Mathematics - 2020-02-13 (page 1 of 3)

1:55 AM

Is a sequence of points in an open neighborhood around the origin of the topologists sine curve convergent without the closure?? I just took a break since I am learning about chow rings / Schubert calculus next....

anakhro

2:20 AM

@7einadh the topologist's sine curve is closed normally. Are you asking if you take the graph of $\sin(1/x)$ then a sequence converging to $(0,0)$ is convergent?

@LeakyNun if $\sigma$ is an automorphism, what does $\alpha^\sigma$ usually mean? I vaguely remember this being notation for something.

$\alpha$ is an element in the domain of $\sigma$.

Is it just application?

Leaky Nun

@anakhro I guess

7einadh

3:00 AM

@anakhro Yes, only with the sequence test for compactness.

@anakhro wait I mean, is there anything else the sequence may converge to, like a finite sequence....

anakhro

I do not understand what you are asking, @7einadh

7einadh

4:21 AM

@anakhro What does "take a sequence" mean in terms of limit points ^ _^

@anakhro or is an open neighborhood around the origin sequentially compact?

@anakhro

@anakhro in the usual subspace topology on topologists sine curve, without 0 x [ -1 , 1]

Nûr

7:20 AM

hello

Calvin Khor

hi Nûr

Nûr

I have a problem to show a basic thing : $U(n)$ is homeomorphic to $SU(n) \times \Bbb S^1$

Could you help me ?

I wanted to consider something like $M \mapsto ( \frac{1}{(\det M)^{1/n}}M, (\det M)^{1/n})$.

But to have continuity I should choose, say, the first n-th root all the time. But then it doesn't map to all the unit circle...

$M \mapsto ( \frac{1}{(\det M)^{1/n}}M, (\det M))$.

Is this one ok ?

And I choose the first n_th root for all

dsm

2:17 PM

@AkivaWeinberger Since we talked about some representation theory yesterday, perhaps you'd be able to help me out with my post here

Peter

2:29 PM

Anyone wants to factor (3^269+2^268)/292969606523244618961426978730022229 ? It has 93 digits and is composite. The quadratic sieve will be best since ECM did not give a factor yet.

anakhro

2:44 PM

@7einadh I am still not understanding what you are asking. Could you maybe state your question as if you were reading it from a math textbook (so formally and rigorously)?

anakhro

2:57 PM

@LeakyNun I dunno if you know some field theory, but what sort of things can we do to address finding the degree of the splitting field for $x^N - 1$ over some $\mathbb F_p$ where $N$ is large and $p\mid N$?

Leaky Nun

the degree of $x^N-1$ is $N$

anakhro

Sorry, I forgot some words. :P

Leaky Nun

:P

anakhro

Immediately, those Fermat's little theorem problems come to mind.

But I don't know how helpful it is.

Leaky Nun

firstly $x^p-1 = (x-1)^p$

so you can immediately remove all factors of $p$ from $N$

and end up with a number coprime with $p$

so let's say $N = p^k M$ with $p \nmid M$

then you only need to care about $x^M-1$

since $(x^M-1)^{p^k} = x^N-1$

and now a bit of finite field theory

there is only one extension for each degree

and that is $\Bbb F_{p^n}$, the splitting field of $X^{p^n-1}-1$

so you need to find the smallest $n$ such that $M \mid p^n-1$

that is, the order of $p$ mod $M$

anakhro

3:01 PM

Oh many I didn't even think about Frobenius.

Leaky Nun

from Euler's totient theorem, $M \mid p^{\varphi(M)} - 1$

so $n$ divides $\varphi(M)$

unfortunately computing $\varphi(M)$ is expensive

(as expensive as factoring $M$)

and there is no way to get around this

but once you obtain $\varphi(M)$ (which is usually a very composite number)

you can start dividing by its prime factors

etc

@anakhro ok?

anakhro

Let me digest this.

So why exactly can we reduce to $x^M - 1$? I get that we can use the Frobenius endomorphism to write it as $(x^M - 1)^{p^k}$, but why can we just look at the smaller one

Leaky Nun

because their splitting fields are the same

$f^b$ splits iff $f$ splits

anakhro

Ah, that makes sense.

Okay, maybe it might help just to pick an $M$. So how about $M=10$ and $p=7$.

So I am looking at $x^{10} - 1$ over the finite field of order 7.

Leaky Nun

ok

so you need to find the order of 7 mod 10

phi(10)=4

and 7^2 = 9 mod 10

so the order is 4

so the splitting field has degree 4

anakhro

3:11 PM

4 is that smallest $n$

Yes, great.

Okay, now I need to work on understanding the why.

Leaky Nun

$\Bbb F_{p^n}$ consists exactly of $0$ and the $(p^n-1)$-st roots of unity

so $x^b-1$ splits in $\Bbb F_{p^n}$ iff $b \mid (p^n-1)$

the category of finite extensions of a given finite field is equivalent to the categorified poset of $\Bbb N_{>0}$ under division

@LukasHeger ^ :P

anakhro

"there is only one extension for each degree and that is $\mathbb F_{p^n}$"

Each degree meaning each $M$?

Leaky Nun

degree referring to degree of extension

anakhro

Oh, I see. So $K=\mathbb F_{p^n} $ is the unique solution to the "equation" $[K: \mathbb F_p] = n$?

Leaky Nun

right

anakhro

3:20 PM

But this is just the splitting field of $x^{p^n-1} - 1$, as you said, and our polynomial is of that form.

Leaky Nun

well

not every number can be written as $p^n-1$

anakhro

Indeed, but we hope we can find a solution to $M=p^n - 1$?

Thorgott

Finite fields are nice.

anakhro

@LeakyNun but how does finding an $n$ such that $M \mid p^n-1$ amount to finding a solution to this?

Leaky Nun

@anakhro no, you can't find a solution to $M=p^n-1$

as the case $M=10$, $p=7$ demonstrates

21 mins ago, by Leaky Nun

so $x^b-1$ splits in $\Bbb F_{p^n}$ iff $b \mid (p^n-1)$

@anakhro $7^4=2401$

so $\Bbb F_{7^4}$ consists of $0$ and the $2400$-th roots of unity

so $x^{10}-1$ splits in $\Bbb F_{7^4}$

Thorgott

3:40 PM

The existence of an $n$ such that $M\mid p^n-1$ is equivalent to $p\nmid M$, so the initial factoring was actually important.

anakhro

I am struggling to see this.

I feel like I am missing something simple that makes this obvious.

Thorgott

Maybe that $x^b-1\mid x^{p^n}-1$ iff $b\mid p^n-1$?

anakhro

I want $x^{10}-1$ to split. It splits in when I have all the $10$-th roots of unity. The set of all $k$-th roots of unity contains all $10$-th roots of unity when $10\mid k$?

Leaky Nun

yeah

take a primitive 2400th root of unity $\zeta$

then $\zeta^{240}$ is a primitive 10th root of unity

Thorgott

If $\mu_n$ are the $n$-th roots of unity, then $\mu_n\cap\mu_m=\mu_{\mathrm{lcm}(n,m)}$

anakhro

3:48 PM

Okay, great. I think that was what I was missing.

anakhro

4:07 PM

So the reason the splitting field has to be $\mathbb F_{p^n}$ is because $\mathbb F_p$ is not extended by a field of any other form?

That is, the only way we get an embedding is if it is a finite field of characteristic p.

Thorgott

A field extension of $\mathbb{F}_p$ is necessarily a vector space over it, so if it's finite, it has cardinality $p^n$ for some $n$. For each $n$, exactly one such field (up to isomorphism) exists and it is the splitting field of $x^{p^n}-1$ in an algebraic closure of $\mathbb{F}_p$.

anakhro

That makes sense, thanks!

Leaky Nun

splitting field of $x^{p^n-1}-1$

geocalc33

> You assume a given family of non-intersecting analytic functions $\in\Bbb R\cap(0,1)^2$ are geodesics on some $2-$ manifold. You know that your family of analytic function obeys $f(0)=1$ and $f(1)=0.$

I think due to the fact that the geodesics converge to each other at two points implies that the manifold has global positive curvature and the metric is not flat.

I think to verify that the family of analytic functions are actually geodesics on some manifold you could test if they are solutions to the geodesic equation and/or try to express the family of analytical functions as solutions …

maybe the functions actually have to be projections of geodesics

anakhro

4:27 PM

@LeakyNun @Thorgott any good recommendations for a book on field theory or Galois theory? Or is D&F the way to go?

It's not a very stimulating read.

Leaky Nun

Ian Stewart

> "Pistols at 25 paces!" Bang!

Thorgott

I haven't read any books on the subjects, so can't recommend any. Keith Conrad has quite a few expository papers on field and Galois theory up on his website though; they're probably not a good primary source, but definitely a good supplementary source.

anakhro

Yeah Conrad is a great writer.

Edward Evans

@Thorgott Conrad's blurbs are incredible lol

Balarka Sen

Morandi

I started reading Szamuely and then stopped, I should continue eventually

Thorgott

4:34 PM

Yeah, I enjoy them a lot.

anakhro

Did Szamuely do a basic Galois theory book, or are you talking about his "Galois" theory book? :P

Balarka Sen

His first chapter is a quite lucid intro to basic Galois theory

I gave some talks amongst friends on the first chapter and covered more than my course has covered so far surprisingly

Oh also read Lang

That has two chapters, one on fields and one on Galois. Both very good.

anakhro

Like his grad algebra book?

Balarka Sen

Ya

Edward Evans

@Balarka I'll probably start reading Szamuely more for algebra this semester

so hmu

Balarka Sen

4:41 PM

Damn nice

I am up for it anytime

Edward Evans

boi

Balarka Sen

I should actually read Galois theory now

Edward Evans

I have algebra 2 and non-commutative algebra this semester

Balarka Sen

I will finish up my field theory assignment real quick and read Galois

Noice

Edward Evans

altho I imagine alg2 will just be homological algebra

Thorgott

4:42 PM

so both commutative and non-commutative algebra

Edward Evans

aye lol

the non-commutative algebra seminar will be some rep theory, Brauer groups, Galois cohomology

anakhro

I like the organization of the stacks project

the slogan tag thing is super nice

Balarka Sen

Oh yeah Stacks project is great for field/galois as well

I learnt separability and trdeg from there

anakhro

I am just having problems with field/galois theory for some reason.

Perhaps I am just not thinking about it enough

Thorgott

I have no clue what Galois cohomology is, but it sounds amazing

anakhro

4:45 PM

I wonder if Galois could see what his name has been attached to, if he would be happy or very scared.

Thorgott

@Balarka are there some cool applications of transcendence degree?

Balarka Sen

@Thorgott Every affine variety is birationally equivalent to a hypersurface in some affine space.

NoName

Galois scared? The guy died in a duel, for goodness sake!

Balarka Sen

Transcendence degree of function field of a variety being $n$ is basically having a finite map to $\Bbb A^n$.

anakhro

There has to be some mathematical pun about Galois and duals.

Balarka Sen

4:49 PM

@BalarkaSen Oh I don't need trdeg for this, just primitive element theorem.

Thorgott

I don't understand any of those words except "affine". Guess I'll have to do more algebra.

Balarka Sen

@anakhro He'd stop thinking about the damn girl he was after for sure at least.

Absolute idiot

anakhro

Heh

$K(\alpha) = K[\alpha]$ when $\alpha$ is algebraic, is it only when, as well?

Thorgott

Yes, you get $K(\alpha)\cong K(X)$ for transcendental $\alpha$.

And $K[\alpha]\cong K[X]$, analogously.

Balarka Sen

$1/\alpha \in K[\alpha]$ gives a polynomial over $K$ satisfied by $\alpha$, if you want to be concrete.

Working inside $K(\alpha) \supset K[\alpha]$

anakhro

5:03 PM

I am trying to wrap my brain around finding a minimal polynomial in some wacky looking field.

I have $\alpha = x^p - x$ and $\beta = y^p - x$, then I define $K = F_p(\alpha,\beta)$. Then I want to find the minimal polynomial of $y$ in $K(x-y)$.

Balarka Sen

You can always use rational canonical form to find minimal polynomial (in an algorithmic fashion) of an element in an extension: can you tell me how?

Thorgott

You always have a hom $K[X]\rightarrow K(\alpha)$. If $\alpha$ is algebraic, this descends to an iso $K[X]/(p)\cong K(\alpha)$ ($p$ being the min poly) by the universal property of the quotient. If $\alpha$ is transcendental, it lifts to an iso $K(X)\cong K(\alpha)$ by the universal property of the fraction field. So $K[X]$ lies "between" the algebraic and transcendental extensions somewhat analogously to how $\mathbb{Z}$ lies between positive characteristic and zero characteristic prime fields.

Balarka Sen

Oh huh

Strange field

anakhro

I thought so, too.

Balarka Sen

I don't get it. What is $x, y$?

Thorgott

5:06 PM

free variables?

Balarka Sen

You have $\Bbb F_p(x, y) \supset \Bbb F_p(x^p - x, y^p - y)$?

anakhro

Free variables, yes.

So $\alpha,\beta\in F_p(x,y)$

I noticed $\alpha-\beta = (x-y)^p$...

Didn't seem useful at this point, though

Balarka Sen

I'll leave this to someone else, gotta go

If nobody did it I'll come back in 10 and try

Thorgott

Guess I'll try and give this a think

anakhro

Like I mean you don't have to.

:P

Thorgott

5:09 PM

I have to prepare for my algebra exam either way, so this is probably good practice

Is $\beta=y^p-x$ or $y^p-y$?

Guess it doesn't make a difference if we're also adjoining $x-y$

anakhro

-x

Thorgott

This starts looking like Artin-Schreier to me

anakhro

That came up in another problem I was doing. I had not heard of it before.

I had to show that $x^p - x + c$ splits over a field of char. p, or it is irreducible with cyclic Galois group. It was cute and not so difficult.

Thorgott

$y$ gets annihilated by $T^p-T-(y^p-y)$ ($T$ being the new variable in the polynomial ring over $K(x-y)$)

so it suffices to check this has no zeroes in $K(x-y)$

anakhro

Sorry, what is $T$ exactly?

Just the indeterminant?

Thorgott

5:19 PM

yes

$T^p-T-(y^p-y)\in K(x-y)[T]$

anakhro

Okay, I see.

Balarka Sen

So, minimal polynomial of $y$ in $\Bbb F_p(x^p - x, y^p - y, x - y)$?

$y^p - y - (y^p - y)$ - oh that's what you wrote

Thorgott

This comes down to saying $K(x-y)$ contains no $p-1$st roots of unity, but I'm not seeing that rn

anakhro

To find that you guys are just trying to find a small degree polynomial that has y as a root, right?

Balarka Sen

Ya

Thorgott

5:26 PM

@Balarka What were you referring to earlier with the rational canonical form btw? I know you can get the minimal polynomial as characteristic polynomial of the left-multiplication map, but this seems to be neither practically useful nor what you were alluding to.

anakhro

Also I should stress it is y^p - x, not y^p - y

Thorgott

$(y^p-x)+(x-y)=y^p-y$

Balarka Sen

Ah, I misread, thanks, @anakhro

@Thorgott Observation: $T^p - T - y^p + y = (T - y)^p - (T - y) = S^p - S$ is what the polynomial looks like in $\Bbb F_p(x, y)$.

anakhro

@Thorgott yeah I just noticed when I was scrolling up that Balarka had wrote it -y, so I was making sure we were all on the same page.

Thorgott

and $S^p-S=S(S^{p-1}-1)$, so to conclude there are no roots we need to exclude the existence of $p-1$st roots of unity or I'm not sure this helps much

Balarka Sen

5:30 PM

The roots of $S^p - S$ in $\Bbb F_p(x, y)$ is exactly the subfield $\Bbb F_p$, right?

Because degree $p$, and everything in $\Bbb F_p$ satisfies that polynomial.

Thorgott

oh, true

Balarka Sen

"the fixed field of the Frobenius acting on $\Bbb F_p(x, y)$ is $\Bbb F_p$"

anakhro

$y+k$ is a root

Balarka Sen

So $T - y \in \Bbb F_p$, but $T$ is not of the form $y + \alpha$ where $\alpha \in \Bbb F_p$ because $y \notin \Bbb F_p(x^p - x, y^p - y, x - y)$.

Well, we have to argue that I guess, but that shouldn't be too hard.

Thorgott

if $y$ is in there, it's the entire thing

Balarka Sen

5:34 PM

Ya

Thorgott

but that also has to be excluded still

arguing by degree would lead to an exercise in circularity..

Balarka Sen

So finally all we need is to show $\Bbb F_p(x, y) \supset \Bbb F_p(x^p - x, y^p - y, x - y)$ is a proper extension, just to make sure we're on the same page.

Thorgott

agreed

if you send $(x,y)$ to $(x,x)$, you get $\mathbb{F}_p(x^p-x)\subsetneq\mathbb{F}_p(x)$

Balarka Sen

I think you can argue by degree. If $y = f(x^p - x, y^p - y, x - y)/g(x^p - x, y^p - y, x - y)$ for some polynomials $f, g$ in three variables, then $y g = f$ says degree of $y$ on one hand is $1\pmod{p}$ and on the other is $0 \pmod{p}$ -- oh that's a way better argument

anakhro

$T^p - T + c$ splits over the field or is irreducible and has cyclic Galois group

Balarka Sen

5:39 PM

Yeah that's true

Good call, everyone

Strange exercise

anakhro

I don't really follow on what you guys have shown, though.

Balarka Sen

To write it up precisely; We want to find minimal polynomial of $y$ as an element of the extension $\Bbb F_p(x, y) = K \supset F = \Bbb F_p(x^p - x, y^p - y, x - y)$. We claim $T^p - T - (y^p - y) \in K[T]$ is the minimal polynomial, and we see $T = y$ is indeed a root.

anakhro

We could just then show that $y\notin F$, right?

And then it follows by the fact I mentioned above.

Balarka Sen

Yeah, exactly, that's all.

Or note that $T^p - T - (y^p - y) = (T - y)^p - (T - y) = S^p - S$ and the roots of $S^p - S$ over $K$ are just $\Bbb F_p$

So $T^p - T - (y^p - y) \in F[T]$ splitting in $F$ would force $y \in F$.

anakhro

How would one show $y$ is not in $F$?

Just linear algebra?

Balarka Sen

5:45 PM

But if $y \in F$ then $K = F$, and Thorgott argues by considering the set map $K \to \Bbb F_p(x)$ sending $x \mapsto x, y \mapsto x$. Under this, $F$ goes to $\Bbb F_p(x^p - x) \subset \Bbb F_p(x)$.

If $K = F$, then $\Bbb F_p(x^p - x) = \Bbb F_p(x)$, contradiction.

Thorgott

the exercise seems rather convoluted, because the variable $x$ is essentially unnecessary

Balarka Sen

Yeah

anakhro

There are more parts to the question if you were interested.

Thorgott

not that it matters for the argument, but the map $K\rightarrow\mathbb{F}_p(x)$ is a hom

anakhro

(i) Find the minimal polynomial for x over $K$, and y over $K(x)$. Then (ii) prove that $F_p(x,y)$ is normal over $K$ and compute the order of $Aut(L/K)$. Then
(iii) describe the orbits of $x$ and $y$ under this automorphism group, and (iv) express the extension as a separable extension of a purely inseparable extension, and a purely inseparable extension of a separable extension.

That's the rest of the question.

I am willing to bet x is not unnecessary for at least one of those questions. :P

Balarka Sen

5:53 PM

@Thorgott How? In $\Bbb F_p(x, y) \to \Bbb F_p(x)$, $x - y$ maps to zero.

Homomorphisms between fields don't have kernel

Thorgott

oh wait, you're right

it's a hom between the polynomial rings, but it doesn't extend as a hom to the rational function fields

Balarka Sen

Right.

Anyway, being a homomorphism is completely unnecessary for us

@anakhro Oof, that looks like an annoying exercise.

Thorgott

Ok, so what's the thing about computing minimal polynomials algorithmically?

anakhro

6:08 PM

@BalarkaSen tell me about it.

I don't even know what a purely inseparable extension is, so I don't know why this exercise is being recommended.

qwertyguy

hi chat, i have a small doubt: if $f_n\rightarrow f$ in $L^2(\mathbb{R})$, can we say $f_n\rightarrow f$ in $L^2((a,b))$

anakhro

Maybe something about bounds on the number of roots.

qwertyguy

(where $(a,b)$ is any interval on the real line)

anakhro

@qwertyguy what doubt do you have?

Thorgott

The perhaps laziest explanation is that a purely inseparable extension is one with separable degree $1$. Equivalently, an extension $K/F$ is purely inseparable if $K^{p^n}\subseteq F$ for sufficiently large $n$. This may or may not be too unnatural of a definition if you consider that an irreducible polynomial is separable iff it is zero or a polynomial in $x^p$ in characteristic $p$ (inseparability is a purely positive characteristic phenomenon).

Turns out any finite extension can be put in form of a tower with one separable and one purely inseparable step, which justifies the terminology.

Lukas Heger

6:19 PM

@LeakyNun no

poset categories don't have non-trivial automorphisms, but objects in the category of finite fields do

qwertyguy

my gut feeling is that it is correct, but I can see $\mathbb{R}$ as a countable union of intervals, thus the integral over $\mathbb{R}$ is bounded by the series of integrals over the intervals, which a priori could not be convergence
but also this passage looks to me quite wrong, dunno I'm a bit confused

Thorgott

@qwertyguy $(a,b)$ is a subset of $\mathbb{R}$, can you compare the integrals of a non-negative function over two sets when one is a subset of the other?

qwertyguy

yeah obv, monotonicity

Thorgott

apply this to $|f_n-f|^2$

qwertyguy

i see

anakhro

6:25 PM

@Thorgott your argument says under this map $F\mapsto \mathbb F_p(x^p - x)$, but that $K\mapsto \mathbb F_p(x)$?

qwertyguy

thanks guys, and sorry for the stupid question, i often get lost around woods

anakhro

@qwertyguy it's not a stupid question!

Lukas Heger

@Thorgott there's a slight subtlety here: for infinite exensions $K/F$ can be purely inseparable, but $K^{p^n} \not \subset F$ for all $n$, take $F=\Bbb F_p(t_1^p,t_2^{p^2},t_3^{p^3}, \dots)$ and $K=\Bbb F_p(t_1,t_2,t_3, \dots)$

qwertyguy

@anakhro you're right, as long as you don't know the answer everything is non-trivial

Lukas Heger

it's still correct to say that $K/F$ is purely inseparable iff for all $k \in K, k^{p^n} \in F$ for some $n$ though

@anakhro geometrically it's obvious that the automorphism group is $\mathrm{PGL}_3(\Bbb F_P)$ :P

anakhro

6:31 PM

@LukasHeger I do not see how that is obvious or geometric. :P

Lukas Heger

oh wait I thought you wanted $\mathrm{Aut}(\Bbb F_p(x,y)/\Bbb F_p)$

Thorgott

@anakhro yes

@Lukas thanks for the catch, I was being sloppy

anakhro

@Thorgott how is it that $K \mapsto \mathbb F_p(x)$?

I thought it would just map to $\mathbb F_p(x^p - x)$ as well.

Thorgott

any polynomial in $x$ is also a polynomial in $x$ and $y$ (essentially, you have $\mathbb{F}_p(x)\subset\mathbb{F}_p(x,y)$)

anakhro

$K = \mathbb F_p(x^p - x, y^p - x)$.

Thorgott

6:35 PM

Balarka is using $K$ to denote $\mathbb{F}_p(x,y)$

anakhro

oh.

So then this doesn't work for my problem?

Thorgott

it does, Balarka was just using different notation is all

Mats Granvik

chatoverflow

chatematics stack exchange

anakhro

@Thorgott oh! I see now.

Thanks.

I am just bad at reading.

@Thorgott I guess the analogous argument and polynomial for $x$ and $K$ works. This time you do not need x-y since you do not have to deal with the y^p - x thing.

Thorgott

actually, I don't like that argument anymore

here's a better one

you have $\mathbb{F}_p(x,y)\supseteq\mathbb{F}_p(x^p-x,y^p-x,x-y)$. The RHS is fixed under mapping $(x,y)\mapsto(x+c,y+c)$ for some $c\in\mathbb{F}_p$, the LHS is not, so the containment is proper

this is still Artin-Schreier in disguise, but a bit more concise, I feel

Lukas Heger

6:51 PM

@anakhro about the automorphism group: purely inseparable extensions don't change the automorphism group: $K(x-y)/K$ is purely inseparable, so the automorphism group of $\Bbb F_p(x,y)/K$ is going to be the same as the automorphism group of $\Bbb F_p(x,y)/K(x-y)$ Now $\Bbb F_p(x,y)/K(x-y)$ is actually Galois and we know that the degree is $p^2$, so we're looking for a group of order $p^2$.

anakhro

@Thorgott definitely more concise.

Is that like saying $L$ is purely inseparable over $K$ if $Aut(L/K) = \{1\}$? @LukasHeger

Lukas Heger

@anakhro that's a special case of the thing I mentioned yeah

actually it's equivalent

anakhro

No need for normality or anything like that?

Lukas Heger

I mean yeah probably

but $\Bbb F_p(x,y)/K$ is normal (at least we're supposed to show that)

Thorgott

$\mathrm{Aut}(\mathbb{Q}(\sqrt[3]{2})/\mathbb{Q})=\{\mathrm{id}\}$, but this surely isn't purely inseparable or am I missing something?

anakhro

6:57 PM

Maybe finite + normal?

@Thorgott the minimal polynomial of $y$ over $\mathbb F_p(x^p-x,y^p-y,x)$ is simply $T^p - y^p$?

Thorgott

purely inseparable is equivalent to $\mathrm{Hom}_K(L,\Omega)$ being trivial

where $\Omega$ is some algebraic closure of $K$

Lukas Heger

@anakhro it's correct if you add normal

anakhro

Or because I can set that as $(T-y)^p$, does it split? I need $y$ in there, then...

Thorgott

and this is equivalent to the automrphism group being trivial in case of normality, because the image of embedding a normal extension into an algebraic closure stays within that normal extension

Lukas Heger

what I wanted to say is this: Let $L/K/F$ be a tower of extension such that $K/F$ is purely inseparable, then $\mathrm{Aut}(L/K) \subset \mathrm{Aut}(L/F)$. Conversely, suppose that $\sigma \in \mathrm{Aut}(L/F)$, then as $K/F$ is automatically normal, $\sigma|_{K} \in \mathrm{Aut}(K/F) = \mathrm{id}$, thus $\sigma \in \mathrm{Aut}(L/K)$

Thorgott

7:01 PM

@anakhro I don't think $y^p$ is in that field

Lukas Heger

the minimal polynomial is going to be $T^p-T-(y^p-y)$

Artin-Schreier again

Thorgott

but this also works with $x$ and $y$ switched and that's how you should get that overall degree $p^2$

anakhro

@Thorgott $y^p - x$ and $x$ are in the field.

OH

I wrote it down wrong this time.

Lukas Heger

you wrote $y^p-y$ above

anakhro

Yeah, sorry. $\mathbb F_p(x^p - x, y^p - x, x)$

Thorgott

7:04 PM

I did an exercise a while ago asking to show that $F(x,y)/F(x^p,y^p)$ is purely inseparable of degree $p^2$. This feels like the uglier sibling thereof.

Lukas Heger

yeah then it's going to be $T^p-y^p$

and the minimal polynomial of $x-y$ over $\Bbb F_p(x^p-x,y^p-x)$ is likely $T^p-(x^p-y^p)$

so we get separability degree $p^2$ and degree $p^3$ overall

so what I said is wrong, the automorphism group has degree $p$, so it's cyclic

no something's wrong here

anakhro

story of my life

Lukas Heger

I was confused about the $x-y$ thing

the total degree is $p^2$ the separbility degree is $p$, the automorphism group is generated by the map $x \mapsto x+1, y \mapsto y+1$

anakhro

7:22 PM

The same trick doesn't work to show that $y\notin\mathbb F_p(x^p-x,y^p-x,x) = \mathbb F_p(x,y^p)$.

Lukas Heger

the minimal polynomial of $x$ over $K(x-y)$ is $T^p-T-x^p-x$, so $\Bbb F_p(x,y)/K(x-y)/K$ is an extension of separable by purely inseparable

@anakhro $\Bbb F_p(x,y^p)$ is the quotient field of $\Bbb F_p[x,y^p]$. $T^p-y^p$ is Eisenstein wrt to the prime element $y^p$

anakhro

Oh Eisenstein!

Thanks!

Lukas Heger

we still haven't proved normality. The following abstract argument works: suppose we have $E_1/E_2/E_3$ such that $E_2/E_3$ is purely inseparable and $E_1/E_2$ is normal, then I claim that $E_1/E_3$ is normal. Let $\sigma \in \mathrm{Aut}(\Omega/E_3)$ where $\Omega/E_3$ is some algebraic extension such that $E_1 \subset \Omega$, then we have in fact $\sigma \in \mathrm{Aut}(\Omega/E_2)$ and thus $\sigma(E_1)=E_1$ by normality of $E_1/E_2$

note that $\Bbb F_p(x,y)/K(x-y)$ is Artin-Schreier, thus Galois and $K(x-y)/K$ is purely inseparable as $(x-y)^p=x^p-y^p=x^p-x-(y^p-x) \in K$

here I used this previous result:

27 mins ago, by Lukas Heger

what I wanted to say is this: Let $L/K/F$ be a tower of extension such that $K/F$ is purely inseparable, then $\mathrm{Aut}(L/K) \subset \mathrm{Aut}(L/F)$. Conversely, suppose that $\sigma \in \mathrm{Aut}(L/F)$, then as $K/F$ is automatically normal, $\sigma|_{K} \in \mathrm{Aut}(K/F) = \mathrm{id}$, thus $\sigma \in \mathrm{Aut}(L/K)$

Thorgott

how do we show that $x-y\not\in K$

anakhro

So you are proving normality of $\mathbb F_p(x,y)/K$ via using the tower $\mathbb F_p(x,y)/K(x-y)/K$?

Lukas Heger

7:32 PM

yes

note that normality is not transitive in towers

but my argument above still works

because purely inseparable is much stronger than normal

@Thorgott I mean, just for proving normality, we don't need that. The trivial extension is purely inseparable

Thorgott

true, but we need it for the degree

Lukas Heger

we could use another tower for that $\Bbb F_p(x,y)/K(x)/K$ seems easier as you already gave a nice argument that $x \notin K$. To see that $y \notin K(x)=\Bbb F_p(x,y^p)$, use the Eisenstein argument I gave above

btw those two towers solve another part of the exercise

Thorgott

ok, and this forces $K(x-y)/K$ to have degree $p$, but I feel like there should be a direct argument

anakhro

There probably should be an argument for why F_p(x,y)/K is normal without appealing to purely inseparability, if that is what you mean.

Lukas Heger

I mean yeah what I did was probably overkill

still works, at least

Thorgott

7:40 PM

there should be a direct argument for $x-y\not\in K$, but a fixed field argument won't work, because of the pure inseparability

anakhro

I assume $F_p(x,y)/K(x)/K$ is a separable extension by a purely inseparable extension?

Lukas Heger

yes

because of the minimal polynomials we computed

and $\Bbb F_p(x,y)/K(x-y)/K$ is purely inseparable by separable

anakhro

What is the associativity in these statements, I am not quite understanding.

Let's call $F_p(x,y)=:L$

Lukas Heger

okay

$L/K(x)$ is purely inseparable

$K(x)/K$ is separable

$L/K(x-y)$ is separable

$K(x-y)/K$ is purely inseparable

technically, you only need to prove three of those four, the other follows by degree + separability degree arguments

if you want to be efficient

I think we basically solved all of the exercise

though we don't necessarily have the most direct arguments for each step

Thorgott

it feels convoluted, but so does the exercise

Lukas Heger

7:50 PM

yeah

I think $L/K$ is the splitting field of $T^p-T-(X^p-X)$ and $T^p-(X^p-Y^p)$

so that's a more direct approach to normality

anakhro

Okay, I think I am unclear on something.

Purely inseparable means that for every $\alpha\in L$, the minimal polynomial of $\alpha$ over the field we are extending has only one root in a splitting field.

I want to show that $L/K(x)$ is purely inseparable...

Lukas Heger

yes, that's one characterization

if the extension is simple, you just need to check the minimal polynomial of a generator

anakhro

Hold on and let me see if I can apply the minimal polynomials to any of the four things.

Can't do it for L/K(x) apparently.

Lukas Heger

the minimal polynomial of $y$ over $K(x)$ is $T^p-y^p$, as you noted

anakhro

And it only has one root, y.

Which is in F_p(y), the splitting field.

Lukas Heger

7:58 PM

the splitting field over $K(x)$ is $L$

I mean you can consider that polynomial over $F_p(y^p)$ if you want, but that's not really relevant to our situation. It still has only one root nonetheless

anakhro

Oh, because it has x in it already, it can't lose things.

Lukas Heger

right

anakhro

But isn't it then inconclusive whether it is purely insep.?

It just agrees with the definition, so far.

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