groups-rings-fields

[mighty kiln]

You just wrote it

x → y, y → x, z → w, w → z

[ornate tiger]

why it's in S_4 it also could be in S_5 right ?

like (xy)(zw)(g)

[mighty kiln]

Usually we use these notation in the context of one specific Sn

Otherwise we'd specify which Sn it belongs to

[ornate tiger]

i didn't understand what u said now

[ornate tiger]

is this true ?

[mighty kiln]

It's the same notation, but it's not the same element

[ornate tiger]

one in S_4 and the other in S_5 so it's possible to have the same look

or form

[mighty kiln]

So we need to specify whether we're talking about an element in S4 or S5

[ornate tiger]

what if we are talking in S_5 there is an element such that (xy)(zw)

[mighty kiln]

Yes

[mighty kiln]

It is this one

[ornate tiger]

ok but 5! or 4! are a lot do i need to write each one

[mighty kiln]

Is there an exercise asking you to write all of them

[ornate tiger]

Find all possible orders of elements in S_7 and A_7 ? . like here S_7 omg what they are talking about

do i need to write all of them

[mighty kiln]

Oh

All possible orders is considerably shorter

You just need to look at the different possible forms yea

[ornate tiger]

that's alot

[mighty kiln]

It's not a lot

Well at least not 7!

[ornate tiger]

7!/2 ?

oh like we are looking at the forms

?

like (abcdefg) is 7

(abcdef)(g) is 6

like this ?

[mighty kiln]

Yea

[ornate tiger]

oh ok

thanks

[ornate tiger]

Can someone give me a simple example so i can understand 100%

[crystal vale]

Since (123) = (13)(12), then they want to say that if we write (123) in another transposition then it contains even number of transposition

I think it is not a good example

[ornate tiger]

what could be the other transposition

[crystal vale]

(23)(13) this work?

[ornate tiger]

no

[crystal vale]

Why?

Actually I am using right to left notation

If you want left to right then you can use (123) = (12)(13) and (123) = (12)(23)

[ornate tiger]

Are you sure?

[crystal vale]

Tell me what will be (12)(23) ?

[ornate tiger]

What

[crystal vale]

After computation of (12)(23) what will be the answer you get

[ornate tiger]

(123)

[crystal vale]

Yes

So there are two different products of transposition of (123) and both have the same number of transposition

Important is now you can define now length of permutation by using the number of transpositions because it is unique

Maybe this is non-general terminology

[ornate tiger]

ok thx

[long obsidian]

The rational functions Q(x,y) in two variables is a field and so are the rational functions in three variables Q(x,y,z)

I'm pretty sure there is a field embedding G:Q(x,y)->Q(x,y,z) given by G(1)=1
G(x)=x
G(y)=y

Why is it enough to specify the image of 1,x,y ? It's just because these elements 1,x,y generate Q(x,y) as a ring right? Also is there some concrete way to say how much bigger Q(x,y,z) is as a field compared to Q(x,y)? It has another variable but it's an indeterminate and I feel like this has something to do with transcendental field extensions which I don't know anything about

[rocky cloak]

It is in fact enough to specify where x and y is mapped, since 1 is always mapped to 1. And yes this is because x and y generate Q(x, y) as a field.

There are at least two ways to measure the size of an extension. The degree of Q(x, y, z)/Q(x, y) is infinite, and the trancendence degree is 1.

I guess the trancendence degree would be the most useful measure here.

[long obsidian]

Thank you! Yeah transcendental degree is probably the most useful.

[tribal moss]

But (13)(23)(12)(13) is also (123).

[crystal vale]

Yes still it has an even number of transpositions

Oh got it my mistake

I mean partiality unique

[long obsidian]

If you have a ring hom f:R-> S and R is generated by R generated by a1,...,an as a ring and f is injective on a1,...,an does it follow that f is injective?

[coral spindle]

Consider any map Z[x] -> Z. Here x generates Z[x].

This doesn't even work for groups. Take any n>0 and consider the map Z → Z/nZ

[long obsidian]

So for instance f:Z[x]->Z given by f(x)=5 is a ring hom and x and 1 generate Z[x] a ring but for instance f(x^2-5x)=25-25=0. So clearly f isn't injective?

[coral spindle]

Yes

Or more simply, f(5) = f(x) = 5.

[long obsidian]

What about if I have something like G:R[x,y]-> R[x,y,z]/<yz=1> given by G(x)=x and G(y)= y, is it clear that this ring hom is injective since the ideal is generated by elements outside the image of G?

[coral spindle]

What is "the ideal"?

Oh, you mean the ideal <yz-1>, sure.

No I don't think this is obvious. For example k(x, y)/<y> is the trivial ring, so looking at generators is not sufficient.

[long obsidian]

But is the reasoning basically something like the inclusion I:R[x,y]->R[x,y,z] is injective then the kernel of the canonical projection pi:R[x,y,z]->R[x,y,z]/I is exactly I so I just need to check that the image of the inclusion I does not intersect J? Does this sound reasonable for this case?

[coral spindle]

Yes

[south patrol]

In this case it is sort of obviouos as it is the inclusion R[x,y]->R[x,y][y^-1] and the kernel is stuff which is killed by a power of y (i.e. 0)

Oftentimes with these sorts of things you can show injectivity by finding a left inverse, though that is not possible here (as y has no inverse in R[x,y])

[ornate tiger]

is (12)(1253) == (253) ?

(143)(23)(24) == (14)(23) ?

(1423)(34)(56)(1324) == (12)(56) ?

[long obsidian]

I see okay. Maybe this is a bit of a different question but what about checking the map f:k[x]->k[x] given by f(x) =x^2 is this obviously injective? I know k[x] is a k-algebra and there is a basis of the algebra/module given by the monomials. It seems like it's enough to argue this is injective just by observing that it maps a basis element to a basis element and is injective. Does this module basis argument seem convincing?

[coral spindle]

Yes, you can argue by seeing k[x] as an infinite dimensional vector space, sure

[ornate tiger]

[slim kayak]

With algebraic structures with a zero element you can also show that the kernel can only contain 0, which happens to be easier most times

[tribal moss]

You can check all of these yourself by working out the permutation on each side in two-line form.

[ornate tiger]

i did and got these answers am asking if what i got is wrong or right

like i had the left side

then reached the right side

[long obsidian]

An injective on the basis argument works even over something like Z[x] with Z[x]->Z[x] with f(x)=x^2 even when the ring Z is only a module?

[slim kayak]

Uh, if it fixes Z sure

The ker = 0 arguments works in any structure with addition and in groups tho, there you just observe that since any non-constant polynomial gets mapped to a non-constant polynomial, the kernel can only contain constants functions

but the map is clearly just the canonical inclusion restricted to the constant functions

[long obsidian]

I see okay I think somehow I'm using that this is an integral domain thank you.

[coral spindle]

More specifically, you will be using the fact that degrees of polynomials behave well.

[errant wedge]

Ah alright

Also lmao funny seeing u here

Done the assgn?

Wait no u took it with yuru already

[ornate tiger]

if we had a group G and a subgroup H do we get from multiplying a coset from the left or right a subgroup of group G because as i can see we don't get back H again we get sometimes something new

[coral spindle]

A (left) coset in G of H is a set of the form gH for some g in G.
There is only one coset of H which is a subgroup, which is H.

When we multiply a (left) coset (on the left) by an element of G, we get another coset, which may or may not be H.

[ornate tiger]

and it's a subgroup of G

[coral spindle]

H is a subgroup of G.

Like I said:

There is only one coset of H which is a subgroup, which is H.

[ornate tiger]

so what's the point of finding this another coset

[tribal moss]

The other cosets don't contain e, for one thing.

[coral spindle]

If you are just now beginning to learn about cosets, rest assured that they turn out to be useful.

[ornate tiger]

ok i will see how they are going to help

[tardy hedge]

The number of different ways a group G can act on a set S is given by the number different homomorphisms from G to Sym(S) right?

Sym(S) is permutation group of elements of S

[rocky cloak]

Yes, an action of G on S is exactly equivalent to a homomorphism G -> Sym(S)

[tardy hedge]

Does ring actions have an “equivalent” definition like that? Im just beginning to study modules

[rocky cloak]

Yes, if M is an abelian group and R is a ring, then a (left) module structure on M is exactly a ring homomorphism R -> End(M)

[tardy hedge]

Oh ok, I will probably see this when i study further?

Ive seen End(M) but dk what it means yet

[rocky cloak]

Yes, End(M) is just the ring of endomorphisms of M.

So group homomorphisms M -> M with point-wise addition and composition as multiplication

[long obsidian]

Can you say a rational function such as f(x,y)/g(x,y) is not constant just by observing that deg(f) is not equal to deg(g)?

[rotund aurora]

no

well maybe yes idk

ah, no

[coral spindle]

Over an infinite field, I think this works

[rotund aurora]

counter-example: ||x^p-x in Z/pZ||

[rotund aurora]

yes, you can reduce this to the case that P(x)=0 for some polynomial of positive degree

[rocky cloak]

I guess "constant" should mean equal to an element of k, not induces a constant function k^n -> k

[long obsidian]

What id like to say is if I have coprime polynomials f,g in Z[x,y] so that f/g is a non constant in Z(x,y) then the rational functions (f/g)^2 is also non constant element of Z(x,y).

Doesn't something like this follow from the coprimality? I think in the case of elements of Z if a and b are coprime then so is a^k and b^l.

I think this coprime argument still works for elements in Z[x,y] . Then in that case if f/g is non constant with f and g non constant and coprime then (f/g)^2 is also a non constant element.

I was thinking I could do this with just arguing by the degree of f and g but I guess not

[rocky cloak]

Yeah if f and g have different degrees, then f^2 and g^2 does aswell.

But f/g can be non-constant even if f and g have the same degree

Your coprime argument should work though

[long obsidian]

Thank you!

[gusty swallow]

Hi i'm watching a series on abstract algebra to learn it and he's doing an example on the dihedral group, he wrote this on the board:
\begin{align*}
N &= { e,r^2,s,sr^2 } = \overline{e} \
rN &= { r,r^3,sr,sr^3 } = \overline{r}
\end{align*}
where $N = \langle s,r^2\rangle \trianglelefteq D_4$. Is it true to say that $rsr^2 = sr^3$? I thought $D_4$ was noncommutative. If i use the order-changing rules i get $rsr^2 = sr^3r^2 = sr^5 = sr$. I'm so confused lol

[cloud walrus]

BigBoyConst

[rocky cloak]

rsr^2 does indeed equal sr, but what's written on the board doesn't contradict that

[gusty swallow]

how come?

[rocky cloak]

Sets are not ordered, two sets are equal if they have the same elements

[gusty swallow]

no but when he's doing the left coset, one of the elements will be $rsr^2$ (by mutliplying $r$ with the last element of $N$), but he replaced that with $sr^3$ which i'm unsure if it's correct or not

[cloud walrus]

BigBoyConst

[gusty swallow]

i dont really get what set equality has to do with this

[coral spindle]

It was rs = sr^3, not rsr^2.

[rocky cloak]

re = r
r r^2 = r^3
r sr^2 = sr
r s = sr^3

[gusty swallow]

OH HE CHANGED THE ORDER

[rocky cloak]

So rN is the set described

[coral spindle]

Yes, like jagr said, order does not matter in sets.

[rocky cloak]

They probably just wrote it up without paying much attention to the order

[gusty swallow]

ye ik but i assumed he would keep it in the same order to show the correspondence

[rocky cloak]

It would probably have been more pedagogical to do so yeah

[gusty swallow]

well either way thanks for answering my question :D

[chilly ocean]

how

[south patrol]

Basically he played it for 19889 days straight

[gusty swallow]

:D

i slightly like the game

[rocky cloak]

I thought Celeste was only released 2,333 days ago. Puzzling 🤔

[gusty swallow]

a true conundrum indeed

[rocky cloak]

Alright, I think I cracked it. , is not a thousands-seperater but a decimal point, and really you only played for 28640 minutes

Got'em!

[gusty swallow]

AH

ya got me

[errant wedge]

So ik for $N\geq 2$, if $n_1\cdots n_k$ is a factorization such that the $n_i$s are coprime, then $\Z_N\cong \Z_{n_1}\times\cdot\times\Z_{n_k}$. Is it true that for $\Z_{m_1}\times\cdot\times \Z_{m_l}$ where there exist $i, j$ such that $i\neq j$ and $\gcd(m_i, m_j)>1$, we have $\Z_{m_1}\times\cdot\times \Z_{m_l}\not\cong Z_M$ where $M=m_1\cdots m_l$?

[cloud walrus]

Sara

[coral spindle]

Yes

[errant wedge]

Ty

How would I approach proving this? Just a place to start pls

[coral spindle]

Start with some very small examples. For instance, why is Z_2 x Z_2 not isomorphic to Z_4. Hint for this small case: look for elements of order 4.

[errant wedge]

ty

[next mango]

What is an unresolvable polynominal?

[tribal moss]

Context?

[dull ginkgo]

Returning to some silly group stuff

[slate fulcrum]

did it with ma kinnon, yuru for 348

[limpid ferry]

I am attempting to prove that this set is a field using subfield test, but struggling on how to find the inverse an element.

[dull ginkgo]

What are the conjugates of each element?

[limpid ferry]

a -b sqrt(2) - c sqrt(3) -d sqrt(6)

[dull ginkgo]

In particular, let’s say we have x = a + b sqrt(2) + c sqrt(3) + d sqrt(6)

Then it’s also

(a + b sqrt(2)) + (c + d sqrt(2)) sqrt(3) right

[dull ginkgo]

Let x = (a + b sqrt(2)) + (c + d sqrt(2)) sqrt(3), y = (a + b sqrt(2))- (c + d sqrt(2)) sqrt(3)

Then what is xy @limpid ferry

[limpid ferry]

it is x^2 -y^2

[cobalt heath]

(Alternatively, you can use that for a field F, F[a] is a field if a is a root of polynomial in F.)

[dull ginkgo]

For absta ||the idea is to use this product to go down to Q(sqrt(2)), and then further use that conjugate to finally get down to Q to have an inverse||

[dull ginkgo]

What

[limpid ferry]

my bad, wait 30s

(a + b sqrt(2)) ^2 - (c + d sqrt(2)) sqrt(3) )^2

[dull ginkgo]

(a + b sqrt(2)) ^2 - 3(c + d sqrt(2))^2

But close enough

So that’s in Q(sqrt(2)) right

[limpid ferry]

I see, thanks

[dull ginkgo]

So what you can do, is then find that product’s conjugate, call it z

Then xyz = r for some r in Q

Then xyz/r = 1 so yz/r is the inverse in your field :3

If you expand yz/r out then you have explicitly found your inverse in terms of the a b c d

Does that help?

[limpid ferry]

Yes, that helps me a lot

[dull ginkgo]

Okay you’re welcome lol

[cobalt heath]

Tbh this kind of computation is quite messy

[dull ginkgo]

But it still works and it seems like an introductory field exercise when a lot of it hasn’t been explored yet

[cobalt heath]

I guess you can use determinant instead later on.

I see.

[dull ginkgo]

So at first it’s computation heavy

Until the theory is developed

[limpid ferry]

Sure thing

at least to me

[dull ginkgo]

Like this will be gotten to at some point but after like polynomial rings and quotients and shit are explored

[cobalt heath]

You don't really need quotients.

[dull ginkgo]

I mean conjugates are helpful yeah

[cobalt heath]

What it uses is just that every element of F[a] is algebraic.

[dull ginkgo]

The inverses would be the product of the remaining conjugates over the minimal poly’s constant term in the base field

[cobalt heath]

Yeah, but you can instead do like

a^2 + pa + q = 0, then
a(a + p) = - q

We take it irreducible so q /= 0, then a^(-1) is some multiple of a + p.

[dull ginkgo]

True

[cobalt heath]

When a is a root of quadratic, basically any element of F[a] is root of quadratic - then you can use this to prove that F[a] is a field.

[dull ginkgo]

But yeah I think the idea of using basic quadratic conjugates is enough for the basic exercise

To go down the chain of fields

[cobalt heath]

Hmmm

[dull ginkgo]

Like this was similar to one of the first exercises for fields in Jacobson

Without the later theory developed

So you shouldn’t a priori have knowledge of all the extra shit

It would be way less obvious with, like cbrt(2) per say

Or something like that

[cobalt heath]

Hmm, I see.

[errant wedge]

Ik it's true that for rings R and S, if R has a unit r and S has a unit s, then (r, s) is a unit of R x S with inverse (r^{-1}, s^{-1}). Are all units of R x S characterized like this? I wanna say yes but would like a sanity check 🙏

Acc wait no it's gotta be true

Sorry, another sanity check, $\Z_4[i]\cong \Z_4\times\Z_4$ right

[cloud walrus]

Sara

[errant wedge]

Wait I'm taking $\phi\colon\Z_4[i]\to\Z_4\times\Z_4$, $\phi(a+bi)=(a, b)$. But $\phi(1+i)\phi(1+i)=(1, 1)(1, 1)=(1, 1)$, while $\phi((1+i)(1+i))=\phi(1+2i-1)=\phi(0+2i)=(0, 2)$. Does this not work as an isomorphism, am I misunderstanding smth, or are these rings not isomorphic?

[cloud walrus]

Sara

[lapis latch]

Yeah don’t think thats gonna work, if you had a ring isomorphism then 1 would go to (1,1) and so 3 goes to (3,3) and then whatever i goes to say (a,b) would need to square to (3,3) but no a,b in Z_4 satisfy that

[errant wedge]

mhm

Yeah idt its true ur right

Thanks

Ive already gotten this thru bash but im tryna figure out if there's a clean way that doesnt involve making a table

[lapis latch]

Well drawing inspiration from the complex numbers I’d look at when a^2 + b^2 is a unit and when it’s a zero divisor, then you can do stuff with the ‘conjugate’

[errant wedge]

ty

[errant wedge]

(3+i) is a unit with inverse (2+i) but 3^2+1^2=10=2 mod 4 which is a 0 divisor

[lapis latch]

(3+i)(2+i)=(6-1) +(3+2)i=1+i no?

[errant wedge]

wait

oh im dumb lol 1 is still the identity

[lapis latch]

2(3-i)(3+i)=0 so 6-2i is a zero divisor

For that

[errant wedge]

ty

[errant wedge]

I see the pattern between a^2+b^2 being a zero divisor but dunno how to figure out the explicit connection/outline for a proof

[lapis latch]

Like for if a^2+b^2 is a zero divisor then a+bi is as well?

[errant wedge]

yeah

[lapis latch]

(a-ib)(a+ib)=a^2+b^2 so if it is a zero divisor in Z_4(say c annihilates it) then c(a-ib) annihilates a+ib, now you do have to handle the case that c(a-ib)=0 but it implies c(a+ib)=0 anyway

[errant wedge]

ty

[errant wedge]

wait sorry this bit i got

the other direction im having trouble with

a+bi 0 divisor implies a^2+b^2 is a 0 divisor

Since I need to find all 0 divisors

[delicate bloom]

multiplication is associative, I'd use that to see it

[lapis latch]

Well if you show that a^2+b^2 unit implies a+ib unit then that’s sufficient, since in a finite ring it’s one or the other

[untold basalt]

Let G be a group of order 357. Prove that the centre of G contains the Sylow 17-subgroup.

How can I prove this? I know that for a p-group H with a normal subgroup K one would have Z(H) \cap K \neq {1}, but G is not a p-group.

[rocky cloak]

Did you not get anywhere with these hints? @untold basalt

[errant wedge]

by one or the other, u mean unit or 0 divisor?

[lapis latch]

Yeah

[untold basalt]

sorry I missed them

[rocky cloak]

Np

[errant wedge]

oh 💀 im dumb its just $(a+bi)(c+di)=0$ with both nonzero implies $(a-bi)((a+bi)(c+di))=(a-bi)0$ implies $((a-bi)(a+bi))(c+di)=(a^2+b^2)(c+di)=0$ where the former has to be nonzero and thus a 0 divisor

[cloud walrus]

Sara

[untold basalt]

I think I got it. Proving that the 17-sylow H is normal is easy with sylow's theorems; this means that the action by conjugation of G on H is well defined. We now wish to show that this action \phi: G \to Aut(H) is trivial. In order to do so, observe that |Aut(C_17)|=16 and, if \phi is to be a homomorphism, then |g| must be a multiple of |phi(g)|, which leaves only one option for \phi, namely the trivial map g \to id for all g \in G.

Next I have to prove that G is solvable: form sylow's theorems we get that there are one 17 sylow H and one 7 sylow K, both normal, so we may consider the subgroup series {1}, HK, G. HK is abelian and |G:HK|=3, therefore the quotients are abelian.

Is this correct?

also HK is normal in G because both group are normal.

Well ok this is always true for a subgroup that is a direct product.

[errant wedge]

just the units/zero-divisors of F, no?

[rocky cloak]

Not quite

(there are more units)

[errant wedge]

finite series would amount to polynomials and ik the units of F[x] are the units of F, so ig I should be considering the infinite series?

[rocky cloak]

Just because something is not a unit in F[x] doesn't necessarily mean it's not a unit in F[[x]]. For example 2 is not a unit in Z, but is a unit in Q.

But yes, in a pair of inverses at least one would have to be infinite

[errant wedge]

this is what i meant sorry

[rocky cloak]

One example you might have learned in high school: the inverse of ||1 - x|| is ||1 + x + x^2 + ...||

[junior badge]

What's the motivation for studying valuation rings?

[cobalt heath]

O do we learn this in highschool?

[rocky cloak]

well I certainly learned about geometric series in the 11th grade, dont know about you

[cobalt heath]

Ah, right. Forgot about geometric series

[bitter locust]

can the classification of finitely generated abelian groups be re-imagined as looking at prime ideals in some ring?

[cobalt heath]

You mean classification of module over PID?

[bitter locust]

or that

maybe my question would make more sense if i asked "is there some ring out there in which its elements behave like abelian groups"

or is there some kind of equivalence of categories, the category of abielian groups and some single-object enriched category that represents some ring

[mighty kiln]

The category of Abelian groups is not equivalent to any single-object category

The "smallest" category any category is equivalent to is its skeleton

Where the objects are isomorphism classes

[untold basalt]

Prove that there exists a non abelian group of order 125.

[bitter locust]

i see, but is there some other notion of equivalence, or some other kind of mapping which would make sense of "mapping prime ideals in some ring to simple abelian groups"

[mighty kiln]

Well Abelian simple groups are just Z/p so they correspond to prime ideals of Z ig

Except (0)

Oh hm then simple modules over a cring are probably maximal ideals

[dull ginkgo]

Maximal ideals have submodules, namely it’s constituent ideals

If we have a group of order p^n, p odd prime, then can we find an isomorphic group that consists of direct product or semidirect product compositions of Z_p^k for k up to n

I think this can be done with finding cyclic subgroups and using Sylow theorems

[chilly radish]

They meant taking a quotient

[dull ginkgo]

Oh then yeah that sounds about right

[chilly radish]

You can do this using Cauchy's theorem

[dull ginkgo]

Cauchy’s theorem says there is an element of order p which seems like the base idea

Like if we have a cyclic subgroup K of order p^k, then we can describe G such that |G| = p^n as
G = K * H or K * H where * is either direct or semidirect product, with the normal factor being the subgroup of most p valuation

And then H would be some group of order p^(n - k)

Seems to hold for p^4 according to this paper

The normal factor seems to be the subgroup of larger p valuation

[chilly radish]

You can use induction to prove a converse of Lagrange

[chilly radish]

• the fact that a p-group has a nontrivial center

[chilly radish]

||This gives you the ability to quotient by an element of order p and use induction||

[rocky cloak]

Also for noncommutative rings maximal (left) ideals give you simple modules. But then two ideals m' and m gives the same module iff m' = mr for some ring element r.

[dull ginkgo]

Isn’t the notion of noncommutative ring modules used to generalize commutative ideas of shit like prime ideals and radicals

Like I forget which radical in particular is generalized as elements annihilating simple modules

[rocky cloak]

The Jacobson radical can be defined as

• intersection of annihilators of simple modules
• intersection of maximal left ideals
• intersection of maximal right ideals
• x such that 1-rx is a unit for all r

All of these work and are equivalent both in the commutative and non-commutative case

[dull ginkgo]

Oh yeah the last is a consequence of left units being elements not lying in any left ideal and if x being in an ideal then 1 + x not lying in an ideal right

[rocky cloak]

You can also define it as the sum of all superfluousleft ideals (ideals N such that N+I = R implies I=R)

[dull ginkgo]

Interesting

Also left units are elements that don’t lie in any proper left ideal right?

I assume you can use left-principal ideals to get to the same point

I am scared of left units that aren’t right units because of the fact that if there’s two nonequal left units then there’s infinitely many

[rocky cloak]

Best to stick to artinian rings

[dull ginkgo]

shartinian rings

[wispy urchin]

artinian well

[dull ginkgo]

Are left artinian rings also noetherian or is that a commutative ring thing

[rocky cloak]

Left artinian implies left Noetherian yes

[dull ginkgo]

Actually I see no issue with using ideal product with M and powers of the Jacobson radical

[rocky cloak]

Using for what?

[dull ginkgo]

Make a module “composition series “

Unsure how much the group theory stuff generalizes

I am mainly just desiring J(R) to be nilpotent lmao

[rocky cloak]

Yeah, so if R is artinian, then M/J(R)M is semisimple. So you can use this to make a composition series

[dull ginkgo]

I am not very well versed in this stuff because it’s not covered by Jacobson in his basic (but still absolutely absurdly terse) ring theory section

A lot of it is me fucking around and trying to connect names for terms with ideas that I’ve used before in exercises and shit

[dull ginkgo]

Is M/J(R) itself semisimple

[rocky cloak]

R/J(R) yes

[dull ginkgo]

Does it need to be a direct sum of finitely many irreducibles or can it be a direct sum over infinite?

[chilly radish]

Semisimple iff artinian with 0 jacobson radical

[dull ginkgo]

I am going off into unexplored territory here for me so sorry if these seem like stupid questions

[rocky cloak]

Well, 1 would need to be in the sum

So contained in a finite sum, but then that's it

[dull ginkgo]

Even for noncommutative?

[rocky cloak]

Yes

[chilly radish]

Especially for Noncommutative

Jacobson radical is boring in the commutative case

[rocky cloak]

I wouldn't say that, but semisimple commutative rings certainly are

[chilly radish]

Sure

[chilly radish]

The theory is much less rich, imo

[dull ginkgo]

I also ask

Does Chinese remainder theorem generalize to noncommutative case with two-sided ideals

just using ideal product instead of intersection

Wait no it doesn’t

[dull ginkgo]

Okay I see because 1 cannot lie in any of the irreducible submodules (ideals) due to unitality

[rocky cloak]

It works just the same yes

[dull ginkgo]

Units being non-ideal (not in any proper ideal) elements is such a useful idea to make so much shit almost immediate and it’s intuitive

Insofar that I often consider that definition first before working with the inverses

Just helps so much when working with ideals

[dull ginkgo]

I haven’t done much with modules yet as I haven’t gotten to them in my books,

But are ideals essentially modules that embed into the ring itself as a module

Left/right suppressed

[coral spindle]

Sure but more simply, they are submodules of the ring.

[dull ginkgo]

Got it

[south patrol]

[with the ring viewed as a left/right module over itself in the obviouos way]

two sided is slightly different but yeah

[dull ginkgo]

Now can you characterize maximal ideals to be ideals such that the quotient module R/M is simple?

[coral spindle]

Yes

[dull ginkgo]

I see

Now I ask, are there simple modules that R can embed into

I assume not at first glance

[coral spindle]

If R is not simple, obviously not, because R will be a submodule.

[south patrol]

correspnodence therem

[dull ginkgo]

Oh that’s a theorem?

[south patrol]

but like note that these are all fields anyway

[dull ginkgo]

Is there more to it or is it just that

[south patrol]

It's a more general thing

[dull ginkgo]

Oh shit you mean Fourth Iso for modules right

[coral spindle]

It is called the correspondence theorem more usually

[south patrol]

If N <= M is a submodule, then the obvious map
{Submodules of M containing N}->{Submodules of M/N}
is a bijection

(and is order-preserving as is clear)

[dull ginkgo]

I just call it quotient lattice iso in my notes lmao

[south patrol]

sure

idk what 4th iso is

lol

but yeah corresponodence theorem is much more standard

at least in comm alg

[dull ginkgo]

Thanks that helps for me actually being able to communicate with others lmao

[south patrol]

what is 4th iso

[south patrol]

oh seems it is this forsmoeauthros

[dull ginkgo]

I am on mobile so I don’t feel like typing it out

[south patrol]

yeah sure

[dull ginkgo]

Now if we have a left R module and a right R module can we define a tensor product

I assume the tensor product is a group though

[coral spindle]

Are you working only in commutative rings for now?

[dull ginkgo]

What do you mean

[coral spindle]

I mean are you only working with commutative rings or are you using noncommutative ones

[dull ginkgo]

Noncommutative just for this moment basically

[coral spindle]

Idk how to make that question clearer

OK

Indeed you need sidendness for your modules for the tensor product to work

[dull ginkgo]

I just think that you can define a tensor product of a left and right module but I think that’d be an abelian group

Idk if that abelian group itself would be a module

[coral spindle]

In general you can only guarantee left or right Z-module structures, so yes indeed you would get a Z-module (left or right, you choose)

But when you have bimodule structure you can get a module over a different ring

[dull ginkgo]

I mean a module over the original ring

Yeah

[coral spindle]

So if M is an (R, S)-bimodule and N is a left S module then M (x)_S N is a left R-module

Same on the other side.

[dull ginkgo]

So can you also characterize it as an adjoint but instead through a weirdly modified hom that also forgets the module structure of the image

[coral spindle]

Weirdly modified? Dunno what you mean

It's just the normal hom, you just need to pay attention to the bimodule structure.

[dull ginkgo]

I am working with noncommutative

[coral spindle]

Yes.

[dull ginkgo]

Like if we have a map bilinear map from right module R, left module L to abelian group A, then would it factor through a group map from the tensor product to A

[coral spindle]

Cba to write it out so just look at this

[dull ginkgo]

Yep that’s what I was thinking

[coral spindle]

The small R and S denote left and right R/S modules.

So nothing changes about the Hom, we just note it's a module.

[dull ginkgo]

So it’s like the classic (commutative) adjunction but instead of being an endofunctor pair it’s between different categories

[coral spindle]

Sure

[dull ginkgo]

Okie cool

[dull ginkgo]

Finally I have another example of a nonendofunctor adjunction

Because I swear to god all the intro ones are that aren’t forgetful/free

[untold basalt]

Discuss the number of Sylow 3-subgroups of a group of order 357.

[dull ginkgo]

Thank you fam

[coral spindle]

If you want another example of an adjunction that isn't working within subcategories, look at the group of units <-> group algebra adjunction

[south patrol]

the definition of a group algebra

😼

[coral spindle]

so true.....................

[dull ginkgo]

OOOOO THANK U

[dull ginkgo]

Actually here’s a question. Let R be a ring with unity, and let K be the set of all left-units of R. Then is K a monoid, and if it’s a group does that imply that all are two sided units

Because if ax = by = 1, then (ba)(xy) = (ab)(yx) = 1

If ax = 1, and K is a group, then there is a y such that xy = 1 implying it’s a two sided unit

Also that if K isn’t left cancellative then I’m pretty sure it’s infinite

It is right cancellative since if xy = xz then axy = axz implying y = z

Actual brain damage :3

[rocky cloak]

For a finite monoid, cancelability from either side means it's a unit.

[dull ginkgo]

Oh yeah true

[cloud walrus]

Heywood Jablome

Heywood Jablome

[untold basalt]

Can someone please have a look at how I wrote this? I think it's correct but I could do with some advice on the writing.

[vivid birch]

Let $S$ be a finite semigroup such that, for all $s,t \in S$ we have: $(s^\omega t^\omega)^{\omega + 1} = s^\omega t^\omega$. How do you prove that for any 3 idempotents $a,b,c$ of $S$: $(abc)^{\omega + 1} = abc$?

[cloud walrus]

Eduude

[errant wedge]

how does injective imply iso here

idk how we have that it's surjective

wait im an idiot and also cant read

that should be im f bar, not im S right

or im f tho ig theyre the same

[cedar maple]

A while back I was messing around with an algebraic structure defined on R^3 with componentwise addition, and multiplication where each component of the product is the sum of the product of the same components and the products of the opposites of the other two components (so (ab)1 = a1b1 + a2b3 + a3b2). In my document, I called it a non-unital ring, but I just realised it's not that because the multiplication doesn't form a semigroup. Is there a name for this kind of algebraic structure? A non-unital ring minus the associativity in multiplication.

[rocky cloak]

Yeah, should be the image of f. Image of S doesn't really make sense

[rocky cloak]

(non-associative) algebra

https://en.m.wikipedia.org/wiki/Non-associative_algebra

[chilly ocean]

Sorry to disturb you now, but keeping in mind of our yesterday's discussion, is this the definition of right cancellativity in this case or is it:

ax = bx => a=b for a,b in K and x in R?

I just want to make sure

[dull ginkgo]

Sorry yeah, the easiest def is that you can cancel x off the right side

[chilly ocean]

Is the last part of |b|≤|a| and |a|≤|b| necessary? Since we have already showed bⁿ=e implying it has same order as a

[ashen heron]

you need to show that n is the least positive integer that satisfies b^x=e

[chilly ocean]

Oh yeah, thanks 👍

[dull ginkgo]

Like 90% of shit pertaining to order has to do with making sure you don’t introduce a smaller n such that x^n = e

[ornate tiger]

U8 is cyclic right ?

<1> is a generator

[coral spindle]

If by U(8) you mean the group of units modulo 8, this is wrong, because <1> = {1}, since 1 is the identity of U(8).

Remember we are multiplying in the group of units, not adding. The group of units is a group under multiplication.

[ornate tiger]

oh thanks i thought we add

[coral spindle]

No, we do not add. 1 + 1 = 2 is not in U(8). 1 - 1 = 0 is not in U(8).

[ornate tiger]

so U8 is not cyclic

[coral spindle]

You have not proved that yet.

Your argument was just wrong, it doesn't mean what you were trying to prove is false.

[ornate tiger]

there is no number in U8 that from it we get back U8

[coral spindle]

OK prove it

[ornate tiger]

ok i will try lets say by contradiction it is cyclic then there is <a> such that <a> = U8 uhh how to continue

i'm thinking about something to do with a^k

can't i say let's take 1 then we rise it for the power we don't get all of U8

[jagged gate]

this just shows <1> is not U8

what about <3>, <5> or <7> ?

[jagged gate]

you can't disprove an existence statement (the group being cyclic) with just one counterexample

[coral spindle]

Here are two approaches:

1. Check every element of U(8) and show that no elements generate the whole group.
2. Find some property of cyclic groups and show that U(8) does not have this property.

Hint for 2: a cyclic group has at most one element of order 2.

[hollow tartan]

i would prefer proving for all 2, 4 p^k and 2p^k p odd prime, i wonder if the proof is illuminating

[coral spindle]

I think that may be a bit far out of Qais' reach

[ornate tiger]

[old spire]

A logician suggested that when trying to prove Feit-Thompson using a machine proof checker, people discovered that "group theory" was really "group-subgroup theory".

Anyway, it makes you wonder: Would abstract algebra be clearer if the "right" category was - instead of let's say, Lie groups and fields - homogeneous spaces and field extensions?

[ornate tiger]

In U8 the identity is 1 right for 3 it has the order 2 O(3) = 2 because 9mod8 = 1 so for 2 we can see this

but the problem ig in 5 beause it's order is also 2

so it's not cyclic

[hollow tartan]

don't worry i think you can do it Qais, i would love to understand it together

[ornate tiger]

No my brain is not braining

i will never reach a good lvl

[hollow tartan]

I also think that, but think of a group like a space, group actions are really illuminating, in fact many things come from group actions

[ornate tiger]

what we get from them

[hollow tartan]

groups

[old spire]

Interesting. Because I was thinking "Really, commutative ring theory takes place in the category of commutative rings and ideals". But you're suggesting instead that it's about rings AND modules. Which is kind of obvious in hindsight, I suppose.

But then fields are really fields-and-vector-spaces.

But now I don't get what my point is...

The origin was trying to formalise the Feit-Thompson theorem. So, maybe it would be good to look up how other hard bits of algebra got done.

[coral spindle]

I couldn't comment on this entirely, but the theory of group schemes is perhaps one example. Instead of seeing an algebraic group as being merely a variety with a group structure, you can see it as a functor k-alg → Grp, which elucidates subgroups of an algebraic group defined over some field extension of k.

[hollow tartan]

thanks i didn't know this

[untold basalt]

Prove that G contains either a normal Sylow p-soubgroup or a normal Sylow q-soubgroup.

I'm having trouble with the case where p^2 = 1 mod q and q =1 mod p

[cloud walrus]

Heywood Jablome

[rocky cloak]

You can do this with a direct counting argument:

If there are p^2 sylow q-subgroups, then ||you have p^2(q-1) = p^2q - p^2 elements of order q||

Then ||you have exactly p^2 elements not of order q, and that's enough for no more than one sylow p-subgroup||

[ornate tiger]

the answer could be a^k such that k∈ 2N or a^j such that j∈ 3N

is what i said correct ?

[coral spindle]

No, the order cannot be (for example) k = 10000000 in 2N.

[ornate tiger]

why i think a ^10000000 will go back to the identity

it should be a^k such that k∈ 6N ?

[coral spindle]

But the order cannot be 10000000

Remember

[ornate tiger]

why

[coral spindle]

The definition of the order is not that a^n = e

it is the smallest n such that a^n = e.

[ornate tiger]

yeah it is

[coral spindle]

So it should now be immediately obvious that if a^24 = e then the order of a cannot be 1000000000.

Or whatever enormous number is in 2N.

[ornate tiger]

so it could be 24 is the order or something smaller

[coral spindle]

Prove it.

[ornate tiger]

i can't

but for example the order could be a^2

2

the order of a is 2

but how to prove this i can't think about anything

[cloud walrus]

Heywood Jablome

[untold basalt]

in the last equation LHS=0 mod |G:H| while RHS =1 mod |G:H|...

Am I missing something?

[coral spindle]

Not necessarily? That removal of a factor of |C_G(x)| might really mess things up, right

Ah no I see

[untold basalt]

the centralizers are divisors of |H|

[coral spindle]

Yes I should've read your setup

Ah yes you only have this equation for x in H, so you don't get the first sum

If this were true for every x in G you would indeed have a contradiction

[untold basalt]

thanks, so that's what I was missing.

[untold basalt]

well I can't really go anywhere from here. Any hints?

From this equation I'm getting |G|=1+n+something, where n is the number of orbits that intersect H

[dim widget]

have you tried looking at the equation and assuming to contradiction that they share a prime factor p?

in particular if $p||H|$ then $p||G|$, but…

[cloud walrus]

Math_Discord_Final_Girl

[tardy hedge]

Whats the best way to think of an R-algebra?

[mighty kiln]

A homomorphism R → S

(into the center of S if S is not commutative for some reason)

[rocky cloak]

As a quiver! (R is an algebraically closed field, your algebra is finite dimensional, and you only care about stuff up to Morita equivalence of course)

[tardy hedge]

Oh yes of course

[mighty kiln]

Let R be an algebraically closed field

Ig one could expect a "ring action on a ring" to be a homomorphism R → Rng(S, S)
But a rng endomorphism of S is just multiplying all of S by the same element in S

[rocky cloak]

You may even let R be not algebraically closed if you loosen your definition of a quiver (and use a species instead)

[rocky cloak]

In case R is not a field I vote monoid in the category of R-modules

(joking, but also not joking)

[tardy hedge]

It seems like the further you go with algebra, even though you learn more abstract concepts, it actually seems to help in understanding all the concepts better

Is that true?

[mighty kiln]

Category makes everything more understandable

[tardy hedge]

I c

It seems like my first encounter with category theory will be later on with modules

Cuz im learning modules now and in dummit and foote theres a section that has something with functors

[mighty kiln]

Indeed R-mod is a common place to start doing homological algebra

[tardy hedge]

What even is homological algebra

I dont have a clue what its supposed to be about

[mighty kiln]

The study of Abelian categories ig (?)

An Abelian category is just a category where you have nice kernels, images, quotients, coproducts, etc

R-Mod is one (some) of them

[rocky cloak]

I'd go as far as saying that homological algebra is the study of categories where homology makes sense (abelian categories, exact categories, triangulated categories), and the study of their homology and exact structure (exact sequences, conflations, distinguished triangles)

[tardy hedge]

What is homology?

[mighty kiln]

In an Abelian category, if the composition of g and f in A -f→ B -g→ C is 0, then the homology at B is ker(g)/im(f)

Intuitively it measures "obstruction"

E.g. if you take the chain of Abelian groups of k-forms on a manifold, under the exterior derivative
Ω0 → Ω1 → Ω2 → …

Then the homology at Ωn measures the failure of closed n-forms to be exact, which turns out to be obstructed by "holes" in the manifold

This homology (called the "de Rham cohomology") therefore counts "holes" in the manifold

[tardy hedge]

Oh wow sounds cool

[south patrol]

I need to see what a quiver is lol

Homological algebra is the study of stable oo-categories

[untold basalt]

I still don't see the contradiction.

[cloud walrus]

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

[dim widget]

from this equation what primes are allowed to divide #G, #H, and [G:H]?

just from basic divisibility rules

[spice whale]

(all of them)

kind of

[chilly radish]

Every abelian category is just a bunch of R-mods stapled together with hot glue

[coral spindle]

Mitchell's embedding theorem is kinda sad man lmao

Like oh here's this cool generalisation of module categories.... could be realllllyyy willdddd... sike they're basically just module categories

[rocky cloak]

I feel like this underestimates how much weirder a subcategory can be than it's ambient category.

Like finite groups are in fact more interesting than just Sn, even though they embed into them

[coral spindle]

You are right, I'm being facetious

I think the embedding theorem says less that Abelian categories are tame and more that module categories can be wild

or just complicated, I suppose

I mean tame and wild in an informal way

[rocky cloak]

Module categories are often wild in a formal way as well

[coral spindle]

Yeah... I regret my choice of words

[rocky cloak]

One day every word in the English language will be made into math jargon

Then no one will be able to speak without making a bad pun

It shall be glorious

Hmm, makes me wonder. What are the abelian categories of interest that are not most naturaly thought of as subcategories of module categories?

There's sheaves, but are there any others?

I know in tau-tilting theory, wide subcategories often arises, that don't necessarily feel very much like module categories. But obviously they are naturally thought of as subcategories, as that's how they're defined.

[wooden fulcrum]

hello if I have the following equivalence relations (defined by the partition classes):
{1, 2} {3, 4} {5, 6} ...
{1, 4} {2, 3} {5, 6} ...
then there is no supremum of the original relations with respect to inclusion if I require all the resulting partition classes to be of the same cardinality right?

idk this isn't really groups rings fields but it's in my algebra II class but I can move it

[rocky cloak]

I'm unsure if I'm parsing your question correctly, but you're interested in the poset of equivalence relations were all equivalence classes have the same cardinality.

Then I agree they don't have a supremum, but there should be many minimal relations containing both.

[wooden fulcrum]

yes that's it thank you, the original question I was doing was whether the poset formed a lattice so I wanted to make sure this counterexample worked

[iron siren]

hodge structures, and commutative finite type gp schemes over a field and variations of that

[tardy hedge]

For M an R-module that is annihilated by ideal I, it says we can make M into an R/I-module by setting (r+I)m = rm .

My question is, how is this well defined? When r+I = s+I, does rm really equal sm in the R-module?

Oh nevermind all good i think

Cause if r+I = s+I then r-s is in I which means (r-s)m = 0 so rm = sm

[tardy hedge]

Viewing klein 4 group as a vector space over Z/2Z … the basis would just be {(0,1),(1,0)}?

I did a group theory question that the solution somehow involved using that vector space perspective. I forget what the question was about

[south patrol]

a fancy way to do this: you can check that providing M with the scalar multiplicatino from R is the same as giving a map R -> End_Z(M) where on the right i mean the ring of group homomorphisms from M to itself

Then this is just factoring that through the ideal I, since it lies in the kernel

[mighty kiln]

Yes

[coral spindle]

Yes that's right

One could even say that V = F_2 (+) F_2

[vivid birch]

I already asked this the other day, but I'm still having some trouble with it:

Let $S$ be a finite semigroup such that, for all $s,t \in S$ we have: $(s^\omega t^\omega)^{\omega + 1} = s^\omega t^\omega$. How do you prove that for any 3 idempotents $a,b,c$ of $S$: $(abc)^{\omega + 1} = abc$?

[cloud walrus]

Eduude

[chilly ocean]

Suppose a,b in G have odd order respectively. Then a² and b² commute iff a and b commute

I would like to verify that in general, one can prove that regardless of order of a or b, if a and b commute then a² and b² commute as well

But the other way around is true only for odd order right?

[coral spindle]

Yes

In general you cannot conclude that if a^2 and b^2 commute then a and b commute. N.b. this is not equivalent to a, b being odd order but this is sufficient.

[chilly ocean]

Alright, thanks

[south patrol]

for example note that elements of order 2 would always be in the centre [i.e. commute with everything] if this were true

which is very far from being the case

[chilly ocean]

True

[sly heron]

why doesn't the definition of a ring homomorphism include $\varphi (0)=0$?

[cloud walrus]

lazynjnja

[lusty marlin]

That follows from the first condition

[sly heron]

ahh right forgout about that

thx

[chilly ocean]

Here's my proof (ab = ba <=> a²b² = b²a²)
Is there any shorter/other way of proving this?

[coral spindle]

Yes. Let 2k-1 be the lcm of the orders of a and b. Then a = a^2k and b = b^2k commute since they are powers of commuting elements.

[chilly ocean]

Neat, thanks!

[tardy hedge]

“A has four distinct C-algebra structures” is just the same as saying there are four distinct ring homomorphisms from C into A?

[coral spindle]

Four distinct ring homomorphisms from C into Z(A), but yes I believe that's right

[tardy hedge]

How do i show the only ring homomorphisms from C to C is identity and complex conjugation?

If im not mistaken?

[wraith cargo]

do you mean ring automorphism?

[tardy hedge]

Oh i guess

[coral spindle]

Well this isn't true iff choice holds (I assumed you meant automorphisms)

[tardy hedge]

What are all the homomorphism?

[coral spindle]

I don't think there's any easy description of those.

[tardy hedge]

Idk why i thought it was simple. I thought cause C is a field then it would be easy

Since kernel can only be 0 or C

[coral spindle]

Do you know any Galois theory? You should know already that just determining the group of automorphisms of a field can be hard.

[tardy hedge]

Not yet

I was studying a bit about fields a while ago but thats bout it

[coral spindle]

Well let it just be known that it's not easy.

[chilly radish]

I wanted to say these are the same but then I remembered C embeds properly as a subring of itself

In infinitely many ways, even

[chilly radish]

I don't think there's any description, since it depends on choice

[tardy hedge]

Can someone give me an example of one

[chilly radish]

In the absence of choice it is consistent that C only has 2 automorphisms

That's impossible, because of what I just said

[tardy hedge]

What does this mean?

Ok yea i have no idea what yall are talking about

[chilly radish]

It means that you can construct a model of set theory without choice in which Aut(C) only has conjugation and identity as elements

This means that you can't describe another automorphism of C without using the axiom of choice

And using choice is inherently nonconstructive, so I literally cannot give you an example

Just say that one exists

And tell you how to show it exists

[crystal vale]

To show the identity element of the permutation matrix always has even number of transpositions.

Now if e = b_1....b_n then we need to prove that n is even.
We know that n≠1 then it is not possible and n=2, then it is trivial.
Let n>2, then they hint that there are three cases such that last rightmost entries b_(n-1)b_n maybe
(ab)(ab)
(ac)(ab) = (ab)(bc)
(bc)(ad) = (ab)(ac)
(cd)(ab) = (ab)(cd)

If the first case occurs then e=b_1...b_(n-2) then by the second principal of mathematical induction n-2 is even.

But now I am not sure about the left three cases.

So if one of them occurs then we can replace the rightmost entries for example if (cd)(ab) is b_(n-1)b_n entries then we replace it by (ab)(cd), continuously we are doing this and after some point we get b_1...(ab)(ab)...(cd) then it length reduced to n-2.

But if we don't get this condition that means a does not fix so it contradicts that b_1...b_n is identity element

Is it correct?

I think there is some mistake in my argument

[tardy hedge]

In dummit and foote it says that C x C has 4 distinct C-algebra structures

[rocky cloak]

Is A a ring? If so I would say this is correct. If A is for example a complex vector space I would think they meant something else

[rocky cloak]

Here's a few fun ones:

Pick a transcendence basis {xi} of C over Q (a maximal set of algebraically independent elements). Then C is the algebraic closure of F = Q({xi}), and any automorphism of F induces an automorphism of C. For example, any permutation of the xi.

Another:
Let X be the set of primes, let U be a non-principal ultrafilter on X, and let Fp denote the algebraic closure of the field with p elements.

Define R = Prod_p Fp and let I be the ideal generated by elements supported on sets not in U. Then R/I is a characteristic 0 algebraically closed field of cardinality continuum, hence isomorphic to C. Applying the Frobenius automorphism on each factor gives an automorphism of R which induces an automorphism of R/I = C

the first one uses axiom of choice to ensure the existence of a transcendence basis, and you might need some form of choice to extend automorphism to normal extensions

The second uses the existence of non-principal ultrafilters, but perhaps more importantly the fact that all algebraically closed field of the same characteristic and transcendence degree are isomorphic

[rocky cloak]

Is there some more context to this?

Like if CxC considered as a vector space, ring, abelian group, something else?

How are they defining C-algebra structure precisely?

Does distinct mean "up to isomorphism"?

My guess would be that CxC is a vector space, "algebra structure" means associative algebra not necessarily unital, and distinct means up to isomorphism.

Because in that case there are 4

[chilly radish]

The second is equivalent to a weaker form of choice

[chilly radish]

Idt dummit and foote does nonunital stuff

Maybe i'm mistaken tho

[coral spindle]

Why would it not be unital? If the embedding of C is correct then the image of the unit of C will be a unit

[rocky cloak]

I'm just trying to guess what definition they're using to make the answer 4

[coral spindle]

I see that

[rocky cloak]

For unital algebras there are only two 2d ones

[rocky cloak]

Are they defining "algebra structure" as ring with embedding of C into center?

[coral spindle]

I can only imagine so

But I agree with you, this seems to be saying that C x C is a C-vector space

and only then asking about algebra structures which respect that structure

[rocky cloak]

If only there was context 😢

[coral spindle]

Btw, how did you figure out there are only four 2d associative unital complex algebras?

Is there some trick to that or did you just happen to know that

I'm trying to see how one might get there. We'd have to look at a basis {1, x} with structure constants a, b with x^2 = a + bx and then somehow get this into some normal form

[rocky cloak]

I just looked it up

[rocky cloak]

Oh, no that I just considered basis {1, x} and x satisfies a degree 2 polynomial

So either repeated roots or not

[coral spindle]

Ah I see, they're isomorphic according to their roots. Nice

[rocky cloak]

Giving either CxC or C[x]/x^2

[coral spindle]

Yeah I see that

[rocky cloak]

After looking up the context, CxC is the ring. And "has four distinct C-algebra structures" means at least 4, not exactly 4.

And structures are not counted up to isomorphism.

[coral spindle]

[cobalt heath]

Hi, may I ask which channel should Haar measure stuff go?

[south patrol]

I guess it depends what you are asking about it

There are measure theoretic things, groupy things, and even functional analysis things for some proofs

[tardy hedge]

Not really sure what this means

The four distinct ones are the fp,q’s the question is talking about right?

What else is there

[tribal moss]

Which question? I tried scrolling up but no mention of any "fp,q's".

[topaz gale]

I'm trying to show that x^2+y^2+x^2y^2 is irreducible in C[x,y]. What's the standard brute force strategy to show this? I tried splitting it up into two factors to derive a contradiction, but I'm unsure which choice of factors will be satisfying. If I choose something like (p(x)y+y)(q(x)y+y), say, I'm missing some x^2.

[cobalt heath]

I am going through its construction

Maybe measure-theoretic?

[cobalt heath]

#real-complex-analysis or #advanced-analysis ? @south patrol

[south patrol]

Probably the latter but depends

but dw too much

[cobalt heath]

Ah, thanks!

[south patrol]

npp

mb

Well go mod x and mod y. You can check that it can't factro as x^2 * 1 mod y, etc, so you wind up with like

(x+Ay+xyp(x,y))(x+(1/A)y + xyq(x,y))

but this is ||x^2y^2 p(x,y)q(x,y) + [stuff of strictly lower degree]||

so ||p(x,y)q(x,y)=1 which means both are constant, so we have (x+Ay+xyB)(x+(1/A)y+(1/B)xy)||

and then expanding out it's clear

@topaz gale

[topaz gale]

Worked it out, the contradiction is clear with the constants (in C). The first equation was enough for me to solve it; thanks a ton @south patrol!

[south patrol]

np!

If you really want to not think abuot it then you can just write as like sum of a_{i,j}x^i y^j and go up the degrees lol

[topaz gale]

lol yeah, I'll remember that trick!

[tardy hedge]

t1(z) = z, t2(z) = z bar

fp,q(z) = (tp(z), tq(z))

[crystal vale]

Anyone?

[rocky cloak]

Well, you could take any of the fpq's and compose with an automorphism of C for example

[tardy hedge]

Yeah i think the missing piece here is how apparently there are many automorphisms of C i didnt know about, but i also think that is above the material that is in dummit and foote at the point im at rn

[pulsar vault]

Hi, I just got the book sphere packing and lattices by Conway and Sloane because I was interested in E8 lattices and leech lattices. I was wondering what are the prereqs to learning lattice theory and sphere packing in general? Thanks

[old spire]

Is there any reason why in a sufficiently well-behaved topological group G, the Borel subgroup B is somehow the largest non-compact subgroup, and G/B is compact?

A Borel subgroup is defined as a maximal connected+solvable subgroup. The correspondence between group notions and topological notions is interesting.

P contains maximal connected+solvable B :: G/P is compact

I don't have an intuition for why. And I don't know what general picture this might be alluding to.

I suppose this has something to do with the Levi decomposition, or Jordan decomposition. Though I can't make precise what the connections are.

It seems that

semisimple :: compact
and
solvable :: non-compact
except, of course, that semisimple groups are never compact over the complex numbers. They do though contain a Zariski dense compact subgroup.

[mighty kiln]

G/B is something you get when you keep quotienting by Abelian subgroups right

[old spire]

It's a coset space. But I guess.

Connected abelian subgroups.

Non-connected would eventually give you the trivial group.

[old spire]

I edited to add connected.

They should be (Zariski-)closed as well.

[mighty kiln]

So they're algebraic groups?

[old spire]

According to Wiki, yeah. It doesn't mention topological groups.

[spiral scarab]

Apologies if this is not the appropriate channel, can anyone recommend me a resource which exemplifies applying the Todd Coxeter algorithm for calculating the order of a group? Preferrably with coset tables.

I couldn't find anything for the life of me, thanks!

[rotund aurora]

maybe Artin's abstract algebra book?

[spiral scarab]

I'm gonna try, thanks!

This is what I need to apply it on:

[spiral scarab]

I found one in Presentation of Groups, D.L. Johnson

His explanation is kind of rushed though

[tardy hedge]

Can you make any abelian group into an R-module for any R? Might be a silly question but im having a hard time seeing the connection

I know that you can make any abelian group into a Z-module … and the same abelian group can have many different R-module structures (F[x] module example)

But i dont know if for some random group and any R, you can put an R-module structure on it

I was thinking about this because i was looking at the example with if px = 0 for a prime p for all x in group, then u can make the group into a Z/pZ - module

Answer is probably no

[south patrol]

No. there are some silly obstructions like every non-zero C-module has uncountable cardinality, so Z/2 can't be a C-module

But more seriously there are many other obstructions

Putting an R-module structure on an ab group M is the same as giving a ring homomorphism R -> End_Z(M) where the RHS is ring of endmorphisms (with multiplication given by composition)

So you can often find obstructions through that

[sly heron]

i've been doing some exercises on coefficient rings and whenever i need to use the correspondence theorem, for me to find some homomorphism whose kernel is part of whatever ideal I'm working with, it usually boils down to finding some coefficients to then find some irreducible polynomial that generates the kernel

however im really confused as how im supposed to come up with these coefficients

like

in the second image, how am I supposed to come up with these coefficients? (x^2 and 3x + 1)

they kinda just pop out of nowhere

[next obsidian]

I don’t know that there’s really a given way to do it

Mostly feel I think

The ring Z[x] isn’t a PID so you aren’t guaranteed that the ideal has a single generator

Which leads me to think that it’ll be hard to describe a process that takes two of the generators and finds their like, gcd or whatever

[sly heron]

well that's annoying

anyways thanks for the clarification

[sly rain]

What book are these exercises from?

[sly heron]

Algebra: Michael Artin 2011 Pearson Education

[sly rain]

thank u

[shell pilot]

Anyone have a good explanation of the Recursion Theorem in Abstract Algebra? I am looking to apply it to the Factorial function.

[placid heart]

Let $R$ be a ring, $S\subseteq R$ be a multiplicatively closed set, $f:R\to R_S$ be the canonical morphism (inclusion), and $J$ be an ideal of $R$. Let $f^: Ideals(R_S) \to Ideals(R)$ be defined by $f^(I)=f^{-1}(I)$.\
Prove that the following are equivalent:\
$J\in Im(f^*)$\
$J = f^{-1}(J\cdot R_S)$\
$s$ is not a zero divisor in $A/J$ for all $s\in S$

[cloud walrus]

Casiel368

[placid heart]

ii implies i is trivial, so I could start there

But i implies iii is too hard so I might prefer i implies ii first

I know that f* is injective but I have trouble using that

[chilly ocean]

Is the map $V^*\otimes W \to \text{Hom}(V,W), \phi\otimes x \mapsto (v \mapsto \phi(v)x)$ injective for all vector space V,W?

[cloud walrus]

eigenpuppet

[void cosmos]

yeah

it is an isomoprhism

[chilly ocean]

In the infinite-dimensional case?

[void cosmos]

no

i only know why this specific map doesn't work though

i dont have any counterexamples in mind

im too weak

[chilly ocean]

It should only surject onto maps of finite rank

[void cosmos]

yeah exactly

but in general ofc that doesn't prove that they are not isomorphic

u need an explicit counterexmaple for V and W

which i do not have in mind

[chilly ocean]

I’m not sure how to show injectivity without choosing a basis

[void cosmos]

yeah u can see why intuitively it doesn't work

for infiite dim

ur isomoprhism is basically just the components

like yeah u

u need a good basis ig whatevers

i just sadly do not have like an explicit counterexample surely someone else does

[old spire]

Consider G/T where G is a linear algebraic group, and T is its maximal torus -- can we say anything about its properties as an algebraic variety?

Let G be a linear algebraic group. What can we say about its properties as an algebraic variety when:

• G is connected and solvable.
• G is connected and nilpotent.
• G is connected and abelian.
• G's maximal connected+solvable subgroup is trivial.
• G's maximal connected+nilpotent subgroup is trivial.
• G's maximal connected+abelian subgroup is trivial.

Property 4 is equivalent to it being a complete variety.

I'm asking because I'm wondering if there's a nice dictionary between algebraic geometry (or topology) on the one hand and properties of (algebraic) groups on the other hand.

[rotund aurora]

@old spire it's better if you don't multipost

[cloud walrus]

quickdoom

quickdoom

[valid quiver]

does anyone know the equivalents for non-commutative rings?

(english is behind slashes)

[rocky cloak]

At least PID, UFD and integrally closed domains you can just add the word noncommutative in front.

[valid quiver]

thought as much hehe

I don't know if there are noncomm euclidian domains

[rocky cloak]

I don't think they come up very often in the noncommutative setting though

[chilly ocean]

To prove it, use the fact that H and K are normal subgroups, and H (intersection) K = {e}

||to get started, work around h1k1h2k2(h1h2k1k2)^-1 a bit||

[tribal moss]

How would unique factorization even work in a noncommutative ring? Usually uniqueness is only up to order, but if the irreducible factors don't commute, that wouldn't work.

[rocky cloak]

You can still require uniqueness up to order.

[coral spindle]

But then surely this would mean that the primes commute?

So in fact everything would commute

[rocky cloak]

No, like the fact that
ab = cd implies {a, b} = {c, d} does not imply that ab = ba

For example the composition factors of a module are unique up to ordering

But not all modules with the same composition factors are isomorphic

[coral spindle]

Ah I misinterpreted what you were saying. So you're actually saying that ab and ba may be distinct elements in general.

OK fine

[chilly ocean]

I don't understand nilpotence yet. Can someone explain it intuitively ?

[valid quiver]

Well, it's a little similiar to order of an element in a group

recall, that order of $x$ is the smallest $n$ such that $x^n=1$

[cloud walrus]

🇵🇸Mína🔆

[valid quiver]

but then, if for an element of a ring happens to exists an $n$ s.t. $x^n=0$, it's nilpotent

[cloud walrus]

🇵🇸Mína🔆

[coral spindle]

(I think that they may be referring to nilpotent groups rather than nilpotent functions)

(Perhaps we can ask for clarification)

[valid quiver]

oh, prolly that,mb

[coral spindle]

Or who knows, it could be nilpotent Lie algebras, nilpotent ideals...

[rocky cloak]

The idea in any case is x^n = 0.

But ^ and 0 might mean slightly different things

[tardy hedge]

Point set Topology seems a lot easier to digest and understand compared to module theory for me

Trying to learn both rn

[vagrant zinc]

ooh nice document, name of book?

[regal zodiac]

Compute the number of additive subgoups H of the field F_p^2 such that | H | = p

I found p+1 such subgroups, is that correct ?

[coral spindle]

Yeah that's right

If I'm not horribly mistaken(!)

[regal zodiac]

OK thanks i struggled a bit ah ah

[coral spindle]

OK well there's a pretty straightforward way to do this.

This is really just choosing a 1d subspace of (F_p)^2. You choose a single generator, and this spans a 1d subspace. Yes, I meant (F_p)^2 since F_(p^2) is a 2d v.s. over F_p, so there's no difference.

Now wlog a generator may be chosen of the form (1, x) or (0, 1). These all produce distinct 1d subspaces and there are p+1 of them.

[regal zodiac]

Yeah i did something like that, thanks

[coral spindle]

Good stuff

[clear ingot]

for the => direction u just rewrite phi(x, y)=phi(x, x)+phi(0, y-x)=phi(1, 1)^x+phi(0, 1)^(y-x)

and analogously for y

right

[south patrol]

Hm idk why you're using multiplicative notation but sure

[spice whale]

yeah if it's not there you get a semidirect product

which may not be a direct product

[lethal cipher]

Okay, so let's say we have that R is a UFD. How would we show that R[x] is also a UFD?
Furthermore, this is gonna sound stupid, but suppose the quadratic F in R[x] is reducible. Can it only be decomposed as a product of linear factors?

The answer to my 2nd question is yes. UFD's are conveniently integral domains, which also implies that if F(x)=P(x)Q(x), the degree of P or Q can't be larger than F(x).

[tardy hedge]

how does that follow?

[lethal cipher]

Because the product of leading terms for non-constant polynomials will never be 0 since neither leading term was 0 in the first place.

Had nothing to do with UFDs really

I'm still stuck on my first question though

[lethal cipher]

Suppose you have an ideal I and you know rad(I) is prime. Does this imply rad(I)=I?

I know that prime ideals are radical.

[delicate bloom]

I think (p^2) would be a counter example

[lethal cipher]

Shoot...you're right

[delicate bloom]

all good 😛

[lethal cipher]

So I'm working on 1.39 part a). And, I was able to show that (t) is an irreducible element, and somehow I think I can use that to show there is no non-trivial prime ideal Q contained in (t).
But... I'm having a hard time seeing how

[errant wedge]

I don't want the answer, but I'm a little confused how to construct $\overline f$. Wouldn't you lose information in the quotient map? Like if two things are mapped into the same equiv class, I'm not sure how you'd differentiate them for $\overline f$

[cloud walrus]

Sara

[lapis latch]

The nice thing is the quotient map doesn’t destroy anything that isn’t already destroyed by f

[errant wedge]

OH bc the kernel is a subring right, if the difference of 2 elements is in the kernel, they're in it too, the equivalence classes are the kernel and one for everything else?

like per elem i mean

wait

its just an ideal i think

but still works

[lethal cipher]

Yep exactly. And so each element in the image corresponds to a coset of the kernel.

[errant wedge]

tyty

[lethal cipher]

Wanna follow up on this briefly so it doesn't get lost

[lapis latch]

Show that in a UFD a non-zero prime ideal contains a prime(using this on Q), then that prime being in (t) leads to things

[chilly ocean]

Sure but these three given statements in the picture is equivalent to saying H and K are normal subgroups that generate G and that their intersection is ${e}$, which is sufficient to define the inner direct product and prove $\alpha$ is a homomorphism:

Let $a := h_1k_1h_2k_2(h_1h_2k_1k_2)^{-1}$

$\implies a = h_1k_1h_2k_2k_2^{-1}k_1^{-1}h_2^{-1}h_1^{-1} = h_1k_1h_2k_1^{-1}h_2^{-1}h_1^{-1}$

Now using associativity:

$a = h_1(k_1h_2k_1^{-1})h_2^{-1}h_1^{-1} = h_1k_1(h_2k_1^{-1}h_2^{-1})h_1^{-1}$

Using normality of H and K, we see that $a \in H \cap K$

$\implies h_1k_1h_2k_2(h_1h_2k_1k_2)^{-1} = e$

$\implies h_1k_1h_2k_2 = h_1h_2k_1k_2$

[cloud walrus]

iceball

[errant wedge]

or sorry, (a-b) in I doesnt imply a in I or b in I, but if (a-b) in ker(f), f(a-b)=f(a)-f(b)=0. If a=b, this is fine bc a~a. But if a/=b, we only have f(a)=f(b), not necessarily zero right?

[errant wedge]

wait im like actually blind

if f(a)=f(b) then there's no distinct information to extract 😭

q(b)=[a], fbar([a])=f(a)=f(b)

So $\overline f([a])=f(a)$ works, injective is straightforward, just have to show uniqueness

[cloud walrus]

Sara

[vocal pebble]

I'm not sure why you'd do this. The commutativity is all that is all that is needed. All the long computations finally rely on the fact that elements of h and k commute with each other. The normality comes from the fact that hk=kh. Using normality to prove elements commute (k1h2=h2k1) is going in a circle when we are already given that hk=kh for all h,k

[lethal cipher]

I do understand this is how to start the problem. But I'm having a hard time coming across the aforementioned things that result from it.

[lapis latch]

Say p is in Q then p is in (t) so there is r in R with p=rt, both p and t are prime and R is a UFD so this implies r is a unit so (p)=(t) which implies Q=(t)

[lethal cipher]

Why does r being a unit imply (p)=(t)? That's the thing I'm not seeing

Also, silly question, but how do we know p is a prime element?

[lapis latch]

p=rt and t=r^{-1}p being multiples imply (p) is in (t) and (t) is in (p) so (t)=(p). If you need justification for that write out what the ideals (t) and (p) are

[lethal cipher]

Yeah, I was able to cross that bridge if p is a prime element

[lapis latch]

We’re using this to get a prime in Q, that’s the p

[lethal cipher]

I see. I misunderstood what you were saying the first time. I'll try that out now

I see. So, if we let q in Q, then since R is a UFD, we can write this as a product of powers of primes. Since Q is a prime ideal, you can iterate through to show that one of those powers of primes is in Q.

And in the end, one of those primes are in Q

[lapis latch]

yeah

[lethal cipher]

That's very insightful. Thank you very much

[errant wedge]

Uniqueness is just $\overline f\circ q = f = g\circ q\implies (\overline f\circ q)(a)=(g\circ q)(a)$ for all $a\in R\implies \overline f([a])=g([a])$ for all $[a]\in R/\ker(f)$?

[cloud walrus]

Sara

[chilly ocean]

Of course, I'm just saying that it's equivalent to H and K being normal and their intersection is {e}. If it's given h and k commute (which is a consequence of normality because that's how inner direct product is defined usually) then yes it implies normality and you can continue with that.

I'm emphasizing on normality because such products are generalised using that condition, by just tweaking it a bit.

In inner direct product, H and K both have to be normal.
In inner semidirect product, any one of them has to be normal.

And so on.

[tardy hedge]

What is the map that shows HomR(M,M) is an R-algebra?

Is it r in R maps to rI with I is identity map?

[rocky cloak]

Yup

[tardy hedge]

Thx

[next obsidian]

Non commutative algebra

[rustic crown]

commutative algebra

[coral spindle]

Non-commutative algebra is like non-spicy curry

[mighty kiln]

Curry that may or may not have spice in it

[coral spindle]

I have maybe said the wrong statement

[summer path]

non spicy curry > spicy curry

(i don't like spicy stuff)

[next obsidian]

Hi bubu

:3c

[summer path]

actually rice without curry > non spicy curry > spicy curry

hi chuchu

[topaz heart]

im a bit confused by this proof. the way you normally proceed from the second line (or at least what ive seen before) is that if $v=(a_1, ... , a_n)$ and $M = b \cdot 1_n - (c_{ij})$ then $Mv = 0$ so $det(M) = 0$ (i.e. $b$ is an eigenvalue of $(c_{ij})$). But the way it is written here to me seems like they are deducing $det(M)\cdot a_j = 0$ for each $j$ directly from the second line and then concluding that $det(M) = 0$.

[cloud walrus]

wiita

[tardy hedge]

Why in the definition of an R-algebra, we want f(R) to be in the center?

[coral spindle]

Bc this is what it means when every element of the algebra is an R-homomorphism

So ok just to use the language of fields here

This would mean that every element's action is R-linear

If the elements of R didn't commute with everything then this would fail

[tardy hedge]

Every element is an R-homomorphism?

Im extreme beginner at this so i already got lost by thay

[coral spindle]

OK the action of* every element is a homomorphism of R-modules

So to make this precise

[cloud walrus]

Boytjie

[tardy hedge]

Yes cool i was just reviewing how R-algebra is an R-module so this seems more familiar to me so far

[cloud walrus]

Boytjie

[coral spindle]

Here I am writing r.x instead of rx, as we might write for algebras over a field, because there is no guarantee that f is injective.

But either way it is the action of R upon A

[tardy hedge]

What you wrote now seems way more promising for me to understand it, thank you. I will read now

[coral spindle]

Another way of seeing this:

• A ring is an Abelian group (Z-module) with a biadditive (Z-bilinear) operation on it.
• An R-algebra is an R-module with an R-bilinear operation on it.

(Ignoring units)

Corollary of course: a ring is a Z-algebra

[tardy hedge]

Lambda_a is an R-module homomorphism from A to A?

[coral spindle]

Yes

lambda is for left

it's the left action of A on itself

[tardy hedge]

So. In general, for A an R-module (and say A is also a ring), to make the map fa: A -> A fa(x)=ax (for some a in A) into an R-module homomorphism, we need the condition that fa(r.x) = a(r.x) = r.(ax) = r.fa(x)

If A is an R-algebra then define the action of R on A the usual way, with r.a = f(r)a, and now this condition there is satisfied

So we need that f(R) to be in centre

Ok…. Then Why do we care or want each a in A to have a module homomorphism associated with it

[coral spindle]

Because it's nice

Maybe you need to see some motivating examples for algebras. The classic one is the group algebra, whose modules are precisely group representations. Another one is an algebra of matrices. It should be clear why each element of these algebras should produce R-linear maps.

[tardy hedge]

Hmm ok yea true probably if i see more examples id get a better idea. At least i think now i understand the technicalities of it better

[coral steeple]

Is the convention for multiplying cycles (written in cycle notation) the same as for function composition, ie factors on the left get applied after the ones on the right?

Actually, it shouldn't matter right? If all one ever ends up multiplying is disjoint cycles, and they commute, it's not a problem nevermind

[coral spindle]

No this is correct, but of course we don't always multiply disjoint cycles.

Unfortunately algebraists also tend to disagree on which way to write function composition.

[barren sierra]

This is the right way to do it but many people are wrong

[coral spindle]

So you should look at the context to see if a particular author writes right-to-left (like much of the rest of the world) or left-to-right

And lo, people have strong feelings on it...

Do I ping Wew & start a fight?

[barren sierra]

They're functions

[coral spindle]

Yes, and many authors write xf instead of f(x) and f o g : x |-> xfg

[barren sierra]

Hence they should should follow the common notation for composition of functions

[coral spindle]

I agree! Don't get me wrong!

But it must be mentioned that some people do it differently

[barren sierra]

NGL I've seen only arguments for why we should write xf instead of f(x)

I've never seen anyone actually do it

[coral spindle]

Isaacs does it in Character Theory of Finite Groups

[barren sierra]

Oh yea you're right

I forgot about that

[coral spindle]

There's another rep theory book that does it but I forget the name

James & Liebeck!

[barren sierra]

Haven't looked at that text ever

[coral spindle]

Weird bc that book is supposed to be more accessible to non-mathematicians

It's a good text they do some cool stuff

[south patrol]

Do they much rip

[coral spindle]

Annoyingly often!

[south patrol]

Tbh I had never seen anyone use composition etc the other way round but yesterday i saw smth cursed

[coral spindle]

I should say that this is the major disagreement that algebraists have. We're quite lucky – when logicians disagree...

[south patrol]

It's like uh say A is a ring and consider A' = End_A(A) [viewing A as a right A module], regard A as a left A' module and then view A as a right A'' = End_A'(A)-module

That made my brain hurt

Lol

[coral spindle]

Oh the double centraliser theorem?

Yeah

[south patrol]

Yes

Is this due to Rieffel

There they phrase it as bicommutant

[coral spindle]

Idk but it's very standard nowadays

[south patrol]

Ah ok

It's pretty cool

[coral spindle]

It's in Isaacs too, for example

yh it's neat

Took me a while to appreciate

[south patrol]

Idk how to think about A''

So End_A(A) can be identified with Z(A) right

[coral spindle]

Yeah

[south patrol]

And elements of A'' uhh

Uhh

Idk how

Anyway I try to phrase it just sounds like A lol

Which needn't be true

[coral spindle]

The way I know it is like so

Let $A$ be a semisimple algebra, and $M$ a simple $A$-module. We will write $A_M$ for the ring of additive functions on $M$ generated by $A$, meaning $A_M = \build{m \mapsto am}{a \in A}$.

Let $D = \operatorname{End}_A(M)$. Then $\operatorname{End}_D(M) = A_M$.

[cloud walrus]

Boytjie

[coral spindle]

This is Isaacs' Theorem 1.16

So it's saying that you don't gain any extra elements of A after double-centralising, but only considering elements of A as functions rather than true blue elements of A

[coral prawn]

so this might be a dumb question, but is there an efficient way to solve equations in monoids in general? in the group of the integers under addition, if we have x+2 = 4, it's immediate that we can do a right subtraction (x+2)-2 = (4)-2 and get x = 2, but in the monoid of the natural numbers there is no subtraction; although it's still trivial to solve x+2 = 4, i'm worried it isn't in general.

also, side question - is there a way to tell from the getgo if an equation in a monoid can be solved? of course the x+2 = 4 example can be solved in the natural numbers under addition, but x+4 = 2 cannot be, and once again it seems like a lot of monoids would have unsolvable equations in general.

[delicate bloom]

I guess since your monoid lies inside a group, you can just solve it in the group and see if the result is not natural, but that's probably not what you want to hear

[coral prawn]

yeah, i was kind of hoping for something that helps solve when it isn't inside a group

[delicate bloom]

I guess I'd focus on examples that aren't monoids inside groups for starters

when can a monoid be extended to a group

[coral prawn]

yeah lol it just occured to me that there MIGHT not be solvable monoids, i should probably find one first

[coral prawn]

well i guess all it would need is invertibility, and all groups are already monoids

[rotund aurora]

I find this question weird, since I don't think you can write down equations that make sense in all monoids. And the equations are likely to be quite boring

[coral prawn]

yeah probably

ill get back to you when i think of an example, i guess it was kind of more a question of existence of unsolvable monoids

[rotund aurora]

is there a "canonical" way of doing this? I mean without choosing a generator of G

I could show it is equal to Z/nZ where n is the order of a generator by choosing one and forming a projective resolution from it

[south patrol]

You seem to be assuming G is cyclic or smth

But anyway you can use a certain long exact sequence

Namely ||0 -> Z -> C -exp -> C^x -> 0||

@rotund aurora

(Often a good idea whenever you see H^k of one thing iso to H^m of anothe)

You still need to show H^1(G,C) = 0, but that follows from some other facts you may have seen before and I can help if you'd like

[coral prawn]

on further investigation it actually looks like solutions to equations of that form ARE possible in general - i wrote up a little (probably terrible) proof for it using the category theoretic idea that monoids are given by functions under composition

that holds for any arrangement of functions of x (at least, when x is present once)

[rotund aurora]

yeah idk why I assumed cyclic lol

also in this case, H^2(G, Z)=H^2(G^ab, Z) but is it a general fact?

[rotund aurora]

this is zero essentially because C is torsion-free

[south patrol]

No or you could conclude the same for Z

The key thing is that |G| is invertible

[rotund aurora]

H^1(G, Z) is zero

[south patrol]

Not for every group