Mathematics - 2020-02-13 (page 2 of 3)

8:00 PM

I agree that we're not done if we directly try to apply the definition of purely insep.

but one can show that a simple extension $E(\alpha)/E$ is purely inseparable iff the minimal polynomial of $\alpha$ has only one root in a splitting field

anakhro

In which case here, (K(x))(y) = L

So L/K(x) would be purely inseparable by that "theorem".

Lukas Heger

yeah

the definition in terms of every minimal polynomial is not really useful to show that an extension has that property (the same holds for normality)

anakhro

Amen to that.

So I only need one direction of that "theorem", that the minimal polynomial of alpha having only one root in a splitting field implies purely separable.

Lukas Heger

actually I think it's easiest is to show that direction indirectly, first show that the minimal polynomial of alpha having onle one root implies that there's only one field homomorphism $E(\alpha) \to \overline{E}$, then use that to show that in fact every minimal polynomial has only one root

Thorgott

a similar characterization holds for, say, algebraicness (algebraicity?) and separability too

I have no clue what noun I'm supposed to use there

Lukas Heger

8:10 PM

in fact, the proof strategy for those three equivalent characterizations of purely inseparable is the same how you show that those three are equivalent:
- $E/K$ is the splitting field of a family of polynomials
- embedding $E$ into $\Omega/K$ any field extension, any $\sigma \in \mathrm{Aut}(\Omega/K)$ satisfies $\sigma(E)=E$
- for every $\alpha \in E$, the minimal polynomial of $\alpha$ over $K$ splits completely

in both cases, it's the easiest to go: some polynomials => field homomorphisms => all minimal polynomials => some polynomials

if that analogy makes sense to anyone

anakhro

Is this $E(\alpha)/E$ thing just like saying that if the only thing we are adding satisfies the definition of purely inseparable, then it is going to be a purely inseparable extension?

So it extends to a subset of things all satisfying the same property?

Like $E(\alpha_1,\dotsc,\alpha_n)/E$ is purely inseparable iff $\alpha_i$ all satisfy that property.

Lukas Heger

yes

in fact, purely inseparable is transitive in towers

so you get that directly from the $n=1$ case by induction

anakhro

Okay, but still yet to do that, so maybe I should sit and do it, now that it makes sense why we would have this anyway.

Thorgott

yeah, the analogy makes sense

anakhro

So when you say it has only one root, do you mean it has only one root in the splitting field, or over all one root?

Balarka Sen

8:16 PM

Separable degree is multiplicative. Purely inseparable is the same as saying separable degree is $1$.

anakhro

Oh wait

that's the same thing

:P

Lukas Heger

@anakhro the splitting field has all roots by definition

anakhro

Ne vermind

Yeah. One of those "doh" moments.

Thorgott

inseparable degree is multiplicative too :)

Lukas Heger

another definition is that a field extension $L \to K$ is purely inseparable iff it is an epimorphism in the category of fields :P

Balarka Sen

8:18 PM

You can completely characterize a finite purely inseparable extension as obtained as a tower $E=F_n/F_{n-1}/\cdots/F_1/F_0 = F$ where $F_i = F_{i-1}(\alpha)$ and $\alpha^p \in F_{i-1} \setminus F_{i-1}^p$.

$p$ being the characteristic.

Infinite you can say similar things because infinite extensions are direct limit of finite things.

Thorgott

I should learn some infinite Galois theory eventually

Lukas Heger

infinite Galois theory is very important for number theory, if you're into that

anakhro

So a map from $E(\alpha)\to \bar E$ must permute the roots of the minimal polynomial.

So it sends $\alpha \mapsto \alpha$

Balarka Sen

Right

And that determines the map completely (by map we mean $E$-homs here)

anakhro

But doesn't necessarily fix all of $E$. Or does it?

Well easily necessarily.

Lukas Heger

8:23 PM

we assume it's $E$-linear

anakhro

Oh.

Balarka Sen

Yeah everything's an $E$-homomorphism.

anakhro

So it does fix $E$ by assumption

Okay, so then I have my unique hom.

So then let $m(x)$ be another minimal polynomial, and I want to show it too only has one root, via this map.

Well if it had more than one, then I could come up with another map?

And that's it?

Lukas Heger

yeah

anakhro

$\blacksquare$?

Lukas Heger

8:25 PM

■.

anakhro

Excellent.

I think I am learning, guys!

Balarka Sen

Nice!

Thorgott

I wanna do some ANT at some point, but it looks like my algebra prof isn't giving that lecture any time soon and will give algebraic geometry instead in two semesters, so that's probably what I'll be doing then

Lukas Heger

@Thorgott if you do alg geo over stuff that is not algebraically closed, galois theory naturally enters the picture as well

Thorgott

sounds good to me

anything to do with Galois theory sounds good tbh

anakhro

8:36 PM

@LukasHeger so because the minimal polynomial $T^p-T-(y^p-y)$ of $y$ over $K(x-y)$ has roots $y+k$ for various $k$, we have that it is not purely inseparable. Right?

Lukas Heger

it's separable, even

anakhro

Yeah, that's the next part I am trying to address.

Lukas Heger

you have $p$ distinct roots for a degree $p$ polynomial

Thorgott

and a simple extension is separable iff its generator is separable

anakhro

Oh, indeed, so that follows still from the $y+k$.

Lukas Heger

8:38 PM

you could also compute the derivative, if you want

hey @Edward

how did you exams go?

Thorgott

a curious fact is that in characteristic $p$, $\alpha$ is separable over $K$ iff $K(\alpha)=K(\alpha^p)$

Edward Evans

Hey @Lukas, I didn't do either of them

how about yours?

Lukas Heger

oh I see

I still haven't had my L-theory exam

Edward Evans

how about adic spaces?

Lukas Heger

but everything else (seminar talks, adic spaces exam) went well

Edward Evans

8:39 PM

nice :D

@Lukas in fact I went to visit Herrn Kasten to give him an ärztliches Attest and both him and Herr Vogel came out of his office at the same time lol

Lukas Heger

convenient

Balarka Sen

@Thorgott Yeah, $K$ is char $p$, $K$ is perfect iff $K = K^p$.

Edward Evans

aye

they just told me to give it to Frau Kiesel but she wasn't in her office, so I left the Atteste in her door

Balarka Sen

That's why you can take the perfect closure by taking a direct limit of the Frobenius endomorphisms

Wait @Edward you are in Bonn yeah

Edward Evans

Heidelberg

Balarka Sen

8:43 PM

Ahh

Alessandro is in Bonn

Edward Evans

Indeed

Thorgott

Hmm, I know that, but I don't immediately see how these are related

Edward Evans

I'm in Heidelberg with @Lukas :P

Lukas Heger

he joined the Heidelberg algebra/NT gang

Edward Evans

the cool gang

Thorgott

8:44 PM

does that gang have more than two members

Balarka Sen

@Thorgott Same proof, essentially, that's all

LOL

lool

rekt

sav

@Thorgott at least four, easily

vesii

8:47 PM

Hi guys, I have a small geometric problem. Let $ABCD$ be a trapezoid ($AB||CD$) where the diagonals meet at $E$. We know that $\angle ABD=34^{\circ}$, $BC=9cm$ and $AC=12cm$. How to calculate $\angle BAC$? I tried the law of sines but without any success.

Can I get a hint please?

Thorgott

a powerful gang then

are you already taking protection money?

Lukas Heger

Hey @Ted

Edward Evans

In place of protection money we make people calculate regulators for us

Ted Shifrin

Hi @Lukas

Edward Evans

Hey @Ted

Ted Shifrin

8:54 PM

Hi @Edward

Thorgott

and here we just regulate calculators..

Balarka Sen

Nice

I wonder if I should apply outside after my bachelors

Too competitive

Edward Evans

@Thorgott haaaaa

Balarka Sen

A two year senior from my uni got in Bonn last year and he's doing algebraic geometry there

Edward Evans

nice

Pretty nice place to be doing alg geo lol

Balarka Sen

8:58 PM

Yeah guess so

Are there any topology crew at all in Bonn or Heidelberg

Lukas Heger

there's only one topologist at Heidelberg

Edward Evans

Banagl

Lukas Heger

but diff geo is pretty big here

Edward Evans

I think you've heard of him @Balarka ?

Balarka Sen

Oh I know Banagl

He does stratified spaces right

Lukas Heger

9:00 PM

yes

Balarka Sen

Cool

Mathphile

Is $x^{1/x}$ the inverse of $f(x)={}^{\infty}x$ for $e^{-e}\le x \le e^{1/e}$?

Thorgott

We only have one topologist and we got him from Bonn lol

Balarka Sen

Oh damn

youtube.com/channel/UCIfzjQH8_-z15ynAQ_TGz8g

I should watch this

Mathphile

${}^{\infty}x$ is infinite tetration

Lukas Heger

9:04 PM

@BalarkaSen Banagl is great lecturer

big on geometric intuition

Ted Shifrin

That should keep you busy for a while, a @Balarka

Rithaniel

My college apparently has an exchange program with Kaiserslautern

Balarka Sen

@Lukas Started watching

Edward Evans

@Lukas one of my Kommilitoninnen said he just draws a tonne of pictures and talks for ages

which sounds great because his accent is incredible

Lukas Heger

@EdwardEvans I mean yeah, but I think he's a great lecturer

he draws the right pictures and talks interesting stuff

Edward Evans

9:07 PM

yeah that's what I meant, but I didn't say it in a way that implied that lol

He'll teach alg top right?

Lukas Heger

yes, not next semster though

next semester he's busy with diff top 2

yeah semester 3

yes

Right

let's just hope alg geo 1, ANT 2 and alg top won't overlap

Edward Evans

9:08 PM

I'll have the "big 3" in Semester 3 then

alg NT, geo, top

aye

hope so

Ted Shifrin

Do they not announce the scheduled hours a term or two ahead?

Edward Evans

one term ahead

Lukas Heger

@TedShifrin just one term ahead

Edward Evans

like at the end of the semester

Lukas Heger

but certain things happen in regular cycles

except when they don't

like ANT2

Edward Evans

9:09 PM

lol right

Ted Shifrin

That's ridiculous. We used to plan ahead far more so that both faculty and students can get organized.

Edward Evans

but at least there's a seminar

Thorgott

I mean, I think plans further than that exist, but they're like the opposite of set in stone

Lukas Heger

do you feel like it was the right decision to come to Heidelberg? @Edward

Edward Evans

Yeah definitely, but my mental health was really poor even before I arrived and then it got worse in January, which is why I moved my exams

But I feel very much like Heidelberg is the place to be for my interests

Ted Shifrin

9:23 PM

Hanging around here is not good for mental health, I guess :)

Edward Evans

lol maybe

Ted Shifrin

It's OK ... you can ignore me :)

anakhro

Hi Ted.

Edward Evans

:D never!

anakhro

@LukasHeger I am a little stuck on computing the order of $Aut(L/K)$. I have found that it is a normal extension so far.

Ted Shifrin

9:31 PM

speaking of ignoring me, hi @anakhro

Balarka Sen

@TedShifrin I hang around here because of my decline in mental health. Correlation does not imply causation!

anakhro

HAVE I IGNORED YOU, TED? D:

Ted Shifrin

Are you sure, a @Balarka? But I'm glad to enjoy your presence again!

Lukas Heger

3 hours 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 $\sigma \in \mathrm{Aut}(L/K)$

Balarka Sen

@TedShifrin Thanks, glad to hear so!

Lukas Heger

9:33 PM

in our situation, we can use the tower $L/K(x-y)/K$

the argument above gives $\mathrm{Aut}(L/K)=\mathrm{Aut}(L/K(x-y))$

but $L/K(x-y)$ is Galois!

so you know the order of that automorphism group

@BalarkaSen I agree with Ted, glad to have you back

anakhro

@LukasHeger is the answer we are looking for $[L:K(x-y)]$, or can we literally find this index?

Edward Evans

@Balarka there is no other who so appreciates my music taste than thou

Balarka Sen

Thanks! Means a lot, @Lukas

Hahah @Edward, I can say the same

Ted Shifrin

Now if we only had robjohn, Daniel F, Pedro, and our former room owner whose anonymous name I've forgotten

Lukas Heger

@anakhro I think we already know that this degree is $p$, right?

Ted Shifrin

9:36 PM

<--- does not appreciate weird musical tastes

Edward Evans

don't JUDGE

anakhro

@LukasHeger perhaps you do, but I am catching up ;)

Edward Evans

@Balarka tbf, speaking of declining mental health, I'm not sure DSBM is a particularly helpful genre

Balarka Sen

@EdwardEvans How much Mastodon have you listened to

Edward Evans

hahaha

a lot

well

the two albums The Hunter and Once more round the sun

I can listen to on repeat

lol

Balarka Sen

9:38 PM

Ah, my favorite is Crack The Skye and Cold Dark Place EP

@EdwardEvans Hah I guess

anakhro

I found a new band recently called The Comet is Coming. It's like a electro jazz trio comprised of a saxophone, drums, and synth/keyboard.

The Comet Is Coming: NPR Music Tiny Desk Concert

►

Pretty high energy. Skip to 1 minute to hear non-noise

@LukasHeger it's $p$ because it's cyclic of order $p$ (being Artin-Schreier?)

Edward Evans

@Balarka mine are, I think, blasteroid and thickening

and the motherlode

Lukas Heger

@anakhro $[L:K]=\mathrm{Aut}(L/K)$ iff $L/K$ is Galois for finite extensions

Edward Evans

especially the video heh

Balarka Sen

I love The Last Baron from Crack The Skye

its like 5 songs

anakhro

9:44 PM

@LukasHeger yes, that's what I am appealing to.

Edward Evans

oh yeah lol

I like oblivion fro mthat album

Balarka Sen

thats a good one as well

anakhro

That leaves one last part to this question.

Orbits of $x$ and $y$ under the group $Aut(L/K)$.

Lukas Heger

$x$ is easy, you know that because of the general Artin-Schreier theory

for $y$, use that description of the orbits of $x$ and the fact that $x-y$ is fixed by all automorphisms, because as I argued $\mathrm{Aut}(L/K)=\mathrm{Gal}(L/K(x-y))$

anakhro

Would be good if I actually knew Artin-Schreier theory. I only know that if I have a polynomial $x^p-x+a$ over a field of characteristic $p$, then it either splits, or is irreducible with cyclic Galois group.

Balarka Sen

9:51 PM

@Lukas @Edward Banagl is so good!

He's so clear

Lukas Heger

@BalarkaSen yes. I head the pleasure to learn alg top from him

Ted Shifrin

Then how come you don't draw pictures, @Lukas? :D

Lukas Heger

I drew pictures in the alg top course :P

now I'm doing more algebra stuff

Ted Shifrin

Blah. :D

Balarka Sen

I have to savagely devour these 10 lectures

Ted Shifrin

9:53 PM

Algebraic geometers draw pictures, too, @Lukas :)

Lukas Heger

yeah

our algebraic geometry course had pictures as well

not so much in the exercises though

it was a bit algebra-heavy I guess (surprising, if you start with schemes in week 3)

Balarka Sen

@LukasHeger Tell me a proof of Zariski's main theorem sometime

Ted Shifrin

I hated the algebraic geometry course I took my last year as an undergrad. It was my 4th course from Mike Artin and he spent a quarter of the course on Zariski's Main Theorem. I just didn't get any of it.

LOL, and look what a @Balarka just wrote.

Balarka Sen

Lmfao

Damn

Lukas Heger

we didn't really do much in the course

Balarka Sen

9:56 PM

I think one of the highly influencial geometric consequences of the main theorem is that normal varieties have connected links at singular points, but the proof is completely dry and pages of commutative algebra

Lukas Heger

just some basic properties of scheme homomorphisms...

Balarka Sen

that's why I want to understand it eventually

Ted Shifrin

Well, I never did. But you can splain it to me.

If my brain still works by the time you understand it :)

Lukas Heger

@BalarkaSen I'd have to learn the proof myself first

but I as a mutant-algebrainlet do sometimes enjoy dry commutative algebra

Ted Shifrin

I think I should relearn some of that variation of Hodge structure stuff :)

Lukas Heger

9:58 PM

@TedShifrin oh yeah, teach me, I'll need it for that seminar

Balarka Sen

@TedShifrin I can in principle try to read this with help from some algebro geometers in my uni around.

Ted Shifrin

Well, we should discuss some, but I'm super rusty.

anakhro

@LukasHeger does it suffice to look at how the roots of $m_{x,K(x-y)}$ and $m_{y,K(x-y)}$ can be permuted?

Because $Aut(L/K) = Aut(L/K(x-y))$

there ought to be a LaTeX command for Aut

Lukas Heger

@anakhro you know it's cyclic, so just try to find a generator

Balarka Sen

$\mathrm{Aut}$

anakhro

10:10 PM

@LukasHeger oh in that case, it's just the +1 map?

Lukas Heger

right

but you need to $+1$ simultanously on $x$ and $y$

or else it won't fix the ground field

anakhro

Ah, that's true.

Balarka Sen

$\newcommand{Aut}{\mathrm{Aut}}$

$\Aut(K/F)$

@anakhro Fixed that for you

anakhro

Dirty tricks, @BalarkaSen

Balarka Sen

Write \Aut

anakhro

10:11 PM

I see what you did.

Balarka Sen

Might wear off if my newcommand thing goes out of the chatbox

anakhro

So then the orbits are just $\{x+k \mid 0\leqq k<p\}$ and similarly for $y$?

How does $\text{Aut}$ compare to $\Aut$?

Balarka Sen

$\renewcommand{\text}{Never\; use \;text\; as\; a\; replacement \; for \; mathrm\; you\; absolute\; noob}$

anakhro

Seems to make sense that way, but I am concerned about the fact that I am also adding to y, and that the orbits are the same.

Well, same style.

Lukas Heger

@anakhro what's concering about it?

Thorgott

10:14 PM

lol

anakhro

@LukasHeger just that a question asks for both of the orbits, but the orbits are near identical in solution.

Balarka Sen

@Thorgott Shhh

anakhro

@BalarkaSen not nice nelly

Balarka Sen

Now you shall never be able to use \text again mwahaha

I wonder when it'll wear off

Or if it will ever wear off

anakhro

$\renewcommand{\text}{Balarka\; is \; not\; a \; nice \; nelly}$

Balarka Sen

10:17 PM

I could define my macros for future use if this effect is eternal

$\renewcommand{\text}{\text}$

Let's fix that.

$\text{Aut}$

Hm, can't

anakhro

Balarka, you broke it. :((((((((

Balarka Sen

Im sorry

I failed everyone

anakhro

Hardly.

You just made it so that I can't use \text anymore.

Which I suppose was your original intention. :P

Balarka Sen

Actually I use \text a lot more than \mathrm

So let's hope this wears off

$\Z$

Yeah it definitely does wear off

Apr 26 '18 at 8:55, by Alessandro Codenotti

\renewcommand{\Z}{whatever you should have typed instead} :P

Oh wait

That's not in TeX

$\ZZ$

Dec 5 '11 at 20:24, by t.b.

test $\renewcommand{\ZZ}{\mathbb{Q}} \ZZ$

Fine

anakhro

finite fields kill me, Balarka

if i die, it was $\mathbb F_{p^n}$...

Balarka Sen

10:25 PM

I feel ya

Lukas Heger

finite fields are the easiest

Thorgott

yeah

by the way, does anyone here know a good reference for elementary aspects of Artin-Schreier-Witt theory?

Lukas Heger

@Thorgott Neukirch ANT gives it as an exercise :P

but I suppose you want to some minimum amount of group cohomology

unfortunately I don't know a good reference

Thorgott

yeah, I tried googling and all I could find was phrased in terms of group cohomology or commutative algebra, neither of which I know

guess that just means I'll have to learn them

thanks anyhow

Balarka Sen

group cohomology is dope

what is Artin-Schreier-Witt theory tho

Thorgott

10:34 PM

it generalizes Artin-Schreier theory to extensions of degree $p^n$ and has something to do with Witt vectors, more I don't know

there was a group cohomoloy lecture this semester, which sadly means we probably won't be getting another one for a while

anakhro

Are all of you as good at analysis as you are at algebra

because sheesh I suck at everything

Lukas Heger

Neukirch has an "abstract Kummer theory" which can be used to derive Kummer theory, Artin-Schreier theory and Artin-Schreier-Witt theory. unfortunately, he leaves the last point as an exercise

anakhro

Less so at geometry, I think.

Thorgott

that sounds really cool

man, there's so much algebra to learn

Lukas Heger

@Thorgott you're doing algebra 1 now, then algebra 2 and then alg geo?

anakhro

10:49 PM

If I have an irreducible polynomial $f\in\mathbb F_p[x]$ of degree $d$ and look at it over $\mathbb F_{p^m}[x]$, can I say something about how many factors $f$ splits into now?

I was helped by Leaky earlier for when $f(x) = x^N - 1$.

But in that case, I knew what $f$ looked like so it was a lot easier, and it played a lot off of that form (in particular, the roots of unity).

Lukas Heger

@anakhro yes you can

consider the roots $\alpha_1, \dots, \alpha_d$ of $\Bbb F_p$

for every $i$, you'll have $\Bbb F_p(\alpha_i)=\Bbb F_{p^d}$

now because $\Bbb F_{p^d} \cdot \Bbb F_{p^m} = \Bbb F_p^{\mathrm{lcm}(d,m)}$, we will have $\Bbb F_{p^d}(\alpha_i)=\Bbb F_{p^\mathrm{lcm}(d,m)}$

Thorgott

yeah, I think so

the lecture I'll attend next semester is called "Commutative Algebra"

but I think it's similar to your algebra 2 based on what Edward told me once

Lukas Heger

this implies that the minimal polynomial of $\alpha_i$ over $\Bbb F_d$ has degree $\mathrm{gcd}(d,m)$

Thorgott

and if alg geo is being offered next winter (as was implied recently), I'll take it

Lukas Heger

so using that this holds for any $i$, one sees that $f$ breaks up into $d/\mathrm{gcd}(d,m)$ irreducible factors, each of degree $\mathrm{gcd}(d,m)$

anakhro

10:55 PM

Is that product the composite field?

That is, the smallest field containing both of them?

Lukas Heger

yes

I meant to write $\Bbb F_{p^d} \cdot \Bbb F_{p^m} = \Bbb F_{p^{\mathrm{lcm}(d,m)}}$

Thorgott

oh, and I'm partaking in the seminar on rep theory next semester too, which I'm excited for

anakhro

Yup, I understood. :)

Balarka Sen

Hey hey hey what are the roots of $x^3 + x + 1$ over $\Bbb F_8$ quick

$\alpha$

$\alpha^2 + 1$ works

What's the third brah

Thorgott

generally, nice polynomials split into irreducible factors of equal degree in Galois extensions, which is pretty great

and by nice I mean separable and irreducible over the ground field

Transcript for

$Mathematics$ Mathematics