements

At least trying to do it the way I was

[cyan marten]

I am confused. What if A = Z and I = (x)? What is A/I in this case?

[next obsidian]

x isn't in Z haha

I would need to be an ideal of A to make sense of this

I'm trying to emulate the familiar statement that A[x]/I = (A/I)[x]

[cyan marten]

I think you should write I[[x]], then

[next obsidian]

Well...

There's even a problem here haha

I want to deal with IA[[x]] so like

Take I, add in all multiples of it by elements of A[[x]]

I don't know if this is the same as power series with coefficients in I

which I think I[[x]] would seem to suggest

But now I'm just honestly really confused as to what I want to be doing

😔

[cyan marten]

I see, so in other words you want I*, which is the extension of I under the hom. from A to A[[x]]

[next obsidian]

Yup

At least... I think so

According to what Matsumura wrote for this exercise

[cyan marten]

Can't you use the fact that completion is exact?

[next obsidian]

I think that's what I want haha

hmmm

Do I need Noetherian hypotheses for that?

[cyan marten]

But here I assumed that I is an ideal of A[x]

[next obsidian]

I think this is okay?

Since like

IA[x] = I[x]

when the left is like extending

the right is polynomials with coeff. in I

So I think you can pass to the extension in A[x] without issue

[cyan marten]

Actually you do

[next obsidian]

fffffffffff

If I had Noetherian I would be done already

hahahaha

[cyan marten]

Yeah 😂

[next obsidian]

I mean injective limits are right exact

[cyan marten]

This emoji is terrible

[next obsidian]

so that might be enough

So like

I did this really fucking horrible calculation

to try and get it using injective limits

So like A[[x]]/I = lim_n (A[x]/(x^n))/I = lim_n (A[x]/I)/(x^n) = lim_n (A/I)[x]/(x^n) = (A/I)[[x]]

I think?

I'm not sure if all those equal signs are kosher tbh

Ah

so I think the questionable one is lim_n (A[x]/(x^n))/I = lim_n (A[x]/I)/(x^n)

To justify this I tried doing this

0 -> I -> A[x]/(x^n) -> (A[x]/(x^n))/I -> 0 is the same as
0 -> I -> A[x]/(x^n) -> (A[x]/I)/(x^n) -> 0

and then if you take projective limits of both you get something like the cokernel of I -> A[[x]] is either

lim_n (A[x]/(x^n))/I or lim_n (A[x]/I)/(x^n)

via right exactness

[cyan marten]

I am really confused about how I is an ideal of every ring that has something to do with A, but I know that it can be made precise.

On a related subject, are all ideals of A[[x]] of the form IA[[x]]?

[wraith obsidian]

What about (x)?

[next obsidian]

I don't think so???

This kind of gets into the question of if IA[[x]] = I[[x]]

[cyan marten]

Oh, yeah. I was thinking whether they are all of the form (I, x), I guess

But that's stupid

[next obsidian]

Turns out lim is left exact so everything I did is moot lmao

[latent anvil]

owned

[cyan marten]

Moot as in unnecessary or wrong?

[next obsidian]

Moot as in wrong

Haha

Unless you know stuff about lim1 in this case haha

And I don’t see any reason why it would have to vanish

[cyan marten]

According to AM, lim is right exact if the first inverse system is a "surjective system"

Which means all homomorphisms are surjective, and they are, in this case.

This doesn't make sense, so let me clarify. If {A_n} etc are inverse systems and we have a SES between them, and {A_n} is surjective, then applying lim gives a SES.

0 --> {A_n} --> {B_n} --> {C_n} --> 0

[hardy vector]

I'm having a little trouble parsing one part of an example of the application of Sylow's Theorem in my lecture notes, specifically the highlighted bits seem to require justification to me but are not justified so I think it must be the case that I'm missing something obvious. If anyone could clarify that for me I'd greatly appreciate it

I've included the relevant sections, the gist of it is that I don't see prop 5.1 as necessitating that the homomorphism is non-trivial which is then used to say that its image is not {1}

it follows immediately if you invoke (2) from thm 5.13 but that is not done; so is it just obvious and I'm missing something?

it feels to me like you should be able to apply prop 5.1 when r_2 = 1 though which would surely be a trivial homomorphism

[mild laurel]

Yes I think you're right that they need to use theorem 5.13

[hardy vector]

ok that's reassuring, thanks for prompt response

[next obsidian]

Oh, huh dope. Maybe this works then, I guess that’s a condition for lim1(A_n) to vanish

I think pretty much universally my A_n was always composed of identity so it’s surjective as you said

What section of AM is this from?

[cyan marten]

This is Proposition 10.2

Wait a second..

Oh nevermind

[next obsidian]

thanks!

[cyan marten]

You're welcome! By the way, by "Matsumura" do you mean "Commutative Ring Theory" or the other book?

[next obsidian]

Yup, the latter

I have a copy of the older one, because there was a Hartshorne problem that referenced it, and the proof was there

and it was literally a basic ass use of Ext

to show a map split haha

but I have no idea how to use Ext so I would've never come up with it hahaha

[cyan marten]

Ahaa

[next obsidian]

Sorry, latter in the sense of "commutative ring theory"

Latter chronologically, former in your message

[cyan marten]

Makes more sense. I thought ot is latter in that sense, too.

[next obsidian]

I don't recall the original having all that many exercises

[vestal snow]

Is there a program where I can check if a polynomial is irreducible over Q?

I think my professor made a typo, but I don't want to email him saying he did and it turns out I'm just too dumb

[mild laurel]

SageMath does this, and you can use cocalc.com

https://ask.sagemath.org/question/34244/irreducibility-of-a-polynomial/

[chilly ocean]

there is also wolfram alpha https://www.wolframalpha.com/input/?i=IrreduciblePolynomialQ[x^2%2B1]

[vestal snow]

It's irreducible

[latent anvil]

What was it?

[cloud walrus]

Have a Banana, Bitch

[chilly ocean]

[mild laurel]

For degree 3, you can just check that it has no rational roots, since if it was reducible, then it must factor as linear * quadratic

[vestal snow]

Yeah but its not easy to show that it has no roots in Q

[mild laurel]

Uh the rational root theorem?

[chilly ocean]

but what if i accidentally calculate (-2)^3-(-2)^2+(-2)+2=0, because i am not good at arithmetic?

rational root theorem drakeno wolframlalpha drakeyes

[vestal snow]

We didn't really go over that theorem in class

[chilly ocean]

at least in the US, we learn rational root theorem in high school

[mild laurel]

I feel like I learned the rational root theorem in high school idk

[vestal snow]

Uhhh

[latent anvil]

I did not

At all

lol

[chilly ocean]

interesting

i did

[vestal snow]

How do you teach that theorem without Gauss's lemma?

[chilly ocean]

it's obviously true

wikipedia has a proof that doesn't use gauss's lemma, but also, things don't get proved in high school

[vestal snow]

Oh wait

[chilly ocean]

i do recall the proof being given in high school, at least mine

[vestal snow]

I just read the theorem

Yeah that's not that complicated

[chilly ocean]

not that we were tested on it, or anything

[vestal snow]

That should probably have been the first that I should have tried lol

I was out here trying Eisenstein with f(x+4)

[mint gulch]

Do you know some Lie Algebra that is semisimple but not simple?

[mint gulch]

Maybe Ruffini's Method?

[crystal vault]

Hello (: I just had a quick question about proving stuff like this

I don't really see how I would go about showing that * restricted to H is a binary operation

The only thing worth showing is that it actually maps H x H -> H, which is guaranteed by assumption

i guess my main problem is just that I find myself saying something like "whatever * does to elements of G, it will do to elements of H, since H is a subset of G". But is there a more precise way to say that? Or is that enough?

[thorn delta]

my understanding is that by definition, a binary operation on H is just a function H x H -> H. So, precisely speaking, all you have to do to show that the restriction of G x G -> G to H x H is a binary operation is that the image of this restriction is contained inside H.

[next obsidian]

I learned it freshman year

[thorn delta]

which is just fancy speak for closed under the operation

[next obsidian]

Re: rational root theorem

Is G a group?

I mean, I guess they want you to check the axioms, but this is pretty much immediate by the assumptions given. I think you might want to think for a second why the identity is in H, but if this isn’t for hw I wouldn’t bother to write anything lol

[next obsidian]

Cant you just take a simple Lie algebra and then square that?

[scarlet estuary]

yeah kirafa you can argue that hyper-formally

like

"take any a, b in H. then a*b is in H as well since H is closed under the binary operation on G"

and a similar argument for the other properties of a group

but... this is pretty silly lmao

its kind of "obvious"

it probably just wants you to think about it to make sure the notion of a subgroup actually makes sense

(in mathematical jargons, we often say "the * operator in H inherits ___ from G" or similar phrasing to capture this idea)

(so an argument that a subgroup of an abelian group is abelian would simply be "the subgroup inherits commutative multiplication from the larger group")

(again, this is jargon-ey, but the term should make sense; we don't suddenly lose local properties of our structure because we're only looking at part of it)

[mint gulch]

Square it in what sense?

The derived algebra +

?

[latent anvil]

Direct sum with itself

[next obsidian]

Yeah like take L x L

[mint gulch]

Ok

[next obsidian]

It’s by definition semi simple

But you can take like the diagonal or something to show it isn’t simple I think

[mint gulch]

Yes, you're right, thanks

L x {0} would be a proper ideal

Wait, if I take L = 0, L is semisimple and not simple, right?

🤔

[crystal vault]

im late to this but thanks to all who helped, i appreciate c:

[shut halo]

I was asked to prove that the intersection of two algebraic sets is an algebraic set. So an algebraic set is given by $V(I) = {p \in C^n | f(p) =0 ,\forall f \in I}$ where $I$ is an ideal in $C[x_1,..,x_n]$.

The way I approached it was to assumed that given two ideals $I$ and $J$ then $V(I) \cap V(J) = {p \in C^n | f(p) =0 ,\forall f \in I \text{ and } J}$. So then I thought that this could be given by $V(I) \cap V(J) = V(I+J) = {p \in C^n | f(p) + g(p) =0 ,\forall f \in I \text{ and } \forall g \in J}$. Since, if $p \in V(I)$ and $p \in V(J)$ then $f(p)=g(p)=0$ so $f(p) + g(p) = 0$. Hence it would follow that $p \in V(I+J)$ as $I+J$ is an ideal.
However, I'm having trouble proving the converse direction, that is given $p \in V(I+J)$, then $p \in V(I)$ and $p \in V(J)$.

Any help would be appreciated.

[cloud walrus]

snypehype

[shut halo]

@chilly ocean oh yes, completely forgot that ideals were additive subgroups

thanks!

[next obsidian]

It follows because they are additive subgroups

[wraith obsidian]

It follows because 0 is mapped to 0 in any ring homomorphism, and Ideals are Kernels

[ivory cosmos]

so uh, I have a proof of this (this is an exercise) but it works whenever k[x] is a UFD, which only requires k to be a UFD

if you have a root, then you have (x - a) as a divisor, so the polynomial is certainly irreducible

if you have no roots, and you consider the factorization of the polynomial into irreducibles, this means that no factors of degree 1 exist

but you can't write 2 and 3 as x + y, without using a 1

(besides 2 + 0, of course)

But this proof seems to work in the more general case where you have a UFD

I don't see where the field assumption is necessary, unless I've made a mistake

🤔

[gritty adder]

I think your argument is correct

but you might be implicitly using Gauss's lemma

[ivory cosmos]

yeah, but that was seen in the last chapter 😛

[gritty adder]

yeah so it holds for any integral domain by passing to the field of fractions

[ivory cosmos]

right yeah, since irreducible <=> irreducible and primitive

So for primitive polynomials in an integral domain, this holds too

good point

[thorn delta]

If $V$ and $W$ are $k$-vector spaces, then $\operatorname{Hom}_k(V,W) \simeq V^*\otimes W$ via a map such that $f \otimes w \mapsto (v \mapsto f(v)w)$.

[cloud walrus]

kxrider

[thorn delta]

hm, there might be a finite dimensionality constraint, but anyway, just evaluating the map in the image on basis elements, it seems to me like these are just rank 1 maps. vectors v map to scalar multiples of w.

[latent anvil]

There is a finite dimensionality constraint

I believe?

Hmm maybe not, sorry

[latent anvil]

Non simple tensors exist

Like, you get linear combinations of rank 1 maps

[thorn delta]

ohhhhh. yeah.... i get it now. thanks

[latent anvil]

Yeah, I believe you need finite dimensionality

But I can't figure out why or find a reference

[thorn delta]

i think its maybe a similar reason to why the dual basis isn't spanning in the infinite dimensional case.

like, the same reasoning you would apply to show the finite dimensional case wouldn't work: there's a sense in which you would need to sum up infinitely many simple tensors to describe a linear map from an infinite dimensional space.

[latent anvil]

Well the reason I'm uncertain is that we are still dualizing V here

Which will do strange things in the infinite dimensional anyways

[cyan marten]

It's always true that dim V tensor W = dim V dim W, right?

Maye we can use a cardinality argument, then..

[magic owl]

well, this does not work in infinite dim

and even if you show the objects are isomorphic this does not mean that the map specified above is an isomorphism

[cyan marten]

Why?
Actually I meant the dimension

I think it works if I am aiming for a negative result. In this case,
dim V* tensor W = dim V* dim W
dim Hom(V, W) = dim V dim W, I think?

[magic owl]

if V or W are infinite dim

[cyan marten]

And it's a theorem that dim V* > dim V if V is infinite dimensional

So if dim W <= dim V, these two can't have the same dimension.

[magic owl]

I am not an expert in cardinal arithmetic but I think the multiplicative stuff is not going to work

i could be wrong

@split cipher am i being dumb

[cyan marten]

Me too. Just to illustrate my point, suppose V has dimension c, and W has dimension |N|. Then the dimension of the RHS is 2^c, while thay of the LHS is c.

I think the dim V* > dim V thing might be wrong

Maybe it's |V*| > |V|

[cyan marten]

How about this?

I think I found the mistake

dim Hom(V, W) = dim V dim W isn't guaranteed

At least, the proof I know uses finite dimensionality.

I see

Is it true that V* tensor W is the space of finite rank linear transformations, and thus in general a proper subspace of Hom(V, W)?

This seems intuitive to me because, after all, tensors are finite sums of rank 1 linear maps.

[latent anvil]

ah yeah, that makes sense to me

[wispy scaffold]

Ryan's book maybe

[latent anvil]

we have a proof above

it's iso to finite rank maps

I wonder if V^* (×) W is the space of compact maps V -> W for V, W banach spaces and taking the projective tensor product

I think so? Because we essentially complete the space of finite rank maps to get compact maps

[wispy scaffold]

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

[latent anvil]

Ah whoops, I was only thinking of the hilbert space case

Didn't realize AP could fail

So yes, assume V, W are banach spaces with AP

Err

I guess this is a condition between them

Yeah exactly

[wispy scaffold]

wdym

[latent anvil]

You want that any compact linear map V -> W is a limit of finite dimensional ones

[wispy scaffold]

i think W should have the AP

ok

this confused me

btw L(\ell^2,\ell^2) doent have the AP

oof

[latent anvil]

You mean this in the sense that there are compact operators L(ell^2,ell^ 2) -> L(ell^2,ell^ 2) that aren't the limit of finite rank ones, right?

[wispy scaffold]

yes

[glossy wing]

if $G$ is a group and $A,B$ are normal subgroups of finite index then $(G:A), (G:B)$ are both finite as well, how do I prove that $(G:A\cap B) \leq (G:A)(G:B)$?

[cloud walrus]

please request a new nickname

[glossy wing]

the equality occurs iff $AB={ab:a\in A, b\in B } =G$

[cloud walrus]

please request a new nickname

[glossy wing]

how do I go on to prove the $<$ part?

[cloud walrus]

please request a new nickname

[chilly ocean]

I would guess that you eg find a surjective map from G/A X G/B to G/(A cap B) or something like that

[glossy wing]

oh thats a good idea

thanks

[carmine fossil]

Kernel of that map is identity

So just use first iso to say H x K =G

If G=H x K means it's a direct product

Ok,not first iso

Note HK is a subgroup of G

And |H||K| /|H inter K|=|HK|

So if H inter K=e and |H||K|=|G| then HK=G

This theorem shows that HK iso to H x K

mb

[magic owl]

to answer your direct question

i dont think that criterion is particularly useful

but i also dont think your suggested contradiction works because i dont think H,K need to commute with eachother for HxK to be G, but i havent had coffee

my guess is this theorem is presented because its used in a future proof in the book

and for no other real reason

drake what r u talking abt

HK isnt direct product

its just a set

a priori

sure

but thats an assumption in the thm

i am actually confused now because I also have no idea why that commutativity assumption

is there

so i will take my own advice

and not think too hard about this

[carmine fossil]

(x,y).(c,d)=(xc,yd)
xycd=xcyd(by homomorphism condition) implying yc=cy for all c,y in G

[magic owl]

ah nice

i didnt bother to check the homomorphism condition

but ofc

use \times lol

oh

i didnt even know cross was a command

[cloud walrus]

mirzathecutiepie

[magic owl]

the commutativity cond is necessary

Im not sure its necessary as you've written the claim

but it is necessary to get the isomorphism to be what you want it to be

so

let f(x,y)=xy

ah lol

I think a counting argument might work regardless in this case to show that you end up getting G?

but its nice to know what the iso is

okay so f(x,y)=xy

f(x,y)f(a,b)=f(xa,yb) for it to be homomorphism

but this requires that

xyab

=

xayb

you know f

what is

f(x,y)f(a,b)=f(xa,yb)

write it out explicitly

[carmine fossil]

This is to make the map a homo

[magic owl]

yeah the multiplication property won't hold if those things don't commute

(this is similar to why if multiplication is a homomorphism the group is abelian)

[carmine fossil]

Yes

Writing it as f(a,b) f(c,d)=f((a,b)(c,d)) might be more useful(f is a hom from H x K to HK )

[magic owl]

no

not quite

f(a,b)=ab is not a map from G to itself

its a map from GxG to G

congrats you have just proven the statement "group objects in the category of groups are abelian"

[chilly ocean]

[unique juniper]

[winter vigil]

I'm sorry but I asked in number theory but got no replies, so here goes as well E(15) is isomorphic to $E(3) \times E(5)$ which is isomorphic to $\mathbb{Z}_2 \times \mathbb{Z}_4$ right ?

[cloud walrus]

Otoro

[hot lake]

never seen the name E(15) being used for the group of invertible elements of Z/15Z

[winter vigil]

It's just the notation from my professor I guess

[next obsidian]

Anyway your statement is correct

You should show that E(R x S) = E(R) x E(S)

then it follows by the Chinese Remainder Theorem telling you that Z/15Z = Z/3Z x Z/5Z

I guess also you need to know that the units of Z/pZ = Z/(p-1)Z

[lavish pike]

so im a bit confused on how to show the left distribute law holding. I have never worked with logs before so am confused on how to use that to my advantage.

cause we can take it along the lines of $(a * b)+ (a * c)=a^(ln b) + (a * c)$

[cloud walrus]

cRaZyNiChOlAs12

[next obsidian]

Log(ab) = log(a) + log(b)

To see this, use the definition of log(x) as the number such that x = e^log(x)

[quaint ivy]

(unrelated: in LaTeX math mode use curly brackets for the exponents, a^{\log(b)})

[next obsidian]

Also omegalol at them denoting multiplication (x) then as (•)

[wicked stump]

not sure how to show the forward implication, i assume I need to use the fact that G is a finite group somehow

Can anyone give me a hint?

Ok, I see that those are the conditions I need to show for H to be a subgroup

What should I be looking for when I take powers of h in H?

I think that the set of powers of h in H would also be finite, but I'm not sure what that means

I see

Ok, I'll give it a try

I am not, we haven't covered that in class yet

Ok, thanks

[lavish pike]

so if i am getting this right then $a^{\log(b)}=log(a)$ and am i allowed to say that $a*c=a^{\log(c)}$

[cloud walrus]

cRaZyNiChOlAs12

[lavish pike]

wait no itd be b=log(a)

[nova plank]

b = log(a) in what context?

What are a and b? What are you trying to show?

[lavish pike]

this

the distributive law, im assuming i would need to do something with logs

[nova plank]

So a and b are just arbitrary real numbers but you are saying b = log(a)?

[lavish pike]

ye that doesnt make much since now does it lol

[nova plank]

No, it does not haha

[lavish pike]

ummm idk im just a bit lost on how to prove the distributive law holds, like what does that mean as the end result

[nova plank]

What is a*(b+c)? And what is (a*b)+(a*c)?

[lavish pike]

well one is $(ab+ac)$ and one is $(2a+b+c)$

[nova plank]

Where did you get that?

I mean in terms of the operations defined there

• is the multiplication and + the addition, I can't make those fancy symbols

[lavish pike]

arent they the same thing then?

[nova plank]

What do you mean?

$b+c = bc$ according to your definition. Then $a*(b+c) = a^{\ln(bc)}

[cloud walrus]

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

[nova plank]

Do the same for the other expression

And then you can see what you have to show are equal

[lavish pike]

can we say that a+c=ac is also applicable to b+c=bc

[nova plank]

Yes

The letters are arbitrary. For any x, and any y, x+y means xy

$\cplus$

[cloud walrus]

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

[nova plank]

$\oplus$

[cloud walrus]

Lunasong

[nova plank]

Damn, I figured it out

$x \oplus y = xy$ and $x \otimes y = x^{\ln(y)}$

[cloud walrus]

Lunasong

[nova plank]

This is for any x and y, so you can make either of them a, b, c, or anything else.

[lavish pike]

ah ok

[nova plank]

So what is $a \otimes ( b \oplus c)? $

[lavish pike]

a$\otimes$(bc)

[cloud walrus]

cRaZyNiChOlAs12

[nova plank]

Get rid of the circle operations completely

Look at the defn of multiplication, now y is bc and x is a

[lavish pike]

well then it is like u said $a^{ln(bc)}$

[cloud walrus]

cRaZyNiChOlAs12

[nova plank]

Yes

Now what is $(a \otimes b) \oplus (a \otimes c)$?

[cloud walrus]

Lunasong

[lavish pike]

$a^{ln(b)}\oplus a^{ln(c)}$ ?

[nova plank]

Yes

[cloud walrus]

cRaZyNiChOlAs12

[lavish pike]

lol i hate texit

[nova plank]

But you need to get rid of the oplus too

It's fine, I can understand, dw about it

[lavish pike]

and then $(a^{ln(b)})(a^{ln(c)})$

[cloud walrus]

cRaZyNiChOlAs12

[nova plank]

Yes exactly

So what you have to show is
$$a^{\ln(bc)} = a^{\ln(b)}a^{\ln(c)}$$

[cloud walrus]

Lunasong

[lavish pike]

since they have the same base (i think that is what its called) cant u just say that exacct thing u just showed?

[nova plank]

What did I show? I didn't show anything

[lavish pike]

well no not showed but what u just said where $a^{ln(bc)}=a^{ln(b)}a^{ln(c)}$

[cloud walrus]

cRaZyNiChOlAs12

[nova plank]

That's what you have to show

That's the whole question

[lavish pike]

so. $a\otimes (b\oplus c)=(*a \otimes b) \oplus (a \otimes c)=a^{ln(b)}a^{ln(c)}=a^{ln(bc)}$

[cloud walrus]

cRaZyNiChOlAs12

[lavish pike]

and that proves the distributive law.

[nova plank]

No, that doesn't prove anything

[lavish pike]

reeeeeeeeeeee

why dont i get this lmfao what is happening

[nova plank]

The order you wrote it there makes no sense

You want to prove
$$a\otimes (b\oplus c)=(a \otimes b) \oplus (a \otimes c)$$

[cloud walrus]

Lunasong

[nova plank]

You don't know this is true, you have to prove it

So it makes no sense to start your proof by saying they are equal already

Now, if you want to know if they are equal, you should look at what the LHS and the RHS are

LHS = a^(ln(bc)) and RHS = a^(ln(b)) a^(ln(c))

Now you need to give an explanation for why those two expressions are equal

[lavish pike]

so can i say that $a \otimes (b \oplus c)=a^{ln(bc)}=a^{ln(b)}a^{ln(c)}=(a \times b) \oplus (a \otimes c)$

[cloud walrus]

cRaZyNiChOlAs12

[lavish pike]

since we are just supposing that those 2 equations are equal i have to prove that they are equal right?

[nova plank]

Yes, but you need to justify why the = in the middle is true

[lavish pike]

oh ok ye see that is what i dont understand cause i have never worked with logs before and i am assuming that i need to do something specific to laws with logs to prove that right?

[nova plank]

Yes, the log rule log(xy) = log(x) + log(y)

[lavish pike]

nd does that apply to the whole ln(bc) in the exponent?

like idk how to apply the log rule

[nova plank]

It applies for any x, y > 0.

And b, c are > 0

[lavish pike]

but i cant just say $a^{ln(bc)}=a^{ln(b)}a^{ln(c)}$

[cloud walrus]

cRaZyNiChOlAs12

[lavish pike]

?

cause that is true by the log rule correct?

[nova plank]

@lavish pike there is a missing step. You can say it, if you can explain to me why it's true

[unique juniper]

dont get the part where let k be the larger of m and n, how can they be sure both a^p^k b ^ p^k are both the identity?

[chilly ocean]

Hello Everyone 🙂 Why does a cartesian product can't form an algebra?
From what i've researched it's not closed under complementation
but I can't figure out why
could someone provide a counterexample or intuition?

[lavish pike]

$a^{ln(bc)}=a^{ln(b)+ln(c)}$ ?

[cloud walrus]

cRaZyNiChOlAs12

[nova plank]

Yes

And then why is that equal to the other thing?

@unique juniper Let's say m is the biggest one, and n the smaller. Then
$$a^{p^m} = a^{p^np^{m-n}} = (a^{p^n})^{p^(m-n)} = e^{p^{m-n}} = e$$

[cloud walrus]

Lunasong

[scarlet estuary]

(For what it's worth, this seems like less of a problem with logarithms and more a conceptual problem with what it means to "prove" something)

[unique juniper]

got it

ty

[nova plank]

Np

[mint gulch]

Every set can form an Algebra. I suppose you mean if I have two Algebras A, B, AxB not necessaryly, right?

[chilly ocean]

sorry for double posting we're tackling it already here #help-0 message

[lavish pike]

is there ever a reason to use the 4 step subring test?

as opposed to the 2 step one

[magic owl]

im not entirely sure what you mean

I think there are two decent ways of proving something is a sub-whatever

first just show directly all the operations and whatever are closed

and all the elements needed are there

or two

prove that your subset is the image of a homomorphism

[chilly ocean]

I'm guessing you mean like rs^-1 is in the subgroup for all r, s. I guess the 4 step looks somehow "nicer" but this is not a great argument

[lavish pike]

well one is like just closed under subtraction and multiplication but the other is closed under addition, subtraction, identity, and i think its like a+x=0 has a solution

[magic owl]

i think that most variations of axiom checking will amount to the same basic amount of work

so it probably doesnt make a huge difference most of the time

[cloud walrus]

PandaMan-AMB

[magic owl]

write it in that form or as a scaled sum of things in that form

[stray rivet]

we went through some of the consequences of the homomorphism in class today and im kinda blown away

REALLY powerful results from something (at least in my limited paradigm) so innocuous as the homomorphism

[latent anvil]

yeah!

Algebra is all about getting a ton of structure from a seemingly simple setup

[chilly ocean]

algebra is all about monkey ooo oo eee ooo

[latent anvil]

@next obsidian moment

How was the algebra exam

[chilly ocean]

pretty good

immediately knew the answer to 3/5 of the questions

[latent anvil]

Nice!!

Manifesting this for me

I got my analysis final rescheduled so I'm not also giving a talk the same day

But I'm taking it at the time reserved for students in Asia or europe

it will be exciting

[chilly ocean]

wow your analysis exam gets a different section for a different timezone?

fancy

[latent anvil]

Not officially

[chilly ocean]

mine was at 9 am on a monday

[latent anvil]

I mean officially we only have a 2 hour time period for the exam

But the prof is doing a 4 hour exam

She just wanted to schedule another time for students who couldn't make it

Also that's rough

[chilly ocean]

yeah i did horribly

[latent anvil]

I'm doing my dold kan talk at 9:30

[chilly ocean]

[latent anvil]

Group before us is going at 8:30

[chilly ocean]

good luck

[latent anvil]

ty

[chilly ocean]

[latent anvil]

pethecat

[chilly ocean]

well it was a wednesday but look

9 am bad

[latent anvil]

9am is bad

My analysis exam was/talk is next Wednesday

[lavish pike]

is there a way to do matrices on texit

[next obsidian]

$\begin{bmatrix}
0 & 0\
0 & 0
\end{bmatrix}$

[cloud walrus]

Chmonkey didnt get into Columbia

[next obsidian]

You can also do pmatrix instead of bmatrix

Also note what looks like \ is actually two slashes, but one got eaten up by Discord

When I change the row

\bmatrix is just \bmatrix

[chilly ocean]

$test \\$

[cloud walrus]

8da | dumbass

[lavish pike]

reeee

are all the \ double

[chilly ocean]

$\begin{bmatrix} 0 & 0\\ 0 & 0 \end{bmatrix}$

[lavish pike]

$\begin{bmatrix}
a & 0\
0 & a
\end{bmatrix}$

[chilly ocean]

Hm

[cloud walrus]

cRaZyNiChOlAs12

[lavish pike]

ahhhhh

[cloud walrus]

8da | dumbass

[lavish pike]

so prove that the set of scalar matrices is a subring of M(R)

do i just take to matricies in the form of $\begin{bmatrix}
a & 0\
0 & a
\end{bmatrix}$ and do a subring test on em

[cloud walrus]

cRaZyNiChOlAs12

[mint gulch]

Yes

[cyan marten]

Are there interesting topologies on finite groups?

Of course, we have to drop the convention that topological groups must be Hausdorff.

Maybe homogeneity makes this impossible

[latent anvil]

Indiscrete

Not interesting but it shows homogeneity doesn't make this impossible

[eager willow]

trying to understand something from Lang:
We have $L / K$ and $K / k$ are both solvable and Galois.
I don't understand the following:
If $\sigma$ is any embedding of $L$ over $k$ in a given algebraic closure, then $\sigma K = K$ and hence $\sigma L$ is a solvable extension of $K$.

This is only stating that $\sigma | K$ acts as an automorphism, and not that $\sigma$ fixes $K$, right? This I think follows from $K / k$ is Galois. But why is $\sigma K$ not an isomorphic copy of $K$, but $K$ itself? The algebraic closure is not specified, is this just a formality, making the assumption that $K$ is an intermediate field $\overline k / K / k$?

[cloud walrus]

datorangeguy

[latent anvil]

I don't think this is correct as stated

For the reasons you list

Like, you need to fix an algebraic closure containing K to make it work like Lang says

And yeah σK = K means that it maps K into itself, not fixing it

[next obsidian]

What page of Lang is this?

[latent anvil]

So like, because K is galois over k it's the splitting field of some separable polynomial f, and then the splitting field of f inside a fixed algebraic closure is a copy of K (so unlike the nongalois case, there is a unique intermediate field k-bar/K'/k with K' iso to K over k)

So maybe this is what he means

[eager willow]

this is page 291, proof of prop 7.1

[next obsidian]

So it says that K is finite over k in what I’m looking at, implying it’s algebraic

[latent anvil]

It's galois over k Alex

[next obsidian]

So any elemrn of K is a root of a polynomial, so doesn’t the fact that sigma fixes k imply that sigma only shuffles K around. Oh yeah

[latent anvil]

almost

Galois gives you that the galois conjugates of any element in K are also in K

otherwise you could get a root of the min polynomial which leaves the field

but also what you're saying doesn't exactly make sense because K might not be contained in the chosen algebraic closure

"shuffle around K" doesn't make sense when the domain of σ isn't a subset of the codomain

[next obsidian]

Fuck haha

[latent anvil]

I think this is what @eager willow, what does Lang mean

[next obsidian]

I’m so *-pilled i have just physically embedded everytbing

[latent anvil]

And the best guess I have is to take an algebraic closure strictly containing K

[next obsidian]

Yeah, once you shift eveything to sit inside a fixed algebraic closure

Then you get what you want

[latent anvil]

This remark makes me feel more comfortable about it

K is sort of unique

[next obsidian]

Shift = take isomorphic things

[latent anvil]

Due to being galois

[eager willow]

that makes sense

[next obsidian]

🥴

Field theory man

[latent anvil]

I don't think I read Lang at all in 50x lol

[next obsidian]

I read it on a midterm

To prove Krull’s intersection theorem

[latent anvil]

hahaha

[next obsidian]

[latent anvil]

How far you've come

[next obsidian]

Which I didn’t even need

Also that’s not cheating it was a course text

[eager willow]

there are gaps in D&F but I've found between the two there's pretty much everything needed

[latent anvil]

for sure

[next obsidian]

Until you need any like

[latent anvil]

I was able to find everything in d&f but it seems like Julia is spending much longer on field theory

so maybe that's why

[next obsidian]

Real heavy homological algebra / commutative algebra I think those do wel as a combo

[latent anvil]

ah that's true

[next obsidian]

My impression is that Lang actually has pretty deep field theory

Like I think it talks about inseparable and blah blah stuff

[latent anvil]

but we used the ug comm alg book for comm alg

And sandor had good homological algebra notes

[next obsidian]

Eh, I don’t know how much anyone used a book in Sándor’s class

Haha

[latent anvil]

true

[eager willow]

^^ and also filtrations are weak in D&F which was hard for last quarter

[latent anvil]

I used Reid

[next obsidian]

Oof :(

[latent anvil]

Yall did filtration stuff?

oh like normal series?

[next obsidian]

Also yeah

[eager willow]

yeah

[latent anvil]

gotcha

[next obsidian]

Ah

Yeah Lang’s section on filtrations in the graded ring part is good I feel

[latent anvil]

yeah idr if we did like upper/lower central series or anything beyond what's needed for solvability

[next obsidian]

That’s what I used to prove Krull’s intersection theorem

[eager willow]

took me a second to realize I had gotten help from yall on this server before I was accepted to uw haha

[latent anvil]

yah I think I saw that haha

It was funny

[next obsidian]

Yeah haha

Sham told me

[latent anvil]

we both spend way too much time here

[next obsidian]

And I was like “oh shut yeah”

Especially this channel specifically

[eager willow]

yup

[next obsidian]

I think once after class

We both literally instantly checked this

And it was like the Spider-Man meme

Where they’re pointing at each other

[eager willow]

lol

[next obsidian]

I think you should be pretty set for next qtr tho

I remember you were doing commutative algebra stuff from A-M probably

You had a nice way to see something about Ass for a direct sum

And the fact you can just go the other direction

[latent anvil]

What even are you doing next quarter? Is sandor going to try and cover rep theory?

[eager willow]

no idea

[next obsidian]

Oh yeah 😅

Julia is nice but Sándor is the goat

No disrespect to Julia

[latent anvil]

508 had similar problems

with the amount of content covered

[eager willow]

yeah I was self studying last year back when quarantine seemed like a new and exciting opportunity to self improve

[next obsidian]

Hahahaha

[latent anvil]

lol

[next obsidian]

I’ve only recently begun to accept that no, quarantine has affected me

And I’m not just built different

And that that is okay

[latent anvil]

yeah you only finished half of Hartshorne

without quar you would've gotten 70%

[next obsidian]

True I’d be done by now

Except for II.6 lmfao

[latent anvil]

Do you know what you want to focus on @eager willow?

Seems like you're doing pretty well in algebra in analysis

[eager willow]

not really sure, I've liked virtually every topic in commutative algebra that I've come across and kinda just chug through everything else for the exposure. Definitely not as strong in analysis as algebra

[latent anvil]

comm alg is really fun

[next obsidian]

It is!

I didn’t get it in 506

[latent anvil]

you could try to hop in Sándor's class next quarter to get more exposure

[next obsidian]

But over the summer I picked up Matsumura and it’s like broooo

Oh yeah!

509

[latent anvil]

I like analysis but I did not enjoy this class...

[next obsidian]

If you’re familiar with a good portion of A-M you should be fine

Although you might have to take some homological algebra results on faith until you do it ij class

[eager willow]

isn't there a time conflict?

[latent anvil]

I don't think that's possible, since Sándor teaches both classes

unless he has a time turner

This quarter people did 505 and 507 concurrently

[next obsidian]

Sándor just built different

[eager willow]

oh nevermind I was thinking of some seminars

[latent anvil]

Lecturing to two zoom windows at once

oh idk anything about those

[unique juniper]

this proof isnt so clear to me

if there is only 1 distinct prime dividing the order of a

how do we know that |a| = p^x for some int x

fundamental theorem of arithmetic ?

[chilly ocean]

Yeah

[unique juniper]

also

"similary the order of $a^{p^{m}u}$ divide q"

this is confusing me

[cloud walrus]

Yes

[chilly ocean]

You know lagrange's theorem right?

[unique juniper]

ye

[chilly ocean]

That thing raised to q equals identity, so it's order must divide q

[unique juniper]

hm

$(a^{p^{m}u})^q = a^{p^{m}q u} = e$

[cloud walrus]

Yes

[unique juniper]

like that? i got it

[chilly ocean]

Yeah

[unique juniper]

how does that use lagranges theorem tho

[chilly ocean]

Hm, yeah sorry I guess I'm misquoting it

[unique juniper]

np

[wind steeple]

why is a complex X where H^i(X) = 0 for all i =/= 0 is quasi isomorphic to a complex concentrated in degree 0 ? It's probably because we have a morphism X_0 -> Ker d_0/Im d_-1 but I don't see how

[cyan marten]

It must be an extension of Q8 by C2, or (equivalently?) of C2 x C2 by C4.

[hot lake]

what's the normalizer of a subgroup again ?

[rustic crown]

oh no, no... if H is a subgroup of G, then N_G(H) is the largest subgroup containing H such that H is normal in N_G(H)

we might not have N_G(H) normal in G

[cyan marten]

If this was true, then there would be no self-normalizing proper subgroups, which is false.

[rigid cave]

Hello! I am curious about one thing: if F(a) and F(b) are field extensions of F and the degree of F(a) over F is equal to the degree of F(b) over F, then is F(a) isomorphic to F(b)?

It must be case, right?

[wind steeple]

no for instant two galois extensions over C of same degree that do not have the same galois group

[chilly ocean]

Or even simpler (in a way), take something like Q(sqrt2) and Q(sqrt3), one has a square root of 2 and the other doesn't

[rigid cave]

Oh that’s true...

But if a and b are roots to the same irreducible polynomial then?

An irreducible polynomial that is not monic

Otherwise it’s trivial

[chilly ocean]

Er, why does it matter that it is monic? (We are working in a field so monic or not doesn't matter right?)

I think field extensions from roots of irreducible polynomials being isomorphic sounds right, this sounds like a theorem

[rigid cave]

But if it’s monic then there’s a theorem that says that F(a)~F[x]/(f(x)) where f(x) is the minimal polynomial to a

So if this irreducible monic polynomial has tho roots then those field extensions are isomorphic

But what if f(x) is not monic? Is this still the case?

[chilly ocean]

Monic or not shouldn't matter, an irreducible polynomial can be multiplied by a constant and it is still irreducible

[rigid cave]

Okay okay I’m staring to get it

But if you multiply the polynomial by a constant, wouldn’t this change the “nature” of this polynomial?

So what I’m askin is, are the root still the same after the multiplication?

[chilly ocean]

I think this is a matter of definitions (whether an irreducible polynomial must be monic or not), and I can't say I remember these details, but surely the roots should stay the same right?

[rigid cave]

I believe so yeah

You could factor the constant

And then factor the polynomial that is left I believe

[chilly ocean]

F(a)=0 iff c*f(a)=0

For c nonzero

[rigid cave]

Okay I get it now! Thank you so much!

[nova plank]

Assuming you're working in an integral domain

[rigid cave]

Yeah and fields are integral domains

[nova plank]

Oh, yes, I didn't read the whole convo

[shut halo]

Anyone got an idea on how to do this?

I need to show this without ideals or something like that

[ivory cosmos]

consider the polynomial f(x, y) = 1

the empty set is the set of zeros of this polynomial

and is a finite subset

[glossy wing]

question

[chilly ocean]

I guess at least it reduces to the one point case, by multiplying polynomials

[shut halo]

I guess that works but what if I want a non-empty subset?

[ivory cosmos]

I misread

the question is not to find an example of a finite zero set

but to show that any finite subset can be expressed as a zero set

[chilly ocean]

Yeah, this question is worded weirdly, I feel like "any" should be "every"

[ivory cosmos]

I misread it as "Is there any"

actually, you can read "Is any finite subset .."

as asking about existence

[carmine fossil]

Wait, Isn't that clearly true,if {(a,b),(c,d)...} Is the finite set, Consider the polynomial ((x-a)^2+(y-b)^2)((x-c)^2+(y-d)^2)...(mb,over C)

[shut halo]

@ivory cosmos so the question is that all points in a finite subset of C^2 should be expressible a solution of a polynomial constant + ax + by + cxy + dxy^2 + ex^2y + .... = 0?

[chilly ocean]

I think the answer is false, without using any AG. attempt: for a singleton set, we wish to find a polynomial whose zeroset is this singleton. view a polynomial as being in x, then for any fixed y there is a root. But this means there is a root for each y, contradiction. (I think there are holes in this argument but maybe it works)

[carmine fossil]

Over R, That is true right?

[chilly ocean]

Yeah over R I think it works

[shut halo]

btw the next part of the question is this

so I think @chilly ocean interpretation should be correct

[chilly ocean]

Uhh, answer is yes for part ii, right? Just pick like the collection of 1 polynomial consisting of x lol

(is there an AG theorem like over an algebraically closed field the zero set of n polynomials in m variables is dimension m-n? This feels sort of right)

[shut halo]

Yeah I think you can also pick {x+y=0} then you should get an infinite number of solutions.

i.e. (x,-x) in C^2 for all x

[charred pewter]

Can someone help me find the elements of the signed permutation matrix group B_2?

[charred pewter]

This is what my professor gave me to start me off but I'm kinda confused

[lone violet]

If T(G) is the subgroup of G of all elements of G with finite order with G abelian then T(G/T(G))=0

[carmine fossil]

Yes?

[lavish pike]

looking at other peoples work in this chat scares me lol

[cloud walrus]

cRaZyNiChOlAs12

[hot lake]

for any subsets A,B of the group, AB is a well defined subset

then they show that if A and B are cosets then AB is a coset

and actually that proof mirrors the proof you would have to write if you defined the product by "pick x in the first coset and y in the second coset, define the product to be the coset of xy" is well defined

kinda

not really ? you have to show that if you "replace" x with xh1 and y with yh2 you get the same coset, so forall x,y in G and h1,h2 in H then there is h3 in H such that xh1yh2 = xyh3

ah

well for the x slot this is basically the same, and for the y slot it is basically trivial

because you would have to show that if y2 = y1h then xy2 is in the same coset as xy1

so you have to show that xy1 is in the same coset as xy2h

[ivory cosmos]

there's a few subtle steps going on, which do work out

[hot lake]

and that's basically the definition of H-coset

[ivory cosmos]

idk, the way I usually think / prove that multiplication here is well defined, is by showing that if x, x' are in xH, and y, y' are in yH, then xyH = x'y'H

this seems the most straightforward

[mild laurel]

What have you tried?

[lavish pike]

well i figured that $1=ab(x)=ab(ab)^{-1}$ buuuut im not really sure if that helps me at all because idk what to do with that information

[cloud walrus]

cRaZyNiChOlAs12

[mild laurel]

Well, you still haven't used that a is a unit, so maybe try using that fact to rearrange your equation

[lavish pike]

so like 1=b(ax)?

[mild laurel]

Well not quite

you haven't really used the fact that a is a unit there, you've just used the associative property

[lavish pike]

how does a being a unit help here? like i dont understand how to even use it in proving that b is aunit

[mild laurel]

you have that ab(ab)^{-1} = 1

how can you use a^{-1} here to rearrange this equation?

[lavish pike]

well im not really sure but i think you could get say that aa^{-1}=1 so then could u say that b=1?

[mild laurel]

how do you get that?

[lavish pike]

how can u use just a^{-1} to manipulate the equation when the equation is (ab)^{-1}?

[lavish pike]

well the a^{-1} just cancels out the a

[mild laurel]

Except you don't have commutivity

you can't just rearrange like that

[lavish pike]

yea ive got no clue.

[mild laurel]

If I write that

5x = 10

how do you solve for x

[unique juniper]

inspection

[lavish pike]

u divide by 5 lol

[mild laurel]

okay

[unique juniper]

[viscid pewter]

what even is life

[mild laurel]

now what can you do to ab(ab)^{-1} = 1

by using the fact that a^{-1} exists

[unique juniper]

i dont understand equivalence classes apparently

[lavish pike]

what does that even mean lmfao. like you can divide by ab so that it is equal like ab^{-1}=1/ab but what is that relevant like i dont understand what u mean by what can u do to ab(ab)^{-1} since a^{-1} exists.

[viscid pewter]

what do you generally use a^-1 for

[mild laurel]

If I write it like

a(y) = 1 where y = (b(ab)^{-1})

what can you do

[viscid pewter]

i hate how brackets are used for goddamn everything

always makes me think of functions

[lavish pike]

i hate how this is completely going over my head like completely

just a massive brain fart

[mild laurel]

again, how how did you solve 5x = 10 earlier

you divided both sides by 5

this is the same thing as multiplying both sides by 5^{-1} right?

[lavish pike]

yea

[mild laurel]

So what can you do to ay = 1

[lavish pike]

so i can multiply both sides by a^{-1}

[mild laurel]

(you need to be a little more careful here if you don't have commutativity, you need to specify which side you're multiplying by)

So you multiply ab(ab)^{-1} = 1 on both sides on the left by a^{-1}

and what does this get you

[lavish pike]

a^{-1}ab(ab)^{-1}=a^{-1}?

[mild laurel]

Right and can you simplify the left hand side of that equation?

[lavish pike]

b(ab)^{-1}=a^{-1}

[mild laurel]

Exactly

now remember what you're trying to get to

[lavish pike]

bx=1?

[mild laurel]

Yep

And how can you rearrange the equation you have into something of that form?

[lavish pike]

-a^-1+b(a^-1 b^-1)=0?

no clue. there is not a chance that is right lol

[mild laurel]

Well, you want 1 on the right hand side

and you want b(something goes here) on the left hand side

[lavish pike]

so like divide by a^{-1)?

to get b(b^{-1})=a^-1/a^-1?

[mild laurel]

Right, you can multiply by the inverse of a^{-1}

but you probably have seen that the inverse of a^{-1} is just a

[lavish pike]

so just multiply both sides by a rather than dividing by a^{-1}

[mild laurel]

Right (these are just the same thing)

[lavish pike]

yea

[mild laurel]

So you get b((ab)^{-1}a) = 1

[lavish pike]

but then doesnt that a in the parentheses cancel out the a^{-1}?

[mild laurel]

Where is the a^{-1}?

There's an (ab)^{-1}, but this is not necessarily the same thing as a^{-1} b^{-1}

[lavish pike]

ah ok

so just to get this straight. we begin with ab(ab)^{-1}=1 multiply both sides by a^{-1} to get a^{-1)ab(ab)^{-1}=a^{-1} then we multiply by a to get b((ab)^{-1}a)=1

[mild laurel]

Uh, you have a typo in the last equation but yes

But depending on your definition of a unit, you're not quite done yet. Usually (when you don't have commutativity), you define that b is a unit if there is some element y such that by = yb = 1

[lavish pike]

how do you prove commutativity then?

or is that not necessary since the proof doesnt ask for it?

[mild laurel]

You don't prove commutativity

You can still have units even if your ring isn't commutative

[lavish pike]

this test is gonna be roughhhh

i dont get the whole proving of unit divisors or zero divisors

like ik unit divisors are ax=1 and bx=0 where i think its like b and x cannot = 0

[mild laurel]

I'm not sure what you mean by you don't get the point

Units and zero divisors are really important special elements of a ring, and when confronted with a new ring, its basically the first thing we ask about it

[lavish pike]

i get its important lol i just dont get how to do it like how do i prove that like that if ab is a zero divisor then prove that a or b is a zero divisor

like i have to get ab(x) into the form of a(bx)=0? or b(ax)=0?

so does that mean if ab is a zero divisor then a(bx)=0 means that a is the zero divisor or if b(ax)=0 then b is the zero divisor?

[mild laurel]

Again, I don't think you're working with commutativity, so you can't rearrange to get b(ax) = 0

But you only need to show that either a or b is a zero divisor, and you showed that a is a zero divisor by seeing that a(bx) = 0

[lavish pike]

so the question has to state that there is commutativity i cant just assume it?

[mild laurel]

depends on your class I guess

some people define rings as being commutative

[lavish pike]

i think we assume it cause in our notes it talks about rings having commutativity

[mild laurel]

then sure

[unique juniper]

can someone check my answer for the second part

We know that G acts on A transitively so theres 1 orbit by definition. But that means that there exists a g in G that can transform any element of A into b. If we take the intersection of these stabilizers, we get atleast the identity, + those that will fix every element of A

[next obsidian]

By definition the kernel of the action would be \cap_b in A G_b right?

but by the first part yo uknow that every b is of the form ga for some g

so at the very least \cap_b in A G_b < \cap_g in G gG_ag^-1 from the other part about stabilizers

so then you need to show the other inclusion, but this is obvious because any gG_ag^-1 = G_{ga}

I just don't think your answer justifies why \cap_g in G gG_ag^-1 is the set of things which fix each element of A

[ember thorn]

What are some examples of class function other than characters?

[sturdy marsh]

uh

[ember thorn]

Fulton says characters induces an iso from representation ring tensor C to class functions

[sturdy marsh]

if you take the difference of two different irreducible characters

then it's still a class function

[mild laurel]

(these are called virtual characters and come up in a lot of different places)

[ember thorn]

true, i read about it too

[sturdy marsh]

and this is a super neat iso

if your group has order n, then all the characters can be realized over some finite extension of Q

so you get an action of the corresponding galois group

on the character ring

and then using the iso you get an action on the rep ring

[cloud walrus]

Brofibration

[sturdy marsh]

which act as k-th power operations on 1-dim reps

[ember thorn]

sadly they probably won’t appear on my lecture

[cloud walrus]

Brofibration

[ember thorn]

no professor working on representation theory in my Uni

[sturdy marsh]

this stuff was from a topology thing lol

The psi^k are the Adams operations on K-theory

The rep ring is the equivariant K-theory of a point

@chilly ocean a vector bundle over a point is a vector space

[chilly ocean]

yes

[sturdy marsh]

an equivariant bundle over a point is a vector space + group action

so a representation

[chilly ocean]

"equivariant bundle"

okay

[sturdy marsh]

take everything, stick a G in front of it

and ask for everything to commute with the G-action

[chilly ocean]

G(ay) bundle

[sturdy marsh]

and this may sound stupid but has plenty of cool consequences

[chilly ocean]

it sounds like something i'd read on nlab

that's a good thing

[sturdy marsh]

lol

the nlab page on equivariant k-theory isn't bad

[latent anvil]

gay bundle gay bundle

principal gay bundle

classifying space for gay bundles 😳

what if you were a classifying space for gay bundles...and i were a universal gay bundle over you...and we were both guys

[chilly ocean]

[wild sapphire]

why is {e,s} not a subgroup of D4 ?

.. or is it

yeah, e the identity, and s the reflection

no rotation

I feel like it's a subgroup

hmm I'm not sure, something to do with the lack of r, but then again the requirement of r is just a property of dihedral groups right

rotation

[delicate bloom]

sounds like they're talking about it being a subgroup that isn't a dihedral group maybe

I'm not psychic either haha

[wild sapphire]

ah, so since s is its own inverse, the inverse exists and so it's a subgroup

a non-dihedral subgroup

yeah, just needs to be a group with the set it's acting on a subset of the original

makes sense

I looked it up and got stuff talking about normal groups which I think is taking about commutativity, but it doesn't include {e,s} so I was getting confused

[next obsidian]

It's a dihedral group on one vertex tho right

[wild sapphire]

hmm, but doesn't it need to have rotation to be dihedral? And if you say rotation does nothing so it's the identity, then wouldn't that be the same for reflection 🤔

then you have a trivial {e}

[next obsidian]

I mean for {e,s}

idk

I thought dihedral on like

n vertices has 2n elements

like a regular n-gon

[wild sapphire]

why is {e,r,r^3} not a subgroup of D4? Composing r and r^3 gives the identity 🤔

[small bison]

well r^2 isn't in the set

[wild sapphire]

is it because it's not closed? like R^2

ah so just like being closed is inferred from the group definition

my intuition says that any subgroup of a dihedral group that doesn't contain any reflections is abelian

but idk if that is true 🤔

[quaint ivy]

since it's D_4, r is an element of order 4

[wild sapphire]

well closure isn't strictly in the definition of a group but it's probably a consequence of it

?

[small bison]

it's part of the definition of your operation

[quaint ivy]

it's this. A subgroup should be closed under the operation since it is itself a group.

[wild sapphire]

ah okay yeah. part of the definition of the group's operation

[small bison]

the kernel of a group homomorphism is always a normal subgroup

is this what you're proving?

hmm usually you can just prove directly that kernels are normal subgroups

could you post the problem statement?

[mild laurel]

It's not too hard to check that the preimage of a subgroup is a subgroup

they only prove that its normal

maybe they proved that preimages of subgroups are subgroups already idk

don't think so

This starts by assuming that H is a subgroup already so

No?

That's only true if H' is the identity subgroup

[cloud walrus]

mirzathecutiepie

[chilly ocean]

diagram chasing

[next obsidian]

wut

is going on

haha

I am not able to figure out what the proposed statement is

[cloud walrus]

mirzathecutiepie

mirzathecutiepie

[past temple]

I'm having trouble with some galois theory

if we have a field extension K/F

and an intermediate field L in between them

suppose K/L, L/F are galois

is K/F galois?

something to do with the fundamental thm of galois theory, which I'm still struggling to understand

[latent anvil]

I don't think so

I might be wrong sorry, let me think about it

My high level concern is that this is sort of like "if H <= N <= G are subgroups and H <= N is normal and N <= G is normal then H <= G is normal"

And this is false

But it's not exactly that since to apply the galois correspondence we would need to know K/F is galois to start with

[past temple]

right

i can't think of any counter examples though

[latent anvil]

hmm

So an extension being galois is equivalent to it being normal and separable (and for an intro course usually also finite)

Have you seen this?

[small bison]

normal towers aren't necessarily normal

like the standard example is with T^4 - 2

over Q

[latent anvil]

nice

ahh I see. The galois group of the splitting field F = Q(2^1/4,i) is order 8 and a subgroup of S4, so it's D8. D8 = <r, f | r^4, f^2, rfrf> has subgroups H = <f> and N = <r^2, f> with H <= N normal and N <= G normal. By the galois correspondence we have intermediate fields K/L/Q with Gal(F/K) = H, Gal(F/L) = N and K/L galois, L/Q galois, but since H is not normal in G we have K/Q not normal

[small bison]

it should be easier than that right? you have a tower $\mathbb{Q} \subseteq \mathbb{Q}(\sqrt{2}) \subseteq \mathbb{Q}(\sqrt[4]{2})$ and the intermediate extensions are normal since they're quadratic extensions

[cloud walrus]

young_smasher

[latent anvil]

oh right, that's much more explicit

nice!

[small bison]

but $\mathbb{Q}(\sqrt[4]{2})$ isn't a normal extension cause e.g. $T^4 - 2$ has a root but doesn't split

[latent anvil]

I'm much better at group theory than field theory 😛

[cloud walrus]

young_smasher

[latent anvil]

I was trying to think about how we could find an intermediate extension between Q(i) and the splitting field which isn't galois over Q and couldn't come up with one

But that's because there isn't one!

Didn't consider that we have another degree two extension Q(sqrt(2))/Q

@past temple does this discussion make sense?

Q(2^1/4)/Q(2^1/2)/Q is a counterexample

each successive extension is degree two, so galois, but Q(2^1/4)/Q isn't a galois extension

[past temple]

ahh sorry for late response

yeah i tried that example as well and it worked, ty!

im confused on this problem now

it wasnt hard showing that R/F is galois, and that <A> is a subgroup of Gal(R(x)/R) of order 3

but im confused on how to find a u that generates the fixed field of A

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