groups-rings-fields

[trail stump]

Because I think for normal series, it holds, but I am pondering about the case of subnormal

[dim widget]

It is not a semidirect product either.

[trail stump]

What do you mean by that

[delicate orchid]

Z/p^2Z is not a semi direct product of Z/pZ with itself

[trail stump]

Oh wait the inner semi product of Z3 and Z3 is still Z3?!

[delicate orchid]

I don't know what you mean by inner

it's not abelian, unless it's the direct product

that's the main point

infact, Aut(Z/pZ) = Z/(p-1)Z

so the only possible embedding is the trivial map

hence the only possible semi direct product is the direct product, which is what we've already seen

[trail stump]

Yeah ok

I am so confused

I always had the impression that if N is normal to G, then the factor group G/N times N would be G

I don’t know what that “times” exactly is

[delicate orchid]

this is not true

well

for the direct product and natural semidirect product it's not

[trail stump]

What kinda product would then

[wraith cargo]

generally none

in certain cases the semidirect product is the most general product you could want for this to be true

but it's only true when the short exact sequence
1 -> N -> G -> G/N ->1
splits

[trail stump]

I feel destroyed right now

But I am so glad I asked, because it would keep hurting me for idk how long

[delicate orchid]

yeah, and this only happens when G is a semi direct product of N and G/N

kinda circular but it is what it is

[wraith cargo]

I mean it's an if and only if idk if that's really circular

Splitting can also be formulated as a section existing

[delicate orchid]

true

fair enough

[delicate orchid]

the idea I think you want to kinda get at though is that G is an extension of N by G/N

which is true, as you can always construct the exact sequence Mr. Incarnate posted

[trail stump]

I think I am starting to understand why my book put so much effort setting up some really “obvious” and tedious stuff before some serious stuff

[kind jacinth]

field of characteristic 2. Are the only possible orders for this infinity and 2?

[formal ermine]

what about 4? and 8? and 16?

[kind jacinth]

so then the order is 2^n

[tribal moss]

Finite fields always have order p^n where p is the characteristic, and there is exactly one field (up to isomorphism) of each such order.

[hollow tartan]

In the sylow theorem(1st) in the proof the part where it's using correspondence theorem say $f: G \longrightarrow G/Z(G)$ where the order of the center is p, why is it there is some subgroup H in G s.t. $|H|=|Z(G)||G/Z(G)|$

Sry I'm only a latex beginner

[chilly ocean]

backslash longrightarrow

it would have made perfect sense if you had not done it latex anyways

[cloud walrus]

VeridisQuo

[hollow tartan]

Thanks

[clever pollen]

Can anyone tell me what this Aut thing is?

Here S = {1, 2, 4, 8, 16, 15, 13, 9}

I know that Aut(G) is the group of automorphisms on a group or graph G

But I'm not sure when I have Aut(A, B)) what is that supposed to be

[tribal moss]

Probably the group of automorphisms phi that satisfy phi(S)=S.

[white oxide]

can i get a hint for this question pls

@dim widget ?

[tribal moss]

Pigeonhole principle: If |G| is not cyclic, then at least one of the divisors of |G| must be the order of more elements than have that order in the cyclic group.

[white oxide]

hmm ok i'll think about that thx

i was trying to use the fundamental theorem of finite abelian groups

[void cosmos]

good idea

[feral umbra]

I'm reading Gromov's paper on polynomial growth and I'm trying to prove that the family of r-balls in the sequence (Gamma_i) is uniformly compact. Here Gamma_i = r_i^-1*Gamma is the same set but has its metric scaled by r_i^-1. I have attached the definition of uniform compactness and of regular growth.

I think I sort-of understand the idea but haven't quite been able to successfully apply (a) and (b) of the regularity yet :c

[chilly radish]

i'm reading gromov's paper
big mistake

[feral umbra]

yes i have noticed haha

i've written multiple page proofs on his "This is obvious" claims

[formal ermine]

#linear-algebra ?

[dull plover]

i just assumed i should post here since this is in L1 spaces

[formal ermine]

that's analysis lol

[white oxide]

does this equality hold because H* absorbs the intersection of H* and K?

[south patrol]

yh

[white oxide]

ty

why does Hi = Hi(Hi+1 intersect K0)?

[long geyser]

what is K0

[white oxide]

oh K0 is the identity

oops

[clever pollen]

Then it's not a mistake 😄 maybe he did it on purpose to make you learn

[kind jacinth]

"There is a ring homomorphism from Z to the ring (2x2) matrices with entries in Z whose kernel is {-1}" true or false/
this is false because kernel doesnt contain 1, but what about 0?

shouldnt kernel also contain 0 as 0 is the additive identity ?

[lament dawn]

The kernel of a ring homomorphism f: A-> B are those elements which are sent to the additive identity of B.

Is it true that every ring homomorphism sends the additive identity of its domain to the additive identity of its codomain?

[kind jacinth]

i thought that is the case. it is the case for groups right?

[lament dawn]

For group homomorphisms of additive groups yes. Otherwise you need to change terminology around a little, but it's the same idea

[lament dawn]

You should prove this if you don't immediately see why it is true!

It follows from definition of ring homomorphism, you do not need any technology at all.

[kind jacinth]

ah alright i see why that is the case

thanks!

[clever pollen]

I'm wondering how they "observed" this, here \varphi_i is multiplication by i function

[white oxide]

yo does anyone have a diagram for the third isomorphism theorem i can't find any good ones online

or maybe i'm too dense

[chilly ocean]

the third isomorphism theorem is easy to remember. it just says dividing works how it should

[white oxide]

yeah true

[prisma ibex]

there's not much to draw in the diagram, it would just be a tower like N-H-G

[white oxide]

just trying to get like an intuitive picture i suppose

oh wait i'm stupid

[prisma ibex]

compared to the more complicated lattice you get in the second isomorphism theorem

[white oxide]

fraleigh has one

[prisma ibex]

related: who on earth came up with the fourth isomorphism

like good lord how would you think to do this

[white oxide]

isn't that butterfly

wasn't that zassenhaus

[prisma ibex]

it goes by various names

[delicate orchid]

steven

[prisma ibex]

it does not go by all names

[dim widget]

Usually it goes by the forbidden group theory jutsu of the Germans.

[chilly ocean]

if you ever find me irl please call me "fourth isomorphism theorem"

[clever pollen]

Here we're looking for all automorphisms of Z_17 such that f(S) = S. What really bothers me is how do we know its only these multiplication functions and how we can determine this in general.

I've found the statement on stack after some search, should be fine, if anyone is interested I can send a link

[dim widget]

I mean because 1 is in S you know that only these functions can take S to S. Then you check that they actually do work

[frigid lark]

Does anyone know what Lang meant by $\delta_{ij}$?

[cloud walrus]

Parrot Tea

[chilly ocean]

kronecker delta. 1 if i = j and 0 otherwise

[frigid lark]

thx

[clever pollen]

If H is a subgroup of G, and acts transitively on some set X, does G also act transitively on X?

Im looking at some example where this claim is used, but im not sure how to prove it just by looking at orbits

[chilly ocean]

if x and x' are in X, then you can find an h in H with hx = x'. but h is also in G, so...

in terms of orbits, if x is in X, then the orbit Hx equals X. but Hx is contained in Gx. so Gx = X

[clever pollen]

Ah I see thanks!

[hollow mica]

anyone know why “transitive action” takes it’s name

does not seem at all related to transitivity

[chilly ocean]

i can see it coming from reading "transitive" as "transitional", like you can transition from any point to any other by the group action

[hollow mica]

ah transitive in a different sense

[sturdy mirage]

anyone have a hint to this problem? #help-13 message

[solar shore]

good morning

or afternoon

what time is it

[rose lark]

How difficult is dummit/foote?

[lapis trail]

I'd say moderate

[warm ember]

whats wrong with this proof of hilbert basis theorem

Assume for the sake of contradiction there exists an infinite ascending chain of ideals
[
I_1\subsetneq I_2\subsetneq I_3\subsetneq\cdots.
]
We can express $I_1$ as
[
I_1={a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0\mid\forall i: a_i\in G_i}
]
where $G_i=(G_i,+)$ are subgroups of $(R,+)$. Since $I_1$ is an ideal,
[
\forall f\in R,g\in I_1: f+g\in I_1.
]
Set $f(x)=b$ so for all $i$,
[
\forall a_i\in G_i:a_ib\in G_i.
]
It follows that $G_i=(G_i,+,\times)$ are actually ideals. Similarly, all the $I_i$ can be expressed in this form. For all $i$, let the ideal of constant terms of $I_i$ be $R_i$. It follows that
[
R_1\subsetneq R_2\subsetneq R_3\subsetneq\cdots,
]
a contradiction.

[cloud walrus]

Kevin Yang

[warm ember]

the ring is R

R is commutative

[next obsidian]

Why can you write I_1 like that? Furthermore, I don’t see why I_1 is closed under adding arbitrary elements from R. Consider the ideal generated by x, this is simply all polynomials with no constant term. This is not closed under addition by R, consider x + 1

[warm ember]

wait sorry

i mean fg\in I_1

its closed under multiplication

[next obsidian]

Then I agree, but I still don’t see how you can justify I_1 being written that way. I don’t know what G_i is supposed to be here, and don’t feel like trying to prove some statement that I don’t really feel is true?

[warm ember]

I_1 can be written that way since its closed under addition right

[next obsidian]

Please write down a proof

[warm ember]

you just add the coeficients

its trivial

[next obsidian]

Because I have no idea

Take the ideal (x)

what is the subgroup G here?

[warm ember]

there are multiple subgroups

one for each term

[next obsidian]

Oh, I see

Uhhh

[warm ember]

idk i just see all the proofs on wikipedia are tricky

[next obsidian]

Sure

Okay I see

You’re just proving the set of coefficients on n-th degree terms are ideals

Yes this is true

[warm ember]

does the last line follow?

[next obsidian]

I don’t see why this is strictly increasing

Without argument

[warm ember]

i never learned "subrings" but i assume $A\subsetneq B$ if the set $A$ is a subset of the set $B$?

[cloud walrus]

Kevin Yang

[next obsidian]

No, it’s a proper inclusion

Otherwise the Noetherian property is impossible to satisfy

[warm ember]

proper subset*

[next obsidian]

You could just take a single ideal over and over

Then yes

But you have to argue why I_k a proper subset of I_k+1 means R_k is a proper subset of R_k+1

Take for example (x^2) < (x)

The ideal of constant terms in both cases is (0)

[warm ember]

oh

hmm

i can just take the ideal of some x^n term

right

they cant all be the same

[next obsidian]

Well this won’t satisfy the initial hypotheses

This forms a descending chain, not an ascending one

[warm ember]

?

[warm ember]

in this case i can take the ideals for the x^2 terms

wait

no

i mean x terms

[next obsidian]

Okay but then just crank the degree up again

[warm ember]

oh wait

[next obsidian]

I don’t think this method will work, when proving the basis theorem for power series rings you do actually look at ideals of lowest coefficients

But for polynomial rings I don’t think this leads to a proof

[warm ember]

ye oops

thanks

[next obsidian]

I don’t think the proof of the basis theorem is that complicated. It involves a trick for sure, but the trick is looking at ideals of constants

Which is what you were trying to do here, that’s the key idea

[warm ember]

oh

[next obsidian]

The details are a bit more complicated than what your proposed proof tried, but at the end of the day it’s not that much more work

[warm ember]

thank you

[next obsidian]

[ebon pine]

I am reading Judson and he defines orbit of an element of G-set X, ie O(x) as the equivalence class which contains x. From what I remember the natural definition of orbit of x was set where x can be mapped to ie O(x) = G*x

Am I misremembering or are the two definitions equivalent

[chilly ocean]

y is in the equivalence class of x if and only if...

write it out and check that they're equivalent

[ebon pine]

Ah, got it. x would obviously lie in it's equivalence class and rest follows since the classes are mutually exclusive

[kind jacinth]

question :a field cant have characteristic 1 right?

only 0 or prime

also no characteristic can ever have "order 1" right?

which i think in other word it means no field has 1 element

same with order 0, thats not possible right?

[lethal dune]

[frigid lark]

{0}

But most people won't consider this a field

[kind jacinth]

and the order for this is 1?

[frigid lark]

What is an order of a characteristic?

[kind jacinth]

should have sent the question for it to be clearer. 1 second

[kind jacinth]

in this case characteristic 1

what are the possible orders?

[frigid lark]

Oh, order is just size

Yeah char 3 implies at least 3 elements

So order 1 is not possible there

[kind jacinth]

and for char 1 the possible order is just 1 and infinit right?

[formal ermine]

F_1 is my favorite field

[summer path]

I think it's time for you to review your definitions

[tribal moss]

Pretty much every book explicitly defines a field to require 1 != 0, so characteristic 1 is not possible at all.

If it were not for being exclude in the definition, the only 'field' with 1 = 0 would be {0}, because in such a 'field' we would have a = 1·a = 0·a = 0 for every element a. So it couldn't have infinite size.

[coarse musk]

If f(x) is reducible
over F, it does not mean f has a root in F. - examples to show this?

where F is a field

[frigid lark]

(X^2+2)^2 over Q

[delicate orchid]

(x^2+1)^2 over R

[coarse musk]

but x=-2?

[frigid lark]

Oh no

Changed it

[coarse musk]

so it doesn't need to be completely?

where even x^2+1 is reduced

[delicate orchid]

What

[frigid lark]

Q^a

Or C

[delicate orchid]

What, again

[frigid lark]

I think they are asking in what field does x^2 + 1 have roots

[delicate orchid]

I see, Q^a isn’t a field albeit

[coarse musk]

oh wait

never mind think I got it

but then the next part baffles me more - If f is irreducible over F, then f cannot have any root
in F, unless when f is a linear polynomial, when it will definitely have a root. This is the right statement, but how do I go about proving this?

[delicate orchid]

Yeah that’s correct

[frigid lark]

F[X] is a ED

I.e. has division with remainder

[delicate orchid]

So say f has a root z, try dividing f by some polynomial involving z

[coarse musk]

wait so this says that if f is a linear polynomial then it will have a root if it is irreducible?

or am I reading this wrong

something doesn't seem to stick right

[delicate orchid]

Linear polynomials are always irreducible

Simple degree argument suffices

[coarse musk]

oh wait

yes

fair

[delicate orchid]

I’m trying to get you to prove the contrapositive, if that’s what’s confusing you

[coarse musk]

i can prove that, but I'm thinking of proving the first part

[coarse musk]

precisely yes

[sonic coral]

a polynomial in a field F is reducible if it can be written as the product of two elements of F both having positive degree. Linear polynomials cannot be written like that even though they may have a root in F

[grand cliff]

Just learnt wreath products, they are cool objects

I like how their elements & operations have quite a few decent interpretations

[delicate orchid]

they've very cool if you're wreathing with a subgroup of S_n

then you get a certain very nice group of automorphisms of a very nice set of graphs

cyclic wreaths are cool too

appear in lots of places

[obsidian loom]

This is the last chapter of my lecture notes

And somehow it seems like we're going back to chapter 1

Is it supposed to be more general than just binary operations in some way?

I don't see it

[delicate orchid]

well, it's pretty obviously more general than a binary operation

GxG -> G
GxX -> X

[hollow tartan]

I'm not sure why the course starts w this, in lin algebra vector spaces has similar def in the beginning of the course

[delicate orchid]

it's important to understand the basic idea of a group before looking at group actions

[obsidian loom]

I think I understand what a group is

But I don't see how this is more general

(other than the obvious difference you pointed out)

Oh wait so, instead of the group acting within itself (binary operations), it acts on a whole different set?

[hollow tartan]

I think it's more general it explains an evolution of sets from magma to field/modules/VS

[delicate orchid]

yes exactly

although there are of course canonical group actions of a group on itself, like g.g' = gg' and g.g' = gg'g^-1

[obsidian loom]

This seems very abstract, is there any interpretation for it?

[delicate orchid]

there are two very easy examples

*three

we'll stick to two actually

the dihedral group acting on the corners of a shape, the symmetric group S_n acting on the set {1, ..., n}

you may have heard that groups are the "symmetries of [something]" - that set X is the [something]

[obsidian loom]

oh shit that makes sense

[delicate orchid]

number 3 is groups of matrices acting on a vector space by matrix multiplication

fyi

[hollow tartan]

VS act on a field, if they act on a ring then form a module

[obsidian loom]

tbh group theory as the "study of symmetry" is what made me interested in the topic initially

[delicate orchid]

no

a field acts on a VS

[obsidian loom]

so up until now i thought it was only called that because of dihedral groups

lol looking back that was kind of dumb of me

[delicate orchid]

yeah you'll love group actions

have you heard of cayley's theorem

[obsidian loom]

i have yeah

[hollow tartan]

Specially for the 1 line proofs w group actions fu lagrange

[delicate orchid]

yeah so given any group action GxX->X for a (we'll keep it finite for now) set X

[formal ermine]

isn't a field acting on an abelian grape a vs

[delicate orchid]

you get a homomorphism from G to S_|X| by mapping g in G to whatever permutation swaps things in X the same way that the group action of g on X does

[hollow tartan]

Do u know why is the the object w the most structure is first? Or is this not necessary?

[delicate orchid]

a field can also just act on a k-vector space as long as it's a subfield of k

[delicate orchid]

why wouldn't you start with the most well behaved object

[hollow tartan]

And it's usually (always) the least behaved object that wins

[delicate orchid]

they're so easy, especially when they're finite dimensional it's basically just looking at the poset category of N

[hollow tartan]

What's a positive category of n

[delicate orchid]

so you introduce students to the idea of thinking about an abstract object via the nicest one

[hollow tartan]

Poset*

[delicate orchid]

then go to groups+rings which are less nice, and then to modules which are a mess

[delicate orchid]

you also get an isomorphism to GL_|X|(k) through this via permutation matrices

[ashen heron]

yeah, consider the action of the group $S_{m}$ on $\mathbb{N}^{m}$, if we define the map as $S_{m}\times \mathbb{N}^{m}\to \mathbb{N}^{m},~~~(\sigma,\alpha)\mapsto (\alpha_{\sigma(1)},\dots,\alpha_{\sigma(m)})$ then you may verify that this satisfies the definition of group action, i hope you can see what this is doing

[cloud walrus]

pikachupikachu

[ashen heron]

tied to the idea of group actions is the idea of orbit, consider $G$ be a group and $X$ a nonempty set then the orbit (under action of $G$) of any $x\in X$ is the set $G\cdot x$, if we now define a relation $x\sim y \iff x\in G\cdot y$ then it is verifiable that this is an equivalence relation therefore the orbits partition $X$ and we denote this space of orbits $X/G$. this notation is similar to how we use $G/N$ when cosets w.r.t subgroup $N$ partition $G$. an example would be the multiplicative group $\mathbb{R}^{\times}=\mathbb{R}\setminus {0}$ and its action on $\mathbb{R}^{2}$ defined as $(t,(x,y)) \mapsto (t\cdot x, t\cdot y)$, this partitions $\mathbb{R}^{2}$ in terms of lines through origin

[cloud walrus]

pikachupikachu

[ashen heron]

i just learned this the other day, so forgive me if i made a mistake

[pastel cliff]

sebbb studies for fields and modules

[south karma]

I have no idea how to prove why the G-equivariant map of G-set G/H has that property

I know that for G-sets X and Y, G-map f:X->Y should be f(gx) = gf(x)

[dim widget]

Have you tried looking where the identity goes and using G-equivariance??

[south karma]

Uhh...

I think I've got it

[late thicket]

Is there anyone reading ring theory?

[lethal dune]

[late thicket]

I'm having difficulties with understanding it

How to understand ring thory

[lethal dune]

A particular part or the entire thing?

[coral spindle]

If you have specific questions, you should ask them here. I don't think anyone can give a good answer to such an open-ended question.

A more answerable question would be "how do I understand the definition of a ring," but I'm not sure if that's what you're asking.

[cosmic oriole]

What do you call a semigroup that can deal with infinite semigroup operations?

For example, if the operation is + and I want to do x1 + x2 + ...

where ... is infinite.

To be more precise, what do you call a semigroup with an extra function f than can take a sequence (xn) and do f (xn) = x1 + x2 + ...

[south karma]

Hmm.. but why morphism G/H -> G/K has the form 1H |-> gK?

[coral spindle]

Well it's gotta go somewhere

set g = phi(1H)

That's it

Or well, more accurately gK = phi(1H).

After which you can just choose a coset representative g.

[south karma]

But why?

Is that the only able form?

[coral spindle]

That's the form of any function G/H -> G/K

let f : G/H -> G/K be some function

then f(1H) = gK for some g! That's just because {gK | g in G} = G/K

[south karma]

A

🤦

What was I even thinking

Thanks! you helped me a lot

[late thicket]

@coral spindle all the topics. I just started reading ring theory

[formal ermine]

hi boytjie

[coral spindle]

Hey

[coral spindle]

I cannot summarise all the topics of ring theory. It is a truly immense field. Please be more specific.

[terse rose]

I have a question that asks to show that a o group with size n greater than 1 with no proper subgroups must have prime order

[coral spindle]

Hint: look at a cyclic subgroup

[terse rose]

Can I just argue that it is a consequence of Lagrange’s theorem directly?

[coral spindle]

It is not a consequence of Lagrange's theorem directly.

Unless you're using some version of it I'm unaware of!

[formal ermine]

what's good

[coral spindle]

that's a deep question

Does anyone really know what good is?

I'm fine, hope you're well too

[formal ermine]

I got tired of asking "what's up" and people responding with "the sky or the ceiling"

[coral spindle]

One of my pet peeves too

[formal ermine]

hi merosity

[delicate bloom]

i find that to know what something isn't helps us understand what something really is

[terse rose]

SFC there exist a prime order group with proper subgroups, then this subgroup order k must divide the order of G by lagrange’s theorem, which implies that k|n which is a contradiction as n is prime

[formal ermine]

system file checker

[delicate bloom]

what's evil

[formal ermine]

this is wrong

take the trivial subgroup

[coral spindle]

Who said anything about it being prime order?

[terse rose]

Oh my bad, I read it wrong

[coral spindle]

You need another result for that to happen

[terse rose]

Wait it says, show that a finite group G of size n=>2 with no proper subgroups other than the trivial subgroup has prime order

[coral spindle]

Yes

But you assumed it had a subgroup of prime order first

You need to justify that assumption

[terse rose]

I did a contradiction proof though?

[coral spindle]

Oh wait, my bad. I misread your proof. In fact it is wrong for different reasons.

You have proved that if G is of prime order, then it has no nontrivial subgroups.

However, you must prove the converse

[terse rose]

Yes my bad, I’ll reattempt it

[coral spindle]

I thought after skim-reading that you had assumed that a group of non-prime order had a prime order subgroup, which I was asking you to justify.

[terse rose]

The actual contradiction should be there exist a group with no proper subgroups but isn’t prime order?

[coral spindle]

Sure

If you want to go down the route of contradiction, you'd start with that

[coral spindle]

.

[white oxide]

yo my professor prolly meant <x^2 + cx + 1> right

cuz i think the point of this question is to find c such that x^2 + cx + 1 is irreducible and hence <f(x)> is maximal

and hence it's a field

[chilly ocean]

for (x^2 + cx + 1)? both () and <> mean the same thing

[white oxide]

oh wait really

huh didn't know that

[coral spindle]

Both notations are used by different authors

[delicate orchid]

<> for groups () for ideals

[south patrol]

Ye <> for v spaces and submodules etc too

[chilly ocean]

notation was a mistake

we need to go back to the time when everything was explained in clear and perfect english

[dim widget]

We need to go back in time and express everything in a simple and intuitive way

Perhaps we could use tensor products to do so

Since TENSOR PRODUCTS ARE ESSENTIAL FOR THE THEORY OF BLACK HOLES

[valid night]

Hi, could someone please check if my proof is correct?

[formal ermine]

tensor products are for amateurs. real chads use Tor_0

[valid night]

Nevermind I think I found a mistake

[dim widget]

@valid night the answer is 2 though! But yes this proof doesn’t make much sense as stated

[coral spindle]

A non-proof-related comment: homomorphism is not spelled "homomorfism".

Your proof is incomplete. You write down a formula and then do not justify why it relates to the 3rd paragraph. It seems like you are claiming that f((a,b)) = f((c,d)), but you have not justified this, or even mentioned it.

[valid night]

Haha I realised the spelling error but didn't bother to correct it

[dim widget]

You can prove this in this way though, just by restricting to transpositions and seeing what you get from applying the definition of homomorphism to cut things down

[coral spindle]

No more hints! Kronos is almost there

[valid night]

Alright thanks a lot guys

[coral spindle]

As an aside, I have a joke proof for n \geq 5: the image is Abelian, ergo the homomorphism factors through S_n/A_n.

[dim widget]

I mean only the assumption n \geq 5 is a joke, otherwise I think it’s basically the correct proof tbh

[coral spindle]

It's correct but it's overkill which is why I think it's funny

[delicate orchid]

u pick up on this fast

[chilly radish]

Umm it's actually existence, not theory

Fake wodzicki fan

Also I'm stealing this problem for my next pset on cosets and homomorphisms

[wooden ember]

i struggled for way too long on a pretty simple problem but that got me thinking: what kinds of results do we have to relate the factorization of an integer polynomial over C to its factorisation over the algebraic closure of Fp

the only way i can think to approach such a big problem is to study how different primes split in the splitting field of your polynomial, which is more or less handleable when the polynomial is separable

but im wondering if there are other different kinds of approaches

[next obsidian]

How do you know about this

[dim widget]

@next obsidian I have been initiated and blessed with knowledge of the sacred texts.

(Illum was shitposting about it)

[white oxide]

can somebody give me a hint about this question? Would it just be the union of Q[x]/<x^2 + x + 1> and Q[x]/<x^2 - x + 1>?

we know that the splitting field has to contain x + <x^2 + x + 1> and x + <x^2 - x + 1>

[coral spindle]

A union of fields is not a field

[white oxide]

oh really

hm

[coral spindle]

Unless one contains the other :)

[white oxide]

ahh i see where you're going

[elder wave]

Ultrapedant

[chilly radish]

• crazy polish dude, 2019(?)

[coral spindle]

Anyway, maybe you could describe it as a subfield of C. You may find it quite difficult to find a quotient of Q[x] that works.

If that's what the question is asking for, then so be it.

[chilly radish]

I know

But it's funnier to pretend he is

[white oxide]

hey that makes it more fun

but i don't get anywhere tho

but anyways okay i'll try that

thx

[coral spindle]

Given that they are quadratic polynomials, it will be extremely easy.

[white oxide]

i think finding the complex roots is the way to go

yeah

oh so it would just be $\mathbb{Q}(\frac{1 + i\sqrt{3}}{2}, \frac{1 - i\sqrt{3}}{2})$

[cloud walrus]

okeyokay

[white oxide]

well ig i have to describe that explicitly

[chilly radish]

Is that not explicit?

[white oxide]

yeah true

i'm just used to describing it by providing a basis

also i think that's wrong, should be -1 in one of them

[coral spindle]

Doesn't actually matter, as it turns out.

[white oxide]

wait yeah

i'm simplifying it rn

tfw you forget the algebra needed to do algebra

[coral spindle]

Btw, if you use what you've learned by doing this calculation, you may be able to go back and work out the splitting field as a quotient of Q[x]

[white oxide]

hm okay i'll try that

[coral spindle]

Hint hint: (-x)^2 = x^2 :)

[white oxide]

haha got it thank uuu

[tribal moss]

One of those is the other to the 5th power.

[white oxide]

oh shit should it would just be one of them

hmm ok

[coral spindle]

Let $a,b$ be rational. Then $\mathbb Q(a + bx) = \mathbb Q(x)$.

[cloud walrus]

Boytjie

[chilly radish]

Third time's the charm

[coral spindle]

Habits, man...

[tribal moss]

You also have w² = w-1 when w=(1+isqrt3)/2.

[white oxide]

yo how you get your Qs to look like that bro

leave out the {}?

[coral spindle]

I am using a different font.

[chilly radish]

Just add it to your texit preamble

[coral spindle]

CBA

I should tho

I rarely use texit these days

[white oxide]

yo how the fuck am i to start on this question

oh wait

nvm

nvm i got it

nvm

[summer path]

[white oxide]

nvm i got it

conjugate

LOL

[coral spindle]

It's a cube root, not a square root

[white oxide]

nvm

[coral spindle]

Hint: extended Euclidean algorithm

[white oxide]

oh nvm i got it fr fr now

plugged that shit into mathway to simplify the expression

we chillin

not tryna do that type of algebra sadly

[coral spindle]

Bro sully reacting himself 💀

[summer path]

you could also just use difference of cubes

[white oxide]

what's that again

[summer path]

which is pretty similar to what you would do if it was a square root

[coral spindle]

That is an ad-hoc way, true. The method using the extended Euclidean algorithm applies to any finite extension of Q.

[white oxide]

hm ok

[summer path]

$a^3-b^3 = (a-b)(a^2+ab+b^2)$

[cloud walrus]

Tubular Cat

[white oxide]

ah

[summer path]

i mean better to do it the ad-hoc way than googling how to simplify it and completely circumvent the entire problem

[white oxide]

nah they're lost

not tryna do that type of algebra

why are the elements of Q(pi) of this form?

i get it's because it's a field of quotients

[coral spindle]

That is the definition

Well, ok. It depends actually on how you're defining Q(pi).

[white oxide]

yeah i was confused about that

[coral spindle]

If you define Q(x) as the smallest complex field containing x, then you can rederive this easily

[summer path]

in some sense, the definition also gives you something like that for Q(\sqrt(2)), but you can simplify it to just be a + b sqrt(2)

[coral spindle]

certainly it contains 1, x, x^2, ... and finite sums thereof

And then also quotients

this is sufficient because this now forms a field

[white oxide]

oh right so it's basically like pi is the indeterminate

and then you form the field of quotients analogous to forming the field of quotients for F[x]?

where F is a field

[coral spindle]

Yes

[white oxide]

ohhhh ok

[coral spindle]

I will expond on this, in case you're interested.

[white oxide]

oh thanks!!

[cloud walrus]

Boytjie

Boytjie

[white oxide]

Yes that makes sense

[cloud walrus]

Boytjie

[coral spindle]

This is an interesting subtlety in my opinion.

[cloud walrus]

Boytjie

Boytjie

[coral spindle]

As I mentioned, an alternative definition is to let $\mathbb Q(\xi)$ to be the smallest subfield of $\mathbb C$ that contains $\xi$. This is actually a subtle one – why should a smallest subfield exist? Might it be possible to find smaller and smaller subfields, ad infinitum, and never find a smallest?

So this definition requires some work, if you want to show that such things actually exist.

Anyway that's all I have to say about that.

Exercise: show that the two definitions of Q(xi) that I've talked about are equivalent.

(not for credit :p )

[white oxide]

this is the case since xi doesn't have any g(x) in Q[x] such that g(xi) = 0 correct

[coral spindle]

That's right

[white oxide]

i think i get it now

it's all coming together

thank you!!!

[coral spindle]

Forgot to mention: pi is a transcendental number.

You've probably heard this fact before, but just in case

[white oxide]

why exactly is the field of complex numbers an infinite extension over the rationals? that would imply that that there is an infinite basis but i suppose i'm trying to find one

[white oxide]

ye

pi is transcendental over Q but not over R

[coral spindle]

Well, pi is a real number so pi - x is a real polynomial with a root pi :P

[white oxide]

ye

[coral spindle]

it's not saying very much, therefore, to say that pi is transcendental over R

[white oxide]

yeah true

[white oxide]

ig my question is more related to linear algebra

why there doesn't exist a finite basis for the complex numbers over the rationals

kinda rusty on that ngl

[coral spindle]

Hint: pi is transcendental

Alternative hint: use the tower law.

[white oxide]

oh wait, is it because we can't write pi as a linear combination of complex numbers with scalars in Q

wait no

i'm tripping

what's the tower law

this is one of the questions where it's on the tip of my fucking tongue

lmfao

[coral spindle]

Typo, fuck

I meant [K:F] = [K:E][E:F]

[white oxide]

oh shit that's on the next page lmao

reading on that rn

[coral spindle]

OK, try using my hint that pi is transcendental if you can't use the tower law.

[summer path]

tower law

[sonic coral]

isn’t that just applying lagrange twice since |E| cancel out?

[coral spindle]

This is not lagrange.

These are fields, not groups. [K : F] = dim_F K

[sonic coral]

gotcha okay

[coral spindle]

The notation is similar because of the fundamental theorem of Galois theory

[white oxide]

can somebody explain to me the equality right after "for clearly" 😭 it's not really clicking as to why is you remove the brackets both fields are equivalent

is it suppose to be clear or am i hella overthinking

oh wait

this is literally the tower law huh

bc Q(3^1/4, Q) is a finite extension of Q and Q(2^1/3, 3^1/4) is a finite extension of Q(3^1/4)

[summer path]

tower law makes a lot of these calculations quite straightforward

or at least makes them a lot more approachable

[white oxide]

yea i totally forgot the tower law

i saw all those Qs and got lost man

[pastel cliff]

how obvious is the first part of this

actually lemme go back

i is algebraic over Q

so Q(i) \cong Q[x]/(x^2 + 1)

dont think that's useful here but still

i was first thinking like

cube root(2) is algebraic over Q so x^3 - 2 must be irreducibl

but that's irreducibility over Q not Q(i) i think

[tribal moss]

It's a cubic. If it's reducible then at least one factor is linear which means one of the three cubic roots of 2 is in Q[i].

[pastel cliff]

oh that sounds more elementary than i expected

can you elaborate on wym by at least one factor is linear

also is what i said actually inaccurate

[silent oxide]

you mean why is at least one factor linear?

[pastel cliff]

i meant like what that means

i think linear and i think like a linear map

[silent oxide]

just degree of polynomial=1

[pastel cliff]

ooh linear as in constant

[silent oxide]

uh no constant polynomials have degree 0

[pastel cliff]

sorry yeah

i see it

was overcomplicating

[quartz quiver]

can someone help me understand the solution to part c? I don't understand: firstly, parts a and b rely on P being a Sylow subgroup of H, but (c) seemingly rely on P being a Sylow subgroup of G, and I don't understand why if $N_G(P)/N_H(P)$ is nilpotent, there exists a nilpotent subgroup $K$ of $N_G(P)$ so $N_G(P) = N_H(P)K$

[cloud walrus]

Darylgolden

[quartz quiver]

<@&286206848099549185>

[pastel cliff]

im a little confused by this question

what is the relationship/difference between Q[x] and Q(x)

and similarly what is the polynomial ring of the field Q(x/x^3 + 1)

[prisma ibex]

1/x is in Q(x) but not in Q[x]

[warm ember]

$\mathbb{Q}(x)=\text{Frac}(\mathbb{Q}[x])$ which is the field of rational functions with rational coefficients

[cloud walrus]

Kevin Yang

[pastel cliff]

does thiss still agree with the definition of smallest subfield containing x...?

[late thicket]

I wanted to know the steps which can help me to learn ring theory. I am feeling difficult with every line

[chilly ocean]

the typical advice is to open up a ring theory book and do the exercises and stuff

you'll have better feedback if you say more than just "it's hard". math is hard, that's a given

[pastel cliff]

ohhhh i think i see why this is true

Q(x) is the smallest field containing Q and x

so by defn must have inverse of x

[prisma ibex]

right, and then you're forced to throw in the inverses of all polynomials with coefficients in Q as well, and then that's the whole field

[pastel cliff]

but then what about the fact that $\Q(\alpha) \cong \Q[x]/(m_\alpha(x))$

[cloud walrus]

not sebbb not stμ₂dying

[prisma ibex]

what about it?

[pastel cliff]

like how does that work when alpha isnt some element but instead x

[prisma ibex]

I mean, it doesn't

[pastel cliff]

oh x isn't algebraic

kek

that is why right

[prisma ibex]

right, in particular the minimal polynomial doesn't make sense

[pastel cliff]

yeah that's why i was confused

[pastel cliff]

so this isn't algebraic is it?

[prisma ibex]

but if it's algebraic you can talk about minimal polynomial and if this is irreducible the dimension of this quotient is the degree of the polynomial

right Q(x) is not algebraic over Q

it has transcendence degree 1

it's the same as like

Q(\pi)

\pi is transcendental, so it isn't the zero of any polynomial defined over Q

so it might as well be a free variable from the point of view of polynomials

[pastel cliff]

well the question is about x over Q(x/x^3 + 1)

[prisma ibex]

oh sorry I can't read

[pastel cliff]

[prisma ibex]

I mean it should be transcendental for the same reason that adjoining x alone is

[pastel cliff]

that's my gut but i wanna formalize it

i'll post attempt in a sec

[prisma ibex]

I mean you just have to show that this is the same as adjoining x after some suitable change of variables

[pastel cliff]

uhhhh

x^3 +1 is in Q(x)

then so is 1/x^3 + 1

and also obviously x/x^3 +1

is that enough

ah no it's not

[glossy crag]

What's a good advanced (not just the Frobenius map and classification) resource on finite fields and their applications to other areas?

[solar shore]

just doing some light review before an exam tomorrow

if phi is a homomorphism from a group G to another group G' where |ker(phi)| = n, then phi is an n-to-1 mapping

i feel unsatisfied with the proof that the book gave, but maybe its just because im being stupid

[chilly ocean]

what does "ker phi = n" mean?

[solar shore]

oh oops

order of ker(phi)

is n

my bad

oh wait...

never mind i just made sense of it in my head

do we want the set of x's such that phi(xg) = phi(g)

?

then that would mean that x is in the kernel?

[chilly ocean]

it essentially boils down to: if \phi(g) = g', then \phi^{-1}(g') equals the coset g ker(\phi)

[chilly ocean]

that is true

[chilly ocean]

if \phi(x) = g', then x = g(g^{-1}x), and this g^{-1}x is in ker(\phi). conversely if x is in ker(\phi), then \phi(gx) = \phi(g) = g'

since g ker(\phi) has as many elements as ker(\phi) does you have your conclusion

[ebon pine]

Here, the first statement should be $p(x) \in \mathbb{Q}[x]$ right?

[cloud walrus]

numbpy

[chilly ocean]

what makes you say so?

[ebon pine]

cause it googling it on wiki says that gauss lemma is irreducibility of p(x) in Q[x] is same as that in Z[x]

oh wait, nvm I got it

[chilly ocean]

your theorem goes from reducibility in Q[x] to reducibility in Z[x]. what you're reading online is likely the contrapositive (irreducible in Z[x] implies irreducible in Q[x])

[ebon pine]

ah, got it. I was confused that polynomial itself should have coefficients in Q but then ofc we can't talk about reducibility in Z[x]

I missed that alpha(x) and beta(x) are in Q[x]

thanks

[peak marsh]

If anyone is studying abstract algebra for their finals, lmk if you want to do practice problems with me in a discord call cause I feel like its so much better to have ppl to collaborate with

[solar shore]

i might take u up on that offer, i got my final in two weeks

[peak marsh]

for sure can I dm you?

[solar shore]

u should b able t

o

[coarse musk]

hi so I have the question, find the galois group of x^3-1 over q, and I'm completely confused on how to approach it

i get the intuition behind how they work, but how do I find it for a given polynomial as such?

[frigid lark]

You mean the Galois group of the splitting field of X^3 - 1

Well this case is a simple one, does moving one of the roots not in Q induce a unique automorphism of the splitting field of X^3 - 1 over Q?

[coarse musk]

I'm not sure

could you help explain in a little more detail?

[frigid lark]

Well you have 3 roots, say 1, zeta and zeta^2?

Moving zeta to zeta induces the identity morphism

Moving zeta to zeta^2 also induces a morphism

[coarse musk]

hmm yes

[frigid lark]

So that's all of your morphisms

[coarse musk]

but over x^3-1?

[frigid lark]

Wdym over x^3 - 1?

[coarse musk]

the splitting field over x^3-1

[frigid lark]

Yes, that's just Q(zeta)

[coarse musk]

what does this mean?

[frigid lark]

The smallest field containing Q and zeta

[coarse musk]

oh yes okay i get that but

how would u represent the galois group

of x^3-1?

[frigid lark]

Isn't it just Z/2Z

[coarse musk]

how do you get that?

i'm very confused

[frigid lark]

What's the degree of the splitting field of x^3 - 1?

[coarse musk]

3?

[frigid lark]

No

[coarse musk]

how would find the degree of the splitting field?

[frigid lark]

Well your two complex roots are just zeta and zeta squared, and your real root is 1

So we just need to find the smallest field that contains all 3 of those roots

So find Q(zeta) basically

But Irr(Q,zeta) is just x^2 + x + 1

[coarse musk]

so you just attach zeta and zeta squared right?

[coarse musk]

what does this mean?

[frigid lark]

Well any field that contains zeta, also contains zeta squared

The minimal polynomial of zeta over Q

[coarse musk]

oh right

[coarse musk]

how do you get x^2+x+1?

[frigid lark]

Factor x^3 - 1

[coarse musk]

oh right

ohhh okay

okay so i get that

till x^2+x+1

which is the minimal polynomial for q(zeta)?

[frigid lark]

Yeah, so [Q(zeta) : Q] = 2

[coarse musk]

ohh okay

[frigid lark]

That means there are exactly two automorphisms of Q(zeta) over Q, as Q(zeta) is a Galois extension

[coarse musk]

that's my question

one map maps zeta to itself

how do you know if there's another map that maps zeta to zeta squared?

[frigid lark]

You could verify it

[coarse musk]

how do I do that?

if you could help me with just the direction

[frigid lark]

Literally construct a map from Q(zeta) to Q(zeta) that fixes Q and sends zeta to zeta^2

Treat Q(zeta) as a vector space if that makes it easier

[coarse musk]

i get this

i got that idea

how do i proceed?

[frigid lark]

Also note that zeta^2 = -zeta - 1

Show that it's a field isomorphism

[coarse musk]

how?

[frigid lark]

Cause zeta is a root of Irr(Q,zeta)

[coarse musk]

i don't get it

[frigid lark]

What does Irr(Q,zeta) mean?

[coarse musk]

not sure

[frigid lark]

It's the unique monic irreducible polynomial over Q that has root zeta

[coarse musk]

oh okay

[coarse musk]

how do you get this though

[frigid lark]

Cause zeta^2 + zeta + 1 = 0

[coarse musk]

how?

[frigid lark]

What's Irr(Q,zeta)?

[coarse musk]

oh wait

okay i get it

1+x+x^2

=0

[coarse musk]

so you get this from there

[chilly ocean]

I need a small hint

[sharp sonnet]

there isn't too much to work with here, so you will have to consider ||the subgroup generated by a and b|| and use the fact that this is ||finite and abelian (what theorems do you know that apply in this scenario?)||

[chilly ocean]

thanks

[ripe basalt]

lie theory question: does anyone know how to prove that L(2) is isomorphic to the adjoint representation of sl_2. L(2) here is the verma module modulo some invariant sl_2 subspace. This is discussed in Hall's book section 4.6

like i am confused how a module can be isomorphic to an adjoint representation (because the adj rep is just a homomorphism)

[chilly ocean]

can anyone give a hint idk how to begin

[runic hemlock]

In Q_p every element x has an element y with py=x

Which is not true in Q_q

[chilly ocean]

but how does it imply that they are not isomorphic?

[runic hemlock]

py is just y+y+...+y p times

it's defined only using the group structure

so it's a property of the group.

[chilly ocean]

wait

thanks

this was simple but I'm dumb

[dim widget]

The adjoint rep comes from the action of $sl_2$ on $sl_2$, they are saying that the quotient is isomorphic to $sl_2$ as a representation of $sl_2$

[cloud walrus]

Topos_Theory_E-Girl

[terse rose]

Not really an abstract algebra question but why does the multiplicative modulo group 7 only have congruence classes 1 to 6?

How about [0]_7?

[south patrol]

The point is that 0 isn't invertible mod 7

As for any a in Z, 0*a = 0 which isn't 1 mod 7

[coral spindle]

In general, the multiplicative group of the integers modulo a prime p will consist of the classes {1, 2, ..., p-1}.

This is not true modulo a non-prime

If you want to try proving this, you may find Bézout's lemma and the extended Euclidean algorithm helpful

[frigid lark]

Or a bit of basic ring theory

[devout slate]

hi! I was reading eisenstein's proof about how cyclotomic polynomial ϕ_p (x) is irreducible over Q for any prime p. Can someone please explain to me the highlighted sentence?

[silent oxide]

suppose (say) Phi(x) was reducible, ...

[formal ermine]

let y = x - 1

[chilly ocean]

if I have a line and point (in 3d), how do I find a perpendiclar line to said line going through said point?

[solar shore]

exam day today

“cosets and stuffs”

[celest furnace]

This is more suited for #linear-algebra , and there will be infinitely many lines perpendicular to that line/point pair

[void cosmos]

if you can find a factorization for p(x), can you find one for p(x+1)?

[formal ermine]

[void cosmos]

hii

[formal ermine]

hi

[lethal dune]

Yes

[void cosmos]

😠

[lethal dune]

You stated that as a question

[void cosmos]

😠

[formal ermine]

you were sullying yourself

or was that your other personality

[lethal dune]

my fake accounts

[formal ermine]

hi topos

[dim widget]

That wasn't all manifolds are smooth, that was all products commute with coproducts

[lethal dune]

no

[formal ermine]

what was coprod in set again

intersection?

[rustic crown]

disjoint union

[lethal dune]

$\varprojlim \varinjlim = \varinjlim \varprojlim$

[pastel cliff]

[formal ermine]

same thing

[rustic crown]

[formal ermine]

plus minus some elements

[lethal dune]

disjoint intersection

[white oxide]

can somebody give me a hint for this question? I wrote x^2 - 1 \equiv 0 mod p, and i know it has to do something with fermat's little theorem but i'm a little stuck

[lethal dune]

how many solutions does x²=1 mod p has?

[formal ermine]

x^2 - 1 = (x + 1)(x - 1)

[white oxide]

oh wait OOPS

i forgot to factor it LMFAO

[formal ermine]

yeah now use that Zp is an integral domain

[white oxide]

ok bet thx

brain not working

[pastel cliff]

mood

[formal ermine]

hi seb

[pastel cliff]

greetings illum

can i get a hint on this plss

[formal ermine]

galois theory in disguise

[rustic crown]

disguise?

[pastel cliff]

[rustic crown]

hewwo sebb

[pastel cliff]

det

i might fail this final

[next obsidian]

F(a) ≈ F(b) when a and b are roots of the same irreducible polynomial

[rustic crown]

[next obsidian]

You can extend isomorphisms of field extensions up to algebraic field extensions

You’ve maybe seen the tower

Like, F’ > F > k

[formal ermine]

Aut instead of Gal

[next obsidian]

E’ > E > k

[formal ermine]

arrgghhhh I remember doing this for a homework

I forgor how I did it lemme think

[next obsidian]

😑

[rustic crown]

u no know the char, alg closure need not be sep closure

[formal ermine]

oh

I'm stupid

[rustic crown]

nuu

dun say that >.<

https://media.discordapp.net/attachments/796976320067272704/818150000482582528/612933479730249728.gif

[pastel cliff]

this shouldn't need proving right

[next obsidian]

I mean the proof is easy

[pastel cliff]

like that seems kinda obvious...?

[next obsidian]

Oog

You better know how to prove it

[dim widget]

It depends on how you define F(a), it could be by definition.

[next obsidian]

I am now less convinced because you said it just seems obvious

But the proof is

F(a) ≈ F[x]/(m_a(x))

By x -> a

But if b is also a root of that polynomial

You can take the iso x -> b

And now you chain the isos

[formal ermine]

my prof's lecture notes are written in a way that I first need to copy them over to my notes to understand them

his hand writing is uhhhh

interesting

it's hard to decipher it

[rustic crown]

imagine profs encrypting lectures notes for fun

[formal ermine]

this prof at least writes small

[pastel cliff]

smallest field containing F and a

but i see what chmonkey said

[formal ermine]

my other prof writes like 2 sentences per page

[next obsidian]

My NT prof had decent handwriting

[formal ermine]

it's a scrolling nightmare

[next obsidian]

He just likes to stand in front of what he was writing

[formal ermine]

it's readable but it's hard for me to understand stuff

[next obsidian]

I’m noticing another issue with your prof’s notes

It’s in German

[formal ermine]

[next obsidian]

And I can’t read German

[pastel cliff]

borrowed some topology notes from a friend...

[next obsidian]

Lmfao

Is that…

English

[pastel cliff]

i tried plugging it into like an AI image to text thing

it returned gibberish

[ebon pine]

Whenever we say G has a normal subgroup are we talking about proper subgroups?

[pastel cliff]

not necesarily

[ebon pine]

cause trivially {e} and the group itself are always a normal subgroup

[dim widget]

You want a definition such that the kernel of any homomorphism of groups is a normal subgroup, so you should include "improper" subgroups.

[ebon pine]

hmm.. am I missing something here then? The answer key says only (b) is correct

[pastel cliff]

maybe review how your prof defined it

[dim widget]

It should say "nontrivial normal subgroup" but you might have a different definition.

[ebon pine]

This is a competitive exam so there's no way of knowing what they actually meant

It's certainly not mentioned in the notation or terminology section

[formal ermine]

this is what my notes look like

[pastel cliff]

.

[long nebula]

The real problem is you're doing algebraic number theory

[wicked zephyr]

What is $k[t]\otimes_{k[t^2]} k[t]$?

[cloud walrus]

Potitov06

[dim widget]

Have you tried substituting R = k[s] with s = t^2 and trying to write S = k[t] over R?

As a finitely generated algebra

[wicked zephyr]

Like, $k[t]\cong k[s][1,t]/(s-t^2)$, something like that?

[cloud walrus]

Potitov06

[wicked zephyr]

well, forget the 1

[dim widget]

Yes, so what is R[t]/(t^2 - s) \otimes_R R[t]/(t^2 - s)

[wicked zephyr]

R[t]/(t^2-s)?

[pastel cliff]

what is the motivation for talking about separability

[wicked zephyr]

It measures the different ways a field can be embedded in its algebraic closure

At least that is the way I think about it

[dim widget]

We talk about it so that students can learn about it.

[wicked zephyr]

Is this correct?

[pastel cliff]

this a result of the correspondence between roots of polynomials and the different embeddings?

wait that's too vague

[wicked zephyr]

It seems weird

[dim widget]

It shouldn't be that, that would be the tensor product with R

What if R is a field (it is not) could you figure out that one?

[next obsidian]

The separable and embedding into alg closure correspondence is whacky

I don’t have good intuition for it

[pastel cliff]

det button

[wicked zephyr]

If R is a field then it should be k[t]/(t^2-s)

k=R

sorry

Sorry, i am not thinking

[dim widget]

Inseparability is saying that a field has "hidden nilpotents" that you see over the algebraic closure.

I.e. inseparable field extensions geometrically are "thicker" than regular field extensions.

[pastel cliff]

whar

[next obsidian]

This is so insanely hard to justify

Tfw geometric reducedness type beats

The proofs become some insane field theory that I’ve only ever found written down in Bourbaki

[dim widget]

I don't really know on what level you want an answer I guess

[pastel cliff]

level

[next obsidian]

Honestly, until you have a point separability starts to play a role

I think you should just shove it in a corner of your brain

It shows up in tricky situations in weird ways

Like I could start to try to talk about differentials

But it would not mean anything unless you knew why you care about them

[pastel cliff]

the following section of notes is on "thm of primitive element"

[dim widget]

If you're asking about why it comes up in Galois theory: it measures the defect in embeddings in the algebraic closure to fully distinguish elements in the field. This also manifests in the lack of "enough" automorphisms of that field extension.

[next obsidian]

A formal way to see what she’s talking about is that separable iff # of distinct embeddings of F (fixing K) into K-bar = [F:K]

[delicate bloom]

interesting

[next obsidian]

A purely inseparable extension in contrast has only one way to embed F into K-bar

[delicate bloom]

nice to hear this stuff talked about so lucidly, shame I don't have time to think about this stuff for the forseeable future

[next obsidian]

For me separable means not inseparable, but this is not stupid because inseparable to me means

x such that x^p in K, but x not in K

Only in char p

This is bad because when you take a derivative you get pdx^p-1 = 0

But you don’t want that derivative to be 0, it shouldn’t be 0

For field extensions as well, it’s “formally smooth” iff the extension is separable

What does this mean…
I’ll tell you when you’re older

Okay here’s another example

You may not have realized, but tensor products of field extensions can be bad

C (x)_R C isn’t an integral domain, it’s C^2

[dim widget]

I think this is the real correct perspective but is unlikely to be appreciated by someone in a galois theory class. This is why, for instance, inseparable extensions can have infinitely many intermediate extensions.

[next obsidian]

But in general they can be so bad that you can even introduce nilpotence

[dim widget]

@wicked zephyr take note!

[next obsidian]

If F/K is separable though, F (x)_K L will never have nilpotence

In fact this characterizes separability

This is what TTEG was talking about with inseparability being about having hidden nilpotence

Note that this definition about having no nilpotence after base change makes sense even for infinite extensions, so one can define non-algebraic separable extensions

Which happen to be the same thing as “separable generated” extensions when the extension is finitely generated which you may have seen in an algebraic geometry book

This condition is saying that you can write F as a separable algebraic extension over K(x1,…,xn) for some transcendence basis x1,…,xn

Such an x1,…,xn is called a separating transcendence basis, but you can have non-separating transcendence bases even in a separable generated extension

Anyway, the point of me just typing to myself is to say that separability is important, but it’s hard to justify why

And you can only really appreciate it until something pops up that forces you to care about it

[wraith cargo]

repeated roots cause inequalities
Separability causes equalities 😎

[wicked zephyr]

I dunno

I mean, if k is a field, the $k[x]/(f)\otimes_{k} k[y]/(g)\cong k[x,y]/(f,g)$

[cloud walrus]

Potitov06

[wicked zephyr]

or that is false¿

[next obsidian]

Potitov, k[t] over k[t^2] is “adjoining a square root”

How would you write that down for field extensions?

[pastel cliff]

[wicked zephyr]

If k is a field, adkoining a square root of $a\in k$ is k[t]/(t^2-a)

[cloud walrus]

Potitov06
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[next obsidian]

Right

So let’s say k[t^2] = R

And let’s call the variable s for right now

What is your a in this case that we’re taking a square root of?

[wicked zephyr]

So k[t] is R[t]/(s-t^2)

[next obsidian]

Yup

[wicked zephyr]

Yes, I know

I wrote that down before

[next obsidian]

Wait no

It’s R[s]

[wicked zephyr]

wat

[next obsidian]

R[s]/(s-t^2)

[wicked zephyr]

But we are taking the square root of s

[next obsidian]

No

We’re taking the square root of t^2

[wicked zephyr]

yeah yeah sorry

that

[next obsidian]

So now we want to evaluate R[s]/(s-t^2) (x)_R R[s]/(s-t^2)

[wicked zephyr]

But wait

[next obsidian]

This is almost the same setup as evaluating C (x)_R C

[wicked zephyr]

But wait

I dunoo why i wrote but wait two times

You are saying R=k[s]

[next obsidian]

No

R = k[t^2]

[wicked zephyr]

ahhh ok

ok

R[s]/[s-t^2]

[next obsidian]

Yes

[next obsidian]

So we want to evaluate this