groups-rings-fields

[plucky flicker]

if you have proved that Z(D_5) is just the identity then u have nothing else to do i guess

because everything follows from this

[fervent rock]

i know Z(D_5) = {e} its like a fact i can use

[plucky flicker]

then u are done

[fervent rock]

sumn like this>?

[pastel cliff]

to show that SLn is a subgroup of GLn is it enough to use determinants

or well

to show its closure maybe

[thorn delta]

i don't see anything wrong

[pastel cliff]

oh okie nvm then

[plucky flicker]

yes

[fervent rock]

ty boti i love you @plucky flicker

lowkey makes sense

[plucky flicker]

no problem ^^

[fervent rock]

i was confused about the single trivial center case

[chilly ocean]

they show even more than is necessary, it suffices to show AB^-1 is in SL2 which is a one line calculation det(AB^-1) = det(A)det(B^-1) = det(A)det(B)^-1 = 1*1 = 1

so like you do not need to show the identity has det 1 (bc take A =B in the above argument)

[rustic crown]

yea but you need to show that SL2 is non-empty then... :p

[chilly ocean]

ah

even cuter is to use the det homomorphism from GL to R^x

and then SL is a subgroup as it is the kernel

[tribal moss]

That's just the same thing with fancier words.

[chilly ocean]

its not stronger? you show it is even normal

[tribal moss]

(And the meat shoved into the claim that det is a homomorphism).

Okay fair.

[fervent rock]

@plucky flicker is it proper to leave the symbols in the set at the bottom?

or do i just write e, r, etc...

[plucky flicker]

u mean writing phi_e or just simply e?

[fervent rock]

yea

[pastel cliff]

det!

[south patrol]

hm suppose V is a simple kG module (for G a finite group and k any (i.e. not necessarily alg closed or char 0) field. If N is a normal subgroup of G, is it true that V is a semisimple kN-module?

This is fine for the case k=C or smth using Frobenius reciprocity and characters but not entirely sure about the general case

[plucky flicker]

both of them is good

writing simply e,r etc means that you identified the elements of Inn with the given isomorphism

[fervent rock]

cool ty

for a homomorphism, essentially its like a isomorphism but not bijective?

so like could i have a homomorphism k: Z_4 -> D_4 where k(x) = r^x?

Since k(x+y) = r^(x+y) = r^x r^y = k(x)k(y)

[pastel cliff]

it's better to think of an isomorphism as a special kind of homomorphism

[agile mason]

Do you guys know some good resources to use to learn abstract algebra?

[silent oxide]

I think of homomorphism as just a map that preserves group/other algebraic structures

[pastel cliff]

Artin is the common textbook recommendation

[south patrol]

bump dis lol

[lethal dune]

Can’t you use some MacKay like argument here? I have to read these things once more

[south patrol]

Hm the way it's phrased it seems this should have an elementary solution, hm

Like it feels like we should be able to consider the submodules generated by g.V where g represent the different cosets of N in G

but those aren't clearly simple

Oh, lmao, this is just a basic form of Clifford's theorem

Oops

[lethal dune]

[south patrol]

Okay, for the sake of posterity lol, the method is to consider some irreducible subrep $U \le V|{N}$. For each $g \in G$, $g \cdot U$ is a sub $N$-rep of $V|N$, since e.g. $N \cdot (g \cdot U) = g \cdot (N \cdot U) \le g \cdot U$ and indeed these are simple (if $W \le g \cdot U$ then $g^{-1} \cdot W \le U$ I guess). Further, $\sum{g \in G} g \cdot U$ is a sub $G$-rep of $V$, so $\sum{g} g \cdot U = V$ by simplicity of the latter

[cloud walrus]

potato

[south patrol]

And then this implies V|_N is semisimple or apparently you can just define this sum to a direct sum

[lethal dune]

it was (semi) simple all along

nice

[knotty tiger]

what's up guys. I have a programming math problem I would be happy for advice. Let's say I have a million vectors and I need to create 10 bases from them in case wheb the dim is 10 like the example . How would you do it in an efficient way?

[coral spindle]

I'm not convinced. Certainly when your group algebra is semisimple I can see why the sum would be direct (this is standard Clifford theory) but you are eliding the crux of the problem here

[south patrol]

Oop

Wait what exactly aren't you convinced by

[coral spindle]

Why would the sum be direct

[south patrol]

Sorry I mistyped a single letter

refined

To a direct sum

[coral spindle]

I'm still not convinced you can refine it to be a direct sum

[south patrol]

Like there's the theorem that if V is a sum of simple A-modules (and like V is artinian and noetherian etc, like in the context I gave we know that A is a f.d k-alg and V is a fg A mod)

Then V is a direct sum of some subset of them

I just looked it up in a textbook

It's how you prove that an A-module M is semisimple if and only if it is a sum of simples

[coral spindle]

I'm trying to remember how the argument goes

[south patrol]

I can give details if u want

So like

[coral spindle]

OK it's a Zorn argument

[south patrol]

Write V as a sum of submodules V_s for s in a set S

Oh well no zorns needed in this context but in general ye

[coral spindle]

Ofc, finite dimensional stuff is fine

I can see how it would go in general

[south patrol]

then take maximal T subset S such that the {V_t : t in T} are independent

Let W be the sum, equivalently a direct sum by hypothesis

Then for each of the V_s, either V_s is contained in W, or it intersects W trivially

In the latter case, we can just form W + V_s

Which is direct

Contradiction

[coral spindle]

Nice

[south patrol]

Yeah it's cool

And the same argument also can be slightly modified to show that every submodule of V is a direct summand

Rep theory is so cool

[solar shore]

having a bit of difficulty showing that $\alpha : U(st) \to U(s) \oplus U(t)$ given by $\alpha(x) = (x \bmod s, x \bmod t)$ is an isomorphism

[cloud walrus]

blanket

[solar shore]

(s and t are relatively prime)

[vagrant zinc]

Hi guys a question how do I do the product of disjoint cycles I have these.

(1345)(234)

[fleet pelican]

uhh so you like treat it like function composition

[vagrant zinc]

I have tried to make them but nothing

[solar shore]

if you start with 1, then you can see that (1345) sends 1 to 3

(234) sends 1 to 1

[fleet pelican]

sigma tau of x is sigma(tau(x))

[vagrant zinc]

I do not reach the particular solution

[fleet pelican]

sigma times tau means first perform tau, then perform sigma. you do this for each number {1,2,3,4,5}

[vagrant zinc]

THANKS

[vagrant zinc]

Nice :D

[tender wharf]

where are you stuck

[charred crescent]

something is not yet clicking when trying to show a function is well-defined for showing isomorphism

i want to define; phi: H -> aHa^{-1} by phi(h) = aha^{-1};
to show well-defined i tried:
suppose h,h' in H such that h = h'.
then aha^{-1} = ah'a^{-1}. Thus phi(h) = phi(h'), hence phi is well-defined.
is this correct? what am i not understanding?

[long geyser]

unless I am missing something, this map is already clearly well-defined

[charred crescent]

if we imagine that it wasn't obvious, would my solution work?

[long geyser]

sure? but well-definedness is only something to care about when there is a choice to be made in the input, which generally occurs in some kind of quotient structure

[charred crescent]

i guess my question is as follows; in general, i am skeptical of the step where i say that because aha^{-1} = ah'a^{-} => phi(h) = phi(h') because idk if i can say just say that aha^{-1} = phi(h) (and similarly for h') without assuming that it is already well-defined

but it's so simple that i don't see another way to show that it is well defined

if that makes sense

[long geyser]

well you are trying to fight a ghost that is not there

so of course you are getting confused

by definition, phi(h) = aha^(-1) and phi(h') = ah'a^(-1)

you are trying to show a statement of the form A implies A

[charred crescent]

okay so i'm overthinking it

[long geyser]

yes

[charred crescent]

okay thank you

[long geyser]

np

[charred crescent]

with all of that said, would this be a valid proof?

i guess i would have to put the homomorphism part above the bijective part to be able to use the fact that a homomorphism is injective iff its kernel is trivial

[long geyser]

yes, I was going to say that just now

other than that it is valid, but it would be better if the "well-defined" section wasn't there at all, it's just gonna confuse the marker if anything

(if there is a marker)

[charred crescent]

okay fair enough, i suppose my professor wants us to show any mapping we "create on the fly" is well-defined in general, and idk when it is appropriate not to do that

if that makes sense

[long geyser]

ok if you were asked to show it is well-defined then sure

weird request but maybe makes sense in the beginning to make sure students understand what it means?

[charred crescent]

im probably doing it for no reason now that i really think about it

anyways, thank you for your help 🙂

[long geyser]

if you are still confused

you only need to show well-definedness when there is a choice to be made in the input, this occurs in beginning abstract algebra almost always in the context of a quotient group/ring/etc

[charred crescent]

hmm, what do you mean by "choice" ?

[long geyser]

or maybe more generally a map where the input has a coset

and the output depends on a representative of the coset

ok the canonical example is

you know the groups Z_n, integers mod n

[charred crescent]

yes

[long geyser]

there each element is an equivalence class of integers

say, [3] = {..., 3 - n, 3, 3+ n, 3 + 2n, ...}

and you define addition on Z_n by [k] + [l] = [k + l]

so you need to check that this is well defined, because [k] = [k + n]

so I need to check [k + l] = [k + n + l]

because I defined my binary operation with respect to a representative of an equivalence class

I could have made another choice, so I need to make sure a different choice does not give me a different answer

[charred crescent]

mmm okay that makes more sense

[long geyser]

in a map like yours, there is literally no choice to be made

[charred crescent]

so for pi: G -> G/N by pi(x) = xN, we must show something like this is well defined?

so for x = y, xN must be equal to yN

[long geyser]

so if a map has an input that is a set and whose output depends on some element of that set, it is important to see if a different element gives a different result

no actually, you don't need to check that

because again, your input is just an element

is not a set where the output depends on an arbitrary element of the input set

so for example, if you had G/N

and you wanna define a map G/N -> G by f(gN) = gh
or something then you'd have to check well-definedness (in this case it clearly isn't, so our function is ill-defined)

[charred crescent]

okay i think that makes more sense now

so for a group G, and normal subgroups H,N, with N subset of H, the map f: G/N -> G/H by phi(xN) = xH

we would have to show that is well-defined

[long geyser]

exactly

because what if instead of x I pick some other element in the coset, but I think you get the idea

[charred crescent]

right so we need to show that the output is invariant to the choice of representative

[long geyser]

[charred crescent]

thank you again

[long geyser]

🙂

[agile mason]

thank you

[agile mason]

thanks a lot

[white oxide]

what exactly do they mean by the quotient of elements in L? how is the quotient necessarily different? the elements are the same, right?

[white oxide]

bump :)

[woven obsidian]

I can't prove the middle equality. Why is [E(X): F(X)] = [E:F]?

Here F is any field, and E is algebraic over F

x is a transcendental element over F

[formal ermine]

x is not in E and not in F so it's just "an extra element", therefore the degree doesn't change because you add it to both fields

[untold cloud]

Hi, guys, why the shaded part is true?

[elder wave]

It’s using 8.2 n times

👍

[untold cloud]

Yeah… but i did not see how i should use it😥

[formal ermine]

is that the commutator

[untold cloud]

Yes

And we assume it is in the centre

I dont even see how to use it 1 time..

[formal ermine]

I think it's just expanding the commutator

[hot lake]

Look at y^nx [x^-1,y^-1] then look at 8.2 n times

[untold cloud]

Okay i see it is true for n=1

[woven obsidian]

sure but I want to prove it

A generic element in $E(x)$ looks like $\frac{g(x)}{h(x)}$ with $g,h \in E[X]$, but I can't see how to write this in a way so that it can be expressed in a natural basis for $E(x)$ over $F(x)$

[cloud walrus]

AoiKunie

[woven obsidian]

My guess is that the same basis as for $E$ over $F$ should work

[cloud walrus]

AoiKunie

[hot lake]

Is the characteristic 0 ?

[woven obsidian]

No

[formal ermine]

can we assume the degree to be finite

[woven obsidian]

It is given that $[E(x):F(x)]$ is finite indirectly, as $E(x)$ is contained in a larger field $K$ with $[K:F(x)]$ finite

[cloud walrus]

AoiKunie

[woven obsidian]

But the point of the argument is really to show that $[E:F]$ is finite. But I am also curious about the equality

[cloud walrus]

AoiKunie

[formal ermine]

I remember solving something similar to this. the equality would be true if [F(alpha) : F] and [E : F] are coprime

[woven obsidian]

But x is transcendental so $[F(x):F] = \infty$

[cloud walrus]

AoiKunie

[hot lake]

If you take n elements of E that are linearly independent over F, shouldn't they also be independent over F(x) ?

[formal ermine]

what if we like

if both are finite, use the tower law as often as we can on both, then look at the minimal polynomials and argue that they will always have the same degree because x is transcendental over our base field or something?

I haven't thought the transcendental part here through though lmao

was just an idea

[woven obsidian]

That looks true, because if something is zero in $F(x)$ then it must hold as a polynomial in x.

[cloud walrus]

AoiKunie

[woven obsidian]

hmm or wait a minute. ${...,x^{-2},x^{-1},1,x,x^2,...}$ isn't an $F$-basis of $F(x)$ over $F$

[cloud walrus]

AoiKunie

[woven obsidian]

so I can see that this holds over $F[x]$ but I can't see it clearly over $F(x)$

[cloud walrus]

AoiKunie

[hot lake]

if sum fi(x)/gi(x) * ei = 0

you can multiply through by all the gi(x) because they are nonzero polynomials

[woven obsidian]

Yeah I just realized

That should work then

For one inequality

[hot lake]

so without loss of generality you can assume sum fi(x) ei = 0

then by looking at the constant coefficient they must all be 0

so you can divide through by x and repeat

[woven obsidian]

Yeah and if we have one nonzero coefficient somewhere we can't get 0

[hot lake]

that shows that [E:F] <= [E(x):F(x)] I think ?

[woven obsidian]

Yeah I think so (but I always mess up these inequalities)

The other direction seems hard in general

[hot lake]

I would pick any generating family for E over F and try to show it generates E(x) over F(x)

[woven obsidian]

There is a solution on https://math.stackexchange.com/questions/3203527/adjoining-a-transcendental-element-to-an-extension-and-a-base-field , but they use conjugates and I'm not sure if that works out in general

To transform the denominator of a fraction into something in $F(x)$

[cloud walrus]

AoiKunie

[woven obsidian]

But to me it just seems like it lies in the fixed field of $E(x)$ over $F(x)$ and not necessarily in $F(x)$, unless it's a separable extension

[cloud walrus]

AoiKunie

[woven obsidian]

or perhaps even galois

[hot lake]

yeah they use the norm from E(x) to F(x)

[woven obsidian]

Hmm, perhaps I'll just be happy with this inequality

Still the general statement seems reasonable

But you need some way to fix your denominators

[hot lake]

you only need to know that the denominator divides something in F(x)

[woven obsidian]

Yeah

But why is the product thing in F(x)?

[glass vine]

Hey, are there any group theorists currently online, I need some notation help for this paper https://londmathsoc.onlinelibrary.wiley.com/doi/abs/10.1112/blms/14.2.119

Mainly in this part, what is meant by beta_i = beta phi_i? My guess is that it is the image of beta but I am not completely sure

[delicate orchid]

Yes that’s correct, sometimes we write f(x) as xf

[glass vine]

Thank you!

[lethal dune]

I write xf all the time

[kind jacinth]

what does it mean for cosets to form a group? like each individual coset is a group? so lets say N is a subgroup of G, then g1N, g2N each have a group property? or is it that G/N has group property?

[coral shale]

set of cosets

[fleet pelican]

it needs multiplication to be well defined

[coral shale]

quotient group yes?

Check the definition of a quotient set

[kind jacinth]

ok so the set of cosets will have a group property. yeah this makes sense as the identity cant be in all cosets at the same time

[kind jacinth]

i thought this is the set of cosets

[fleet pelican]

you need to guarantee that gNhN doesn't depend on your choice of h and g

[coral shale]

the claim is that this quotient set is a group with the right operation

[fleet pelican]

it will not be a group in general

[coral shale]

To form a quotient set, you need an equivalence relation - see if you can figure out what it is in the specific case of a quotient group

[coral shale]

u also maybe have misunderstood this point

the identity is in only 1 coset

the cosets form a partition

[kind jacinth]

yes this makes sense, and the union of all cosets would result in the group

and the set of the cosets could sometimes behave like a group

[coral shale]

precisely iff your subgroup is normal

thats nice to prove

[kind jacinth]

ohh this is useful thanks

[coral shale]

well technically you need to pick the right binary operation

ofc

and the usual one we pick for quotient groups will fail to be well defined if your subgroup isnt normal iirc

[kind jacinth]

when we say normal subgroup we just mean that doing left coset or right coset doesnt matter right?

[fleet pelican]

gN=Ng for all g in G

[kind jacinth]

alright thanks!

[coral shale]

yh u may need to prove that depending on which defn uve been given

[kind jacinth]

i was given the definition gHg^-1 = H, which i think results in gH=Hg by multipying both sides by g

[coral shale]

yes

there a lot of equivalent defns

proofwiki lists most i think

worth checking and maybe quick proving

[fleet pelican]

the best one is that G/N is a group

[coral shale]

nah best one is N is the kernel of...

[fleet pelican]

I mean, that's just factoring a map through the projection tbh

[lethal dune]

As a side note as to why we desire the subgroup to be normal is to want N as the identity element of G/N, that is (gN)(N)=gN = (N)(gN)=(Ng)(N) where they are treated as cosets. which should give you gN=Ng. Think about it a bit

[kind jacinth]

so normal subgroup is the same as taking the kernel of a function (like such as from f: Z->Z4) then N=ker(f) and in this case N is the identity element of G/N.

[fleet pelican]

As mentioned in this comment, a kernel is always normal

[lethal dune]

yes ker ↔ Normal subgroup

[coral shale]

sorry was a half joke I didn't complete

"N is normal iff its the kernel of some homomorphism phi : G -> H for some group H"

Too broad to be useful usually but

[fleet pelican]

I think ker -> normal is the useful direction

[untold cloud]

Hi, guys, suppose G is a group and H is a subgroup of G, then suppose {y_1,...,y_n} is a right transversal of H, then we have a right group action, Hy_i\to H_y_i x, for x\in G

But how can i translate this in another definition of group action, there is a group homomorphism G\to Sym(right cosets of H)?

[fervent folio]

Let $p\in\mathbb{P}$ and $L:=\mathbb{Q}(\sqrt[3]{p})$, my task is to determine the degree of the extension $L/\mathbb{Q}$ and find all intermediate fields and which of them are normal.

[cloud walrus]

Enoo58

[fervent folio]

With the minimal polynomial $X^3-p$ which is irreducible the degree should be 3. Hence there are no intermediate fields. Is this correct?

[cloud walrus]

Enoo58

[agile burrow]

Right, so the point of a transversal is that H y_i x is equal to H y_j for some unique y_j in your transversal, so this maps i to a unique element of {1, ..., n}. Repeating this for all i gives you the bijection from {1,...n} to itself and this is the element of Sym(n) which x maps to

There are a few things to check here, namely that the map we've defined from {1,...,n} to itself is actually a bijection and that the corresponding map G -> Sym(n) is a group homomorphism

[untold cloud]

Thanks! i think i got it

[lethal dune]

Yes that’s true

[kind jacinth]

why do they have to do a(0+0) for this proof?

why cant u say a*0 = 0?

[spice whale]

because that's not an axiom of rings

you're trying to prove that

[kind jacinth]

ok but then from a0 + a0 how can u conclude 0?

[fleet pelican]

0=0+0

[spice whale]

take away a0 from both sides

surely it says that underneath though

[south patrol]

(Also like, i assume they expect familiarity with groups and in groups we know that if a = a + b then b = 0)

[fleet pelican]

just that -a exists

[spice whale]

well yeah that's basically what i said

[fleet pelican]

its a ring axiom that you have additive inverse

[kind jacinth]

ohh ok i see, and thats possible because for rings there exists a -a0 such that a0-a0 = 0

the (R, +)?

[spice whale]

yeah

[kind jacinth]

and for 2, -ab is inverse of ab for (R, +) so i can prove it by showing a(-b) is an inverse as well

under addition

idk about 3

[vagrant zinc]

guys is this multiplication of products of disjoint cycles correct?

[fleet pelican]

yes

[fleet pelican]

can you formalize that?

[obsidian loom]

Hello

sorry to interrupt

I'm looking at the proof of the fact that A_n is non-abelian for n > 3

They only show that A_4 is not abelian, and since A_4 is a subgroup of A_n, n >=4, they conclude that A_n is only abelian for n <= 3

why is it sufficient to show the case for A_4?

[kind jacinth]

so we know if a,b element of R then ab element of R since its closed under multiplication. then we also know for ab there should exist a -ab such that ab + (-ab) = 0.
now we take a(-b) => ab + a(-b) = ab + (-ab) by associativity of multiplication and this gives 0 by the (R,+)
thus -ab = a(-b) right?

[fleet pelican]

because there is a subgroup of A_5 that is isomorphic to A_4, for example, the elements of A_5 that only shift around {1,2,3,4} but leave 5 untouched, same for any n > 4

[obsidian loom]

so if a group contains a non-abelian subgroup, then it cannot be abelian itself?

[fleet pelican]

If you had ab + a(-b) = ab + (-ab) by associativity of multiplication you wouldn't need a proof, you would already know a(-b)=-ab

[fleet pelican]

Do you recall the definition of abelian group?

[tribal furnace]

What you want to show is that the additive inverse of ab is equal to a times the additive inverse of b (and vice versa)

[obsidian loom]

yes, a commutative group where order doesn't matter

how does that help us?

[fleet pelican]

can you be more precise?

[obsidian loom]

x*y = y*x for any x,y in G

[fleet pelican]

so if there's a nonabelian group that means there's an xy such that xy isnt yx

so if there's a nonabelian subgroup....

[kind jacinth]

a(-b) => ab + a(-b) = a(b-b) = a(0) which gives 0, we know this from (1)

distributivity cz of addition ^

[obsidian loom]

right, i see now

thanks a lot

[fleet pelican]

can you assume 2 and do something similar for 3

[kind jacinth]

for 1 i know it was related to "0" being the identity in addition, for 2 i know it was related to "-ab" being inverse in addition, I am not sure what 3 relates to.

but i can try proving it assuming 2 holds

well using 2 i can replace a with -a then i get (-a)(-b) = -(-a)b. I know (-a)b = - ab so by replacing i get (-a)(-b) = -(-ab)

and from this well that is ab right?

[fleet pelican]

(-a)(-b) isn't (-a)b

[kind jacinth]

woops

[kind jacinth]

no its -(-a)b

[fleet pelican]

you know that -(-a)=a though

[kind jacinth]

so i get ab

thanks!

[kind jacinth]

how would i prove this? do i manually prove its commutative for all elements in F and that they have an inverse?

[fleet pelican]

i guess, yes

[celest furnace]

Maybe try interpreting it geometrically, Z_2 acts on shapes by flipping which might help

Or check the 4x4 table

[lethal dune]

Only need to check 1 elem tbh

[molten viper]

So, I've been asked to find the splitting field of x^4 + 2 over Z_3, constructed as a quotient of polynomial rings. I can factor it over Z_3 as (x+1)(x+2)(x^2+1), so Z_3(i) splits this, do I have that Z_3(i) isomorphic to Z_3[x]/(x^2+1)?

[formal ermine]

or find an isomorphism to another field

[fleet pelican]

what other field?

[rotund aurora]

finite fields are "canonical" lol

[fleet pelican]

ah i was confusing F4 with Z4 and thought they were distinct

[lethal dune]

that's not a field

[molten viper]

"that"

[uneven scaffold]

If A is a ring and p a prime ideal. How many maximal ideals does A_p have? I think we are looking for ideals of A_p such that A_p/I is a field. This is a field if we kill all the non-units of A_p so is this question trying to make us find the non-units of A_p?

[south patrol]

Well

[lethal dune]

do you know the correspondence between ideals of A_p and of A

[south patrol]

Do you know what primes of Ap look like

lol beat me to it

[lethal dune]

No I did, "well" doesn't count

[south patrol]

wait what

[lethal dune]

you can have it

[south patrol]

lol i said you beat me to it

[lethal dune]

lol misread

[uneven scaffold]

I know the correspondence between prime ideals, but I don't recall that there was anything related to maximals?

[fleet pelican]

maximal->prime

[lethal dune]

what can you say about elements of A_p \ pA_p

[uneven scaffold]

I guess pA_p will eventually contain all the non-units of A_p and this is the only one?

[lethal dune]

well yeah

so any proper ideal can't contain elements outside pA_p as they are all units

[uneven scaffold]

My question is perhaps more related on how do I determine which are the non-units of A_p?

[lethal dune]

can you show it

if yes, you already have the answer

say in Z_ab what happens to ab?

[uneven scaffold]

We have a bijection the prime ideals of A_p and the prime ideals of A that are subsets of p, but p is arbitary in our case so this correspondence doesn't do much I think

[lethal dune]

your p is not prime here?

[uneven scaffold]

I mean yes it's prime, but still A could have many prime ideals?

[lethal dune]

they'll all lie inside pA_p

infact any ideal will lie inside pA_p

[uneven scaffold]

i think I don't understand pA_p then properly

[lethal dune]

making it "the" maximal ideal of A_p

pA_p is the ideal generated by elements a/s where a ∈ p and s ∈ A\p

[uneven scaffold]

Why should this contain all prime ideals of A?

[lethal dune]

it doesn't contain all prime ideals of A

only the ideals that are contained in p

[uneven scaffold]

You meant something else with this?

[lethal dune]

I meant all ideals of A_p lie in pA_p

try to see why elements outside pA_p are units in A_p

[uneven scaffold]

So units in A_p are elements a/b where a is in A and b in p such that there exists c/d in A_p for which a/b * c/d = ac/bd = 1

Do we get something on the denominator as both b and d are in p?

[lethal dune]

so elements outside pA_p are r/s where r,s ∈ A\p which by definition are units in A_p so you get r/s is unit

[indigo ridge]

can our sequence of products be less than size n, for example say n=5, then will the number say x= (a_1)(a_2) and or (a_1)(a_2)(a_3) be in our set?

or does it strictly have to be a string of products of size n?

in other words is i=n or $i\le n$

[cloud walrus]

jayzsparrow

[lethal dune]

no n is arbitrary here

infact n can be 0 giving you identity of G

[indigo ridge]

Perfect thanks 🙂

[sonic coral]

the ring Z_p with addition and multiplication mod p is an integral domain since Z_p has no zero-divisors. This is because p is relatively prime with every natural number less than it right?

[south patrol]

How are you defining Zp

[sonic coral]

integers mod p

[south patrol]

No I mean

The way you say it implies you are thinking of it as like {0,1,...,p-1} with some operation right

Or are you doing it as a quotient ring

[sonic coral]

the former i think

[south patrol]

Well this is because if ab = 0 mod p, then p|ab and what can you conclude

[sonic coral]

just use euclids lemma

[south patrol]

yh

[formal ermine]

wait why is that a lemma

isn't that the definition of a prime number...?

[south patrol]

Eh

[formal ermine]

non unit and p | ab => p | a or p | b

[south patrol]

Usually/often primes are defined in school and stuff as basically (positive) irreducibles in Z

[sonic coral]

i wondered that too when i proved it because it seemed like the definition of a prime

[south patrol]

and then Euclid's lemma shows that this coincides with the notion of prime for a ring (up to sign)

[formal ermine]

I see

wait

no

nvm I'm still confused

cuz like

[south patrol]

lol

So like I was taught in highschool that a positive integer p is prime if the only (positive) integers dividing it are 1 and p

and i think that is the normal definition of primes people would give before uni right

[formal ermine]

idk we aren't taught what a prime number is here in high school

[sonic coral]

i was taught a natural number was prime if it’s only divisors are one and itself

[formal ermine]

we don't do any number theory here

we only do like

euclidean geometry

and right now we're doing derivatives

[south patrol]

financial ones?

jk

[formal ermine]

aren't irreducible elements prime elements

or was that only in ufds

ah wait nvm

in ufds

[south patrol]

Yeah you're right

[formal ermine]

prime iff irreducible

otherwise

prime => irreducible

if only domain

[south patrol]

assuming x is non-zero like

x is prime iff (x) is a prime ideal, and x is irreducible iff (x) is maximal among all proper principal ideas

i think

[formal ermine]

isn't the latter only the case in pids

a ufd is a pid iff every nonzero prime ideal is maximal

in ufds prime iff irreducible

[south patrol]

huh

wdym

[formal ermine]

like

[south patrol]

Well what I have said implies that

[formal ermine]

oh wait

I didn't read the

"among all proper principal ideals"

lol

[south patrol]

Ye

ok stuck on some noncomm alg ting

Let $V$ be a semisimple $A$-module of the form $V = \bigoplus_{i \in I} V_i$ where the $V_i$ are \underline{pairwise non-isomorphic} simple $A$-modules. is it true that any submodule is of the form $\bigoplus_{j \in J} V_j$ for some $J \subset I$? and any hints on how to prove that

[cloud walrus]

potato

[south patrol]

Wait I think I had a proof lol

[formal ermine]

fancy latex

[formal ermine]

I don't know that much module theory but these are just my 2 cents on it: ||it can't be something smaller than a Vi because they are simple. if it contains 1 non zero element of a Vi then it contains the entire Vi because the smallest cyclic submodule is the Vi itself||

idk if you want that spoilered

[south patrol]

I mean I already had that

But it isn't quite enough because of the existence of elements like (a,b,c) or whatever

hm

[formal ermine]

but if you look at the columns individually then they must at least contain the submodules generated by each of the nonzero elements in that column

but those are just the entire Vi

[south patrol]

Again sure but not sure how that is enough

Like it must rely on the non-isomorphic bit

Because e.g. { (a,a) : a in C } is a sub C-module of C^2

Oh lol okay I've got it

nice

[formal ermine]

what's the solution

I'm curious

[south patrol]

So

oh lol so it's the same as what I already had lol was staring at me in the face ugh

let $W \le V$ be a submodule. It's also semisimple (standard fact). Suppose $S \le W$ is a (non-zero) simple submodule and consider the projections $S \to V_i$ for each $i$. Now there must be some $a$ such that the map $S \to V_a$ is non-zero (since $S \ne 0$) and thus this map is an isomorphism. So there must be exactly one such $a$, since otherwise we'd wind up with an isomorphism $V_a \simeq V_b$ for $a \ne b$, a contradiction. This means that $S = V_a$.

[cloud walrus]

potato

[south patrol]

so like, W is a direct sum of some of the V_a

[formal ermine]

oh that's cool

didn't know of the first fact

[south patrol]

Yeah i mean so like

[formal ermine]

aaaaactually I didn't know of semisimple and simple modules before this conversation at all

[south patrol]

The easiest way afaik is that a module is semisimple if and only if it is a sum (not necessarily direct one!) of simple modules

then if W is a submodule of V, consider U: = sum of all the simple submodules of W

Then you can show U = W

So actually that is related to the above proof i gave lol

[white oxide]

how is it possible that 0 + a1x + 0x^2 is a polynomial if the ai are all distinct?

for example that would imply that a0 = a3

and they're not all distinct

moreover how does this remedy there being more than one formal sum representing the same polynomial

[tender wharf]

how are they distinct

[tender wharf]

1+2x = 1+2x+0x^2 = 1+2x+0x^2 + 0x^3

[white oxide]

i thought by definition a1, a2, a3, .... an in a ring are distinct

since it's a set

there can't be repeating elements

[formal ermine]

why should they be distinct

it's like

a_0 = 2

a_1 = 2

a_2 = 3

where our ring is Z

why should we not be able to pick 2 twice

[tender wharf]

no set

[white oxide]

i thought by definition a ring is a set with the operations of multiplication and addition and some axioms

[south patrol]

But the polynomial isn't an element of the original ring anyway

[tender wharf]

Z[x]

[delicate orchid]

it is, and there's no reason why 2+2x+3x^2 (the example illuminator gave) can't be an element of that set

[white oxide]

oh wait i'm actually stupid

i was looking at the ai counting up

and not the x^i

thanks!

[vagrant zinc]

Yes or no ?

[pliant forge]

(y x)(y x) : y->x->y => (y x)^2 = ()

[cloud walrus]

walter

walter

[chilly ocean]

What's klein 4-group's basis?

V_4={1,a,b,c}

[noble hedge]

If $G$ is a finite $p$-group of order $p^k$, does $G$ always have an element of order $p^{k-1}$?

[cloud walrus]

(𒀭)

[fleet pelican]

yes.

[noble hedge]

Why?

I feel like it's obvious but my mind is slipping

[fleet pelican]

weird arguments with conjugation

it isn't obvious at all

[noble hedge]

that makes me feel a bit better

What does the proof look like?

[fleet pelican]

it's a very difficult one

wait no, I think I have the existence of a subgroup of order p^k-1 sorry, idk if it's cyclic

[lethal dune]

take NxNxN...N where N is Z_p

[fleet pelican]

S_6 has a subgroup of order 16 and nothing of order 8

[lethal dune]

not a p-group

[fleet pelican]

wdym

[vagrant zinc]

How do you learn to demonstrate group theory?

[long nebula]

No, but it has a subgroup of order p^(k-1)

[barren sierra]

I have a question about tensors

[cloud walrus]

Spamakin🎷

[barren sierra]

The proof began as follows

[cloud walrus]

Spamakin🎷

Spamakin🎷

[thorn delta]

the tensor product is generated as an abelian group by the set of all simple tensors. This means that every tensor is a finite sum of simple tensors

[barren sierra]

ah ok

[solar shore]

$\alpha : U(st) \to U(s) \oplus U(t)$ given by $\alpha(x) = (x \bmod s, x \bmod t)$ where $\gcd(s, t) = 1$

[cloud walrus]

blanket

[solar shore]

could someone help me prove that alpha is a surjection?

im having a huge amount of trouble for it

[rotund aurora]

where do normal operators show up in math?

I think I have never seen an actual application of the spectral theorems

guess they are important in Hilbert spacy things

[pliant raptor]

Is it true that $\left( \mathbb{Z}/p^n\mathbb{Z} \right)/\left( \mathbb{Z}/p^{n-1}\mathbb{Z} \right) \cong \mathbb{Z}/p\mathbb{Z}$?

[cloud walrus]

ImHackingXD

[frigid lark]

Yes

As groups certainly

[tribal moss]

For that even to make sense you need to have in mind a particular way to view Z/p^(n-1)Z as a subgroup o Z/p^nZ.
That embedding is what will be doing the heavy lifting.

[kind temple]

for a hint on the embedding, try figuring out how Z/2Z sits most naturally inside of Z/2^2Z, same for Z/3Z inside of Z/3^2Z and try to generalize.

[proper moth]

Can I get a rough outline for how to prove the existence of an inverse for G, composed of the set of all ordered pairs (a,b) where a,b are real, b is non-zero, under the operation (a,b)+(A,B) = (a+bA, bB) ?

I was thinking of trying to solve for (a+bA, bB) = (0,1) (the identity), but I need to solve bB = 1 which just suggests B (= b^-1) = b ?

[kind temple]

yes, then A = -a/b

and i think you meant B = b^-1 = 1/b

then just check that (a,b) + (-a/b, 1/b) = (0, 1) = (-a/b, 1/b) + (a,b)

[proper moth]

ah I see, yeah

thanks!

[pliant raptor]

Ok, thanks

[grand yarrow]

i had this question on my exam that asked me to prove h^2=h iff h is a subgroup where is a finite set.

the argument i constructed used the fact that h is finite

so that h^2=h implies that every element must have an inverse and hence by subgroup test

h is a subgroup

so i am wondering is there a counterexample for which h^2=2

for an infinite subset h

which is not a subgroup?

[lethal dune]

sorry what? question doesn't make sense

[paper flint]

I think they meant to say a finite (non-empty) subset H is a subgroup iff HH=H

I do not follow the argument though

[grand yarrow]

yeah this

[grand yarrow]

a finite (non-empty) subset H is a subgroup iff HH=H

[lethal dune]

where should there be an identity in H?

[grand yarrow]

cuz since

h^2=h then

all powers of an element should be in it

g g^2 g^3 //

this must repeat

[paper flint]

I think the finite order stipulation helps with that

[grand yarrow]

hence g inverse exists

so gginverse=e

exists

in h

but what if H were infinite?

[lethal dune]

you need finiteness here... hmm

[paper flint]

This would not be true. Think in terms of groups where you do have elements of infinite order.

[grand yarrow]

yeah like say Z

but i can't construct a subset with HH=H

[paper flint]

A simple example would be ||H=Z^+ in (Z,+). Clearly closed under addition, which implies HH=H, but not a subgroup.||

[lethal dune]

yeah not true for infinitee

[paper flint]

Think

[lethal dune]

Z+ u {0} example

lol all of us gave the same example

[grand yarrow]

Oh got it

nvm this was ez

[lethal dune]

it's true if you assume that every element has finite order tho

[paper flint]

All infinite groups are isomorphic to Z 😌

[lethal dune]

that's what we are using actually,

(cyclic)

[grand yarrow]

yeah true

[grand yarrow]

any example?

like where this fails but the other doesn't

so i wanna construct an infinite set H but with every element having a finite order

[lethal dune]

I mean you can just take Z/2 x Z/2 x... countably many times and any infinite subgroup will do

[south patrol]

Fr

[paper flint]

Tarski monster

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

[vague granite]

Right so I'm reading a textbook on ring theory and there's an exercise which asks to prove the universal ring property for homomorphisms on commutative polynomial rings with unity.

It then asks if the theorem is true for rings without unity but I couldn't find anywhere in the original proof that makes use of unity?

And most sources on the internet talk about rings with unity so I feel like I'm getting something wrong.

[kind temple]

what is the precise statement that you have to prove?

[vague granite]

Since if we have a ring homomorphism $\phi: R\rightarrow S$, and extended ring homomorphism $\psi: R[X]\rightarrow S$ with $\psi(X) = a\in S$ then necessarily $\psi(\sum c_i X^i)= \sum \phi(c_i)a^i$.

[fleet pelican]

that follows from definition of homomorphism

[vague granite]

exactly

It's the last sentence that's throwing me off

Also I'm not sure where commutativity is playing a role

[cloud walrus]

Philka

[fleet pelican]

tbh i wouldn't worry much, unless you're actually working in noncommutative geometry or something it's best to assume your rings have a unit, are commutative and are noetherian.

[vague granite]

I guess. But it is a bit frustrating to not know what the author is hinting at

[fleet pelican]

I would imagine a polynomial ring without unit might behave funny, since you can't have a monic polynomial.

[vague granite]

That might be true but in this context it just doesn't show up at all

[hot lake]

I would need to check how they define R[x] to know if x is in it or not

[fleet pelican]

yeah, probably you can't have phi of x then

[vague granite]

but phi(x) is meaningless?

[fleet pelican]

because x isnt in R[x]

but like, you usually wouldn't consider polynomials over a ring without unity. I've used noncommutative rings in lie algebra, but don't think we ever used polynomials, just like adjoint matrices.

really feels like the author is being goofy for no reason imo

[vague granite]

Isn't R[x] always defined up to isomorphism as a set of finite sequences with the appropriate maps?

how would x not be in R[x]

[fleet pelican]

Because R[x] is polynomials with coefficients in R

And 1 isn't in R

So 1x isn't either.

[vague granite]

❗

[fleet pelican]

You only have rx for r in R

[vague granite]

That is probably the crux of it then

[fleet pelican]

Seems to be. But again, I think the author is being a silly goose.

[vague granite]

Because if you take 2Z[X] for example you wouldn't have X in it

but if you define psi(x)=a then it neccesarily contains x...

[fleet pelican]

I think you get it now.

[vague granite]

okay...ill-defined problem

[fleet pelican]

Looks like it was trying to be clever but ended up confusing.

[vague granite]

But couldn't you have a polynomial ring with X but not 1?

Like <2,X>

[fleet pelican]

It comes down to how you define things then.

[vague granite]

I guess the underlying assumption is X is in R[X] but you can't assume 1 is

I think if you allow all that it would still work

[fleet pelican]

This isn't something people tend to worry about, I can't think of any useful/interesting examples.

[vague granite]

Well it's always better to make a universal property more universier I reckon

[fleet pelican]

#category-theory

[vagrant zinc]

Someone knows the guy, I think I've seen him around here.

[formal ermine]

why?

[vagrant zinc]

He handles the same book that I have, and I wanted to ask him something.

[vagrant zinc]

[formal ermine]

I don't understand why the l-adic cyclotomic character is considered a galois representation. it's G_Q surj Z_l^times, and Z_l^times isn't a field. the author of what I'm reading addresses this issue by saying that we can view Z_l^times inside Q_l^times (i.e. GL1(Q_l)). how does this resolve the issue though?

[untold cloud]

Hi, guys, suppose V is a left vector space over a division ring, then why V* is a right vector space?

[formal ermine]

is V* the dual space?

[untold cloud]

Yes

[south patrol]

vector space over a division ring

But uh

[lethal dune]

[south patrol]

Okay so consider how you can put a multiplication on V*

[lethal dune]

||well (f•r)(x)=f(r•x)||

[south patrol]

You have a map f: V -> R for R the division ring

Lol

I was gonna give hints towards it

But ye

[lethal dune]

otherwise you won't get a(b(x))=(ab)x

I mean I can still

[south patrol]

njug

*hug

[formal ermine]

uwu?

[untold cloud]

Oh, I see! thanks!

[untold cloud]

Thanks too

[formal ermine]

hi det

[rustic crown]

hi illu

[ebon pine]

hi det

[rustic crown]

hewwo

[ebon pine]

Are there any good abstract algebra problem books? Googling shows Herstein but I don't think it's a problem book per se

[delicate orchid]

fulton-harris: representation theory has some good exercises

[formal ermine]

abstract algebra is a broad description

what are you looking for specifically

[ebon pine]

basic stuff like group theory, rings, modules, and galois theory

Things usually covered in AA first course

[sly rain]

From first practice sheet of a Intro to Abstract algebra class:

Show that {3,9,15,21} with x*y:= xy mod 24 is a group.

Like is doing a multiplication table by hand and observing + showing that multiplication in z/nz is associative enough or is there some catch/something im missing?

[barren sierra]

It might be good to explicitly note what the inverses of each element are, what the identity is

[formal ermine]

that's still too broad

like even just "rings and modules" is too broad

do you mean introductory exercises? or comm alg? or xy? or xz?

[ebon pine]

Here's the syllabus

[formal ermine]

does it not cover solvable groups lol?

here are some ring theory exercises if you want

[ebon pine]

this is for a competitive exam so the syllabus is slightly reduced

[formal ermine]

Z_n here means Z/nZ and not the n adic integers

[formal ermine]

doing galois theory without knowing solvable groups removes all of the fun imo

like you can't even proof why there's no formula in radicals for the roots of a general quintic or higher then

[ebon pine]

competitive exams are opposite of fun so it matches the vibe I guess

[ebon pine]

Thanks!

[formal ermine]

but yeah you can just pick up any algebra textbook (I recommend artin) and do the exercises in it

[summer path]

Here are a bunch of group theory exercises

[formal ermine]

yeah check the pins for group theory exercises

[ebon pine]

I did the first few chapters of Artin but then I felt that the exercises were either easy or much above my level

[formal ermine]

eg?

[ebon pine]

For example M.14 which asks us to prove that SL2(Z) is completely generated by only 2 matrices

I was stuck there for quite long, then I took few matrices and tried to write them using only E and E'

After a bit of trial and error I finally figured the algorithm but writing it properly again took quite some time

[formal ermine]

see, you were still able to figure it out

[ebon pine]

I mean yeaah but like it takes almost a day, I barely have 2 months for the exam

[formal ermine]

um what

[ebon pine]

what?

[formal ermine]

2 months to learn all of this?

[ebon pine]

yeaah, I'm almost done with group theory so the situation isn't that bleak

[formal ermine]

[ebon pine]

Plus it's not my first time with them

This is more of a revision

[elder wave]

2 months is a lot of time

[formal ermine]

not if it's your first time dealing with those topics

[ebon pine]

not when I also have to revise analysis also

[elder wave]

Even then

[elder wave]

Fair

[ebon pine]

Most importantly, there's no point in talking about if I have enough time or not
I'll try to do my best as much as I can

[formal ermine]

what kind of competition is this btw

[ebon pine]

Thanks tubu

[ebon pine]

NET, qualifying this allows you to do a PhD

[summer path]

I mean it might be useful to still balance your time on the materials and not like say dump all your time on one topic in algebra lol

[rotund aurora]

Dixon problems in group theory. Richard Borcherds acually recommended it. I haven't worked through it, but I plan on doing so when I do more group theory again

[formal ermine]

oh I thought you meant like a real math competition lmao

like those in high school

[ebon pine]

Dixon problems in gt are graduate level though

[rotund aurora]

still abstract algebra

[ebon pine]

I mean, it is a national level exam so it's fair to call it a competition

Not only I need to qualify it but I have do so with good grades to get good rank to get good uni

anyway

Thanks for the help everyone

[rotund basin]

Is this a typo in Fundamentals of Semigroup Theory by John Howie? I believe it should read $K_{a} = { a^{m}, a^{m+1}, \ldots ,a^{m+r-1} }$

[cloud walrus]

TwoStickmen

[summer path]

Yes

[rotund basin]

Cool cool, just had to make sure I wasn't going crazy lol

[white oxide]

can i get a hint for a?

[formal ermine]

this was a stupid question

it's just the codomain being GL1(Q_l) but the image is just Z_l^times

[rustic crown]

t/t

[white oxide]

lmao i just thought of that

gg

wait but how would you show that it's the unity

you can't do r/t multiplied by t/t and just cancel out the ts right

[rustic crown]

you can

[white oxide]

why are we allowed to do that tho

[rustic crown]

becasue by definition (r, t) and (rt, tt) are equivalent

[white oxide]

ah i see that

thanks!

[rustic crown]

potato

[south patrol]

Hi

I am do commutative algebra

[rustic crown]

[molten viper]

We love a good commutativity

[karmic moat]

can i have hint for 4.ii please :) ive already shown that φ is an automorphism; my “educated guess” is that the minimal polynomial of ε in Q is x^n-1; if thats right, not sure how to apply eisenteins

[molten viper]

Are you familiar with the cyclotomic polynomials?

[karmic moat]

havent heard the term before

[molten viper]

Well, take a look, but for prime p phi_p(x) is eisenstein I believe

x^p-1 is a multiple of phi_p(x) is your hint

But x^p-1 isn’t irreducible

[karmic moat]

oh i see

imma think about this ty

[formal ermine]

x^n - 1 isn't irreducible

1 is a root

[molten viper]

I should be clear, the cyclotomic polynomials aren’t themselves Eisenstein but you should check if f(phi) is

[formal ermine]

what is an eisenstein polynomial lol

[karmic moat]

by eisenstein i assume theyre talking about a polynomial which satisfied eisenstein’s criterion?

[formal ermine]

aha

you can't directly apply the eisenstein criterion to phi_p

[karmic moat]

ye thats what galstaff said

wait

[karmic moat]

wdym f(phi)?

[formal ermine]

if f is your minimal polynomial then you should look at f(x + 1)

[karmic moat]

oh

f as in the mapping little phi defined in the img

[formal ermine]

yeah apply phi to your minimal polynomial

[karmic moat]

kk

[formal ermine]

you still need to make an educated guess on the minimal polynomial though lol

[karmic moat]

yeah

[formal ermine]

it's pretty simple

your minimal polynomial divides x^n - 1

obviously

but x^n - 1 isn't irreducible because it has 1 as a root

what happens if you divide 1 out?

[karmic moat]

something like (x-1)(x^{n-1} something something)?

ah wait

seems like i can keep factoring x-1 out

x^(n-1) + x^(n-2) + … + x + 1?

[karmic moat]

n is prime, so this is n-th cyclotomic polynomil

polonomial*

[formal ermine]

yes

[molten viper]

Ding ding ding, now check that it’s irreducible and that epsilon is actually a root

And I think check that Q(e) splits it

[karmic moat]

I suppose epsilon is a root can be checked via induction on [x^(n-1) + x^(n-2) + ... + x + 1]/[x-epsilon]

[molten viper]

Well, you already know epsilon is a root

[formal ermine]

epsilon is a root because it's a root of x^n - 1 and you only divided 1 out

[molten viper]

Yes

[karmic moat]

ah irght

right*

[molten viper]

The question I suppose is how do you know there aren’t other roots not in Q(eps)

@formal ermine do you have something productive to add?

[formal ermine]

your "questions" make no sense

[molten viper]

I’m sorry to hear that you feel that way

If he wants to he can ask for clarification

[sonic coral]

is there any more efficient way to do reduce a polynomial over some field

or am i just supposed to exhaust all the possibilities?

[formal ermine]

uhh there was this one algorithm

I forgot what it was called

[sonic coral]

maybe the rational root theorem?

[formal ermine]

berlekamp theorem

that was it

[sonic coral]

what’s the traditional way to go about it? just use intuition and guess and check?

bc like with a quadratic it’s rather easy since we know that if it’s reducible, it’s going to be two linear factors, but for anything with a degree bigger than 2 we have more cases to consider right

[formal ermine]

for 3 it also has a linear term

[sonic coral]

yes, but it could have three of them, or one and an irreducible quadratic?

[formal ermine]

yea

[sonic coral]

so that’s a lot of cases to check. that’s why i’m not sure if this is the typical way you’d go about trying to reduce a polynomial, or if there’s special things to look for

[formal ermine]

I personally would use a cas lmao

[sonic coral]

see that’s what i was thinking lol

i could code it in python pretty easily

so i suppose we’re more concerned with if something is reducible, not what it reduces to?

[formal ermine]

usually, yes

[sonic coral]

another question, if i have a polynomial f(x) of degree 2 or 3 in some field Z_p, i know that it is reducible iff f(a)=0 for some a in Z_p, but that also implies that (x+a) is a factor in the reduced polynomial right

[south patrol]

ye

Also like this holds in any field

Not just Zp

[rustic crown]

Fp

.<

[sonic coral]

that’s where i got the intuition for it

[rustic crown]

also x-a

and not x+a

https://tenor.com/view/cute-eevee-gif-21737791

[south patrol]

Fun things are eevee

[rustic crown]

[south patrol]

Pokemon is darwinian propaganda

[rustic crown]

[sonic coral]

i thought it was x+a bc the example i did in Z_5 gave me f(2)=f(3)=0, and i reduced to (x+2)(x+3), but since we can shift the remainders of division modulo 5 i got it backwards. idk if that makes sense or not

[formal ermine]

lol

[sonic coral]

,rotate

[sonic coral]

does this last step make sense? idk if i’m allowed to do this or not

[south patrol]

Yes

[molten viper]

Am I correct that GF(p^n) is isomorphic to (Z_p)^n

[rustic crown]

nu

well additively yea

but if you mean product of rings, then the right thing isn't even a field

[molten viper]

Hm

[south patrol]

Nor is it isomorphic to Z/p^nZ in general

[molten viper]

What do the elements of GF(p^n) look like?

[south patrol]

Uh I mean

Depends on ur construction lol

[molten viper]

So, the overarching question is this

I know that E cap F always contains 0 and 1

I'm thinking maybe like, min(e,f) plays a role

[molten viper]

Well, I have no clue how to count them really

So is GF(p) isomorphic to Z_p?

[formal ermine]

yes

finite fields are unique up to isomorphism

[molten viper]

So what the heck is up with p^n

[formal ermine]

you can think of F_p^n as the splitting field of x^p^n - x over Z/pZ

[molten viper]

Hmmm

So I guess what does a subfield look like

[formal ermine]

there's this one nice proposition that a finite field of order p^n contains a subfield of order p^m for each m | n (and no other subfields)

proof: exercise.

the intersection of E and F follows immediately from that

[south patrol]

Proof: gal theory lol

[formal ermine]

frobenius endomorphism my beloved

[molten viper]

I think I'd say I have 0 clue what's really going on here

[formal ermine]

the intersection of two fields is a subfield of both

[south patrol]

Fröb

[formal ermine]

now apply this

[molten viper]

So it's the GCD(e,f)

[formal ermine]

yes

[molten viper]

Ok gotcha that was my first guess but I had my isomorphism wrong

[south patrol]

Girlfriend(p) = Z/(p)

So girlfriend = Z/

[formal ermine]

lol that reminds me of this one person's about me

let me find it

[south patrol]

qed

[pastel cliff]

[agile burrow]

frobenius

[pastel cliff]

i fucking love that name

[south patrol]

Classic devisage

Is it Frob or fröb

I never know

[formal ermine]

@frigid lark

[south patrol]

I thought it was the latter but am now being gaslit

[pastel cliff]

i just learned it the other day, im so happy i get to unironically say the word frobenius

[formal ermine]

I know understand why some people are pedantic about Z/pZ vs Zp

[rustic crown]

potato getting cooked

[south patrol]

Zp has like 3 potential meanings

[frigid lark]

Wait, so it does help?!

[formal ermine]

what is the other one

cyclic, p adic, __?

[south patrol]

Ig localisation of Z at the prime p lol

[formal ermine]

oh right

I still need to reread localization

[south patrol]

But then ud use like

[formal ermine]

only gave it a quick look once

[south patrol]

Well

Or the element p, rather

[formal ermine]

potato what's your favorite math field

[pastel cliff]

probably potato

[frigid lark]

Z/57Z

[south patrol]

Algebraic topology is what I know the most about

But I am more interested in algebra the more I do

Like rep theory

So I'm not entirely sure what I'll end up doing

Hbu

[pastel cliff]

alg topology has been fun this semester

[south patrol]

Nice

[pastel cliff]

probably bc my prof does very handwavey proofs, but the material itself is also very fun

homology confus me rn tho

[formal ermine]

idk haven't tried out a lot so far

[molten viper]

Would it be right to say that E cap F is the largest subfield contained within both?

[south patrol]

Chain complexes took me a while to get jntuition for

[formal ermine]

but I like number theory

and galois theory

[frigid lark]

You know cohomology, so just flip the arrows

[formal ermine]

I've really been enjoying the exposition about galois reps that I am reading right now so far

[rotund aurora]

what are you reading btw

[south patrol]

Sexy

[formal ermine]

https://people.math.harvard.edu/~smarks/mod-forms-tutorial/mf-notes/galois-reps.pdf

this

[rotund aurora]

I still dont know what a Galois rep is

thank you

[formal ermine]

it's a continuous representation of the absolute galois group of Q or a subgroup

[south patrol]

OK these are cool

[lapis trail]

Anyone know about Penrose calculus

[rotund aurora]

lol i thought it was something different

[formal ermine]

continuous rep*

fuck

how do you spell continuous again

WAIT

I SPELT IT RIGHT THE FIRST TRY

LETS GO

[rotund aurora]

like I think I once read something about Gal reps in the fly and the author was mixing it with modular forms and my brain melted probably

[lapis trail]

Lol

[pastel cliff]

continuoouououus

[formal ermine]

yeah it's like

let me get my diagram

[frigid lark]

Ouwuo

[rotund aurora]

nice

[agile burrow]

algebraic nt

that's cool

[pastel cliff]

i wanna learn this

im in a NT class rn

[south patrol]

Stetig

[formal ermine]

me too

[south patrol]

more like

Uhh idk

[pastel cliff]

but it has spent too much time oon elementary stuff that feels like trivia

[formal ermine]

STETIG DIFFERENZIERBAR

[south patrol]

Lol

[pastel cliff]

so now im behind cuz i got bored

[rotund aurora]

I had seen this diagram

[formal ermine]

REPRÄSENTATIONSTHEORIE DER ABSOLUTEN GALOIS GRUPPE VON DEN RATIONALEN ZAHLEN

[rotund aurora]

so much to learn

[south patrol]

Stetig differenzierbare Abildungen von meinem Gehirn auf sich selbst

[pastel cliff]

so real

[formal ermine]

isn't the connection between galois reps and automorphic forms conjectural

[south patrol]

Lol

Braucht mehr Genitiv

[rotund aurora]

mmh I think so

like I think that's literally the langlands program or something

but idk about that

[formal ermine]

@prisma ibex ?

[rotund aurora]

[south patrol]

Repräsentationstheorie der absoluten galois gruppe der rationalen Zahlen

[formal ermine]

lol

[formal ermine]

repräsentationen der absoluten galois gruppe des körpers der rationalen funktionen der ganzen zahlen

[south patrol]

so wahr mein bester freund

WzT

[prisma ibex]

In general yes but known in some cases