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21:00
shoot milnor an email on that issue
"Dear Milnor, you have used hashtag to denote cardinality in your differential topology textbook, please don't use cancerous notation kthnxbai"
"Oi Milnor,
Address him as Jacky boi
Ufcknw0tm8 with the #? Signed Amin Quickmafs"
@BalarkaSen Ohh when I meant "implicitly" I meant that in the proof itself he never stated (or so I thought because I didn't read the word "also" in the second line) that he let $y$ be a regular value for both $f$ and $g$ , so I just assumed he was trying to prove something stronger using a theorem he hadn't proved before (and expected the reader to fill in the details)
@EricSilva Address him as bruv
21:03
Actually no I won't even specify why
Why is the hashtag bad
I'll just be like "woah, ufcknw0tm8? Sincerely, Quickmafs"
@Daminark @EricSilva Stop it, please, stop
It's just strange, like I use |X| for cardinality
But that is bad in places where u use the bars for other things lol
e.g. differential topology
21:06
I use double bars for norm
And from now on I use it for absolute value as well
I don't have a set thing for norms cuz I live in analysis land where u need 20 norms at all times
I use single bar for measures of sets
Hi @mixedmath
Wooouuuuaaaaattt?
Oi Eric, ufcknw0tm8?
@BalarkaSen hiya
I thought Bass was bad with its m(A) business
21:09
@Daminark Says the one who uses thinking emojies to denote Riemannian manifolds
I mean it's pretty normal
But cardinality is just counting measure (if we don't differentiate between different infininties), so if you use single bar for cardinality, you use it for a measure
Only compact ones
@EricSilva like the extension $\Bbb Q(\sqrt[5]2,\zeta_5)$ of $\Bbb Q$?
@Daminark You should make a custom noncompact Riemannian manifold/thinking emoji
21:10
Esp in the math I like: that is, the math which uses measure theory a lot
@mixedmath How's it going
Non-compact Riemannian manifolds get a fidget spinner
@MatheiBoulomenos sure if ur a degenerate who uses the counting measure
Not like me le intellectual
@MatheiBoulomenos oh hey you're here
Mu is my preference for measure lol
21:10
@EricSilva why specify the norm you use, but not the measure?
By degenerate I mean a @Daminark -type
@BalarkaSen I'm doing pretty well. I'm travelling, and I have a time change that I'm getting used to. I'm keeping busy.
\mu(R^n) = \aleph_1 confirmed
@MatheiBoulomenos in stuff I usually do there are only two measures that matter
@mixedmath Ah. Safe journey
21:12
@Eric define the sum of two numbers to be the integral of...
My freshman analysis course defined an absolutely convergent series as a series which is integrable wrt to the counting measure ...
And lol I've seen only 3 measures in my life
Hausdorff, counting, and Lebesgue-Stieltjes
Omg that's horrible
Oh wait there's also Haar
@Daminark hausdorff is like a family of measures tho
21:13
Haar is also a family of measures
I mean I'm being extremely facetious
@MatheiBoulomenos my old measure theory professor wouldn't buy that, she defined integrable as having either its positive or negative part with finite integral
Marianna?
I guess u only have taken 1
Meaning it's still okay for a function to have infinite integral as long as there's no infinity minus infinity business
That's a weird definition of integrable to me
21:15
Yeah Marianna. Soug, on the other hand, just says integrable = L^1
Like most analysts
my one analysis prof actually proved the Minkowski inequality in a more general setting involving two different measures. The usual triangle equality follows from that if you take one the measures to be the counting measure on a two-element set
Oh Lord
i hate that lol
You must have had a fun prof lmao
21:18
my other prof used dominated convergence for the counting measure to prove that you can rearrange the terms in an absolutely convergent series ...
I think you can use DCT to prove that the harmonic series diverges btw
it's like bringing a spear to a knife fight
G embeds into S_n = there is a _ group action from G to a set of n letters
faithful?
yes, faithful
21:19
Yes.
Flippety flip
Sniped
Snipe and be sniped is my motto
what do you call a group action where gx=x implies g=e?
Actually those actions can be quite expensive
lol finally you aren’t sniped
smacks Demonark
21:21
Wait hold on not you too!!!
Leaky can't smack, that power only belongs to Ted.
someone should banish Daminark to the trash
I will never leave the Barn!
Oh lol speaking of
Soug dragged me so hard yesterday
lmao what did he say
get soug'd
21:23
Once he noticed me he was like oh I didn't see you, did you go home over the weekend? I'm like yeah, then he was like "Thank God"
I smack whomever doesn’t smack himself
lol
Luis dragged the entire uchicago undergrad math body
@Daminark how very sougalicious
@LeakyNun so are you smacking yourself?
@Eric for a moment I read "dragged" as "drugged"
and was confused and terrified
21:25
Everyone cracks up, and then later he ran out of chalk and was like "Yeah people say this university is rich but it can't even get chalk. Two rooms down there's a table missing a leg"
He went on to explain that it was because the math department controls the rooms, so while they'd look nice if the University had control, you'd have classes in SSR"
i want all my classes to be in one hallway for all time
"Maybe I'll give you as homework to find a stupid leg and fix the table." glances at me "Wait no you can do it, that'll be your rent"
@MatheiBoulomenos that’s the point
what a meanie
21:27
soug is an ok guy
he just wants to watch soccer crack open a cold one and do PDEs
2
I love him. Like I think we can say we're bros
i might ask him to do a reading course before i graduate
@MatheiBoulomenos degree 5 polynomial S5 with exactly one real root?
Oh that could be fun. I don't think I'll have time for that
And I think I'm more inclined to do one with Marianna anyway :P
either him or luis cause i love luis
21:31
Hi @Ted
Hi @Balarka, Eric, Demonark, Leaky
Hi @TedShifrin
the Argentinian Avenger
hi @Ted
LOL at that description, Eric ... I don't think my students ever quite said that about me :P
Lol Luis' beer skit was funny
21:32
hi @Mathei
How's it going @Ted?
@Daminark his lectures in my elliptic pde class are hilarious because every once in a while he throws potshots at the postdocs and profs in the audience
Doing OK, Demonark. Pondering ways to challenge my best AoPS student.
Lol nice. Anything on Neves yet?
@EricSilva: I had a colleague audit my diff top class once years ago, and I repeatedly had to ask him to restrain himself from getting carried away answering my questions to the students.
21:33
no he mostly doesent pick on the full professors who are sitting
he constantly drags the instructors and assistant profs
Well, I didn't want the professor depriving the students of the pleasure of engaging meaningfully :)
I dunno if that's so wonderful, ERic.
@Ted there's more faculty in the class than students in this case
If you're gonna tease faculty, be an equal opportunity tease.
You shouldn't let off the older guys.
Lol, I've only seen a few prof disses
it's mostly the younger faculty in the class
Neves is sitting in but always shows up late
@Daminark the noncompact ego drag on souganidis was the greatest one of all time
21:36
I had faculty sit in on my grad diff geo class one time, and one guy was irrepressible but made it all about him instead of about the actual students, so I really had to shut him up. The other faculty member made insightful remarks that were definitely helpful to the students.
Any math happening here?
Schlag one time gave what was not even really a diss at Peter May, and one time Peter May was like "I was working with Souganidis on the admissions committee, and, get ready you won't believe this, he was a pleasure to work with"
@Ted the faculty tends to sit in the back, mostly it's me and grad students that answer questions, so that's fine
@Daminark Schlag has called Peter May "He-who-shall-not-be-named" on multiple occasions
Oh oh ... is Leaky trying to usurp my smack powers?
I mean to be fair, he later on said that they didn't disagree on anything, which is extremely rare in general
@Ted when I go home soon I'll probably read some AT
@TedShifrin I had this vague sketch of an idea yesterday while asleep
21:39
@LeakyNun $x^5+x^4+2x+1$ should do it
We were talking about $\Bbb P^1 \times \Bbb P^1$ and $\Bbb P^2$ and how blowing up the latter twice and blowing down once gives the former, right?
@MatheiBoulomenos hmm
where is it prime this time?
Heya @TedShifrin :)
Notice that if you delete the meridian & longitudinal $\Bbb P^1$ from the former and the line at infinity $\Bbb P^1$ from the latter, you get the same object: 4-ball.
@LeakyNun just plug in $2$
you get 53
21:41
right
I miscalculated lol
the argument why the Galois group is $S_5$ is very similar to the last one
Sure, @Balarka. That's pretty much equivalent to the picture I draw.
hi @Perturbative.
@TedShifrin Right, and the picture is this: blowup the former at the intersection of the meridian and the longitude and then blowdown the preimage of the meridian and the longitude (which then becomes -1 curves).
@MatheiBoulomenos hmm?
One fact about subgroups of $S_5$ is that the only ones which have an order divisible by $30$ are $A_5$ and $S_5$
21:43
That's the birational morphism $\Bbb P^1 \times \Bbb P^1 \dashrightarrow \Bbb P^2$
Wait, I want to blow down the proper transform of the line joining the two points.
@TedShifrin You're thinking of going from the latter to the former.
I'm going from P^1xP^1 to P^2.
Oh, right, you're starting at $\Bbb P^1\times\Bbb P^1$ and I'm starting at $\Bbb P^2$.
@MatheiBoulomenos oh, cube root of unity
21:44
modulo 2
If you reduce $x^5+x^4+2x+1$ modulo $2$ you get $(x^3+x+1)(x^2+x+1)$
this tells us that the Galois group contains a cycle of the type (abc)(de)
i.e. an element of order $6$
and then discriminant again :D
21:46
Galois theory is dank
@TedShifrin So now I'm wondering if this phenomenon is more general. I'm 70% sure that if you take a surface $\Sigma_g$ of genus $g$ and look at the $2g$ curves generating $H_1$, and blowup each point of intersection of the pair of curves on each handle, then the curves become $2g$ collections of $-1$ curves, blowing down which would give you $S^2$. (By blowup/blowdown I mean real blowup/blowdown, with RP^2)
it works so perfectly
Galois theory is great
(for polynomials with small degrees and small coefficients
runs away
but please don't ask for a degree $6$ polynomial with Galois group $S_6$
$S_6$ has a lot of subgroups
21:47
Let’s ask for a degree 5 Z5 polynomial
so I fear that similar arguments are unlikely to work
@MatheiBoulomenos: It's not a standard topic in most undergraduate/graduate courses that introduce Galois Theory that you can validly detect cycle structure by reduction mod appropriate $p$. I've forgotten the name for that.
yup that’s it I’m including the galois group into the adjective
Oh, ugh, Balarka, I never think about real blow-ups.
@TedShifrin I don't know any name other than reduction mod $p$.
21:48
@MatheiBoulomenos didn’t you send a link before?
We covered it without proofs in one exercise in my undergrad abstract algebra course
And I covered it with proof in the Galois theory coursera course
it's really useful
we did that in my galois theory course but everything ive learned ive already forgotten
good for you
You need to make sure the prime isn't ramified, though, right, @Mathei?
@MatheiBoulomenos that’s the point. it’s a computational nightmare and isn’t really useful in general
21:49
I would never forget Galois theory
@TedShifrin Well, be my guest to find more examples of similar nature, where you have two complex surfaces and a birational map $f : X \dashrightarrow Y$ which is defined on an open set $U$ of $X$ such that $X - U$ consists of a bunch of curves in $X$ which intersect each other at various points, and in fact it turns out that blowup of $X$ at the points of intersection, followed by blowdown of the proper transform of those curves, is $Y$.
It's in Lang's Algebra. That's where I learned it.
I am pretty sure this holds much more generally.
If we can settle this for surfaces I can ask higher dimension :)
@TedShifrin yes, but if you're already reducing the polynomial and factoring it, that's easy to verify
Oh, that's all over algebraic geometry, @Balarka. For example, you can create the cubic surface by starting off blowing up six general points in $\Bbb P^2$, I think.
The most amazing thing I discovered teaching algebra was, @Mathei, that the generic polynomial in $\Bbb Z[x]$ that is irreducible /$\Bbb Q$ is nevertheless reducible mod every $p$. This follows from Tchebotarev density.
21:51
Ah
@MatheiBoulomenos what should I look for?
@LeakyNun just make sure that the factorization mod p doesn't have multiple factors
@MatheiBoulomenos, @Leaky: Show that $x^4-10x^2+1$ is irreducible/$\Bbb Q$ but reducible mod every $p$.
I did that for $x^4+1$ once
@MatheiBoulomenos but you can’t get that from that
21:52
Right. That's the other standard example. But, amazingly, it's seriously true for a.e. polynomial :P
you had to use multiplicativity of Legendre symbols somewhere
Yes, @MatheiBoulomenos, although I am not a number theorist and never think about Legendre symbols. But, yes, basic group theory helps. :P
@TedShifrin and its Galois group is V4?
Yes, that's true.
Not germane to this discussion, however.
@Ted Do you have a counterexample?
21:53
A counterexample to whom?
To the phenomenon I was mentioning.
Of two varieties that are not birational?
No, no. Let me be more precise.
Hello!!!
Give a birational morphism $f : X \dashrightarrow Y$ of surfaces such that $f$ is defined on a Zariski open set $U$ of $X$ such that $X \setminus U = \{C_i\}$ is a finite union of curves, but nonetheless if you blowup $X$ at each of the various points of intersections of $C_i$, and blowdown the proper transform of each $C_i$, you do not get $Y$.
21:57
Oh, well, there are certainly examples where you have to do higher-order blow-ups.
Ah, to remove the intersections of $C_i$, right? I see
What if I demand the intersections are all transverse?
You just need to blowup once to remove transverse intersections
No, but imagine that you blew up a point on the exceptional divisor after doing one blow-up. That doesn't fit your paradigm.
The usual birational geometry game has blow-ups along blow-ups. ... :)
Oh, but you don't blowup intersections with exceptional divisors. Just the intersections of $C_i$'s. Eg, you simply blowup P^1 x P^1 at the point of intersection of the meridian and the longitude, not further on the intersections with the exceptional divisor.
I want to show that when $0<|a|<1$, that then the set $\{ 1+a+ \dots+a^k+ \dots+ a^n | n \in \mathbb{N}\}$ is bounded.

We have that $\sum_{i=0}^n a^i=\frac{1-a^{n+1}}{1-a}$.

We know that when $|a|<1$, then $\lim_{n \to +\infty} a^n=0$.

Does the fact that $\lim_{n \to +\infty} \sum_{i=0}^n a^i=\frac{1}{1-\alpha}$ help to show the desired result?
@TedShifrin I think that's for higher dimensions, when intersections happen on a subvariety instead of points.
22:01
So you aren't allowing me a birational construction outside of what you delineated?
@Evinda It certainly is bounded above by $\dfrac1{1-a}$ as you have stated
@evinda: You should know how to answer this by yourself.
@TedShifrin Consider $\mathbb{Q}(\sqrt{2})/\mathbb{Q}$ and $\mathbb{Q}(\sqrt{3}/\mathbb{Q}$ these are both Galois extensions of degree $2$ We also have that $\mathbb{Q}(\sqrt{2}) \cap \mathbb{Q}(\sqrt{3}) = \mathbb{Q}$ (For example by comparing discriminants).
As $\operatorname{char}(\mathbb{Q}) = 0$, this implies that $\operatorname{Gal}(x^4-10x^2+1) \cong \mathbb{Z}/(2) \times \mathbb{Z}/(2)$. The roots of $x^4-10x^2+1$ are $\pm \sqrt{2} \pm \sqrt{3}$ (all four combinations of signs). Using the isomorphism from the product theorem for Galois groups, we see that the Galois group acts transitively on these roots, so $x^4-10x^2+1$ is irreducible
@MatheiBoulomenos lol what
you hardly need Galois theroy to prove that it is irreducible
22:03
@MatheiBoulomenos: I can certainly argue irreducibility /$\Bbb Q$ in a totally elementary way, but, yes, you can infer it from Galois theory.
but that's eye-opening for me as well
I'm too lazy to do it in an elementary way :P
LOL
@Balarka: I never really thought much about birational geometry.
@TedShifrin: I don't really have a lot of examples to play with, which is why I wondered if you perhaps thought about it before.
I can think about real blowups but it appears to be true in most of the examples I looked.
I'm just way too far removed from this stuff to have any intelligent comments off the top of my head.
22:06
That's fair. Thanks, though.
Did you look at the Whitney umbrella?
Maybe I'm trying to prove something like the zig-zag theorem for low dimensions, I dunno :P
I dunno this stuff.
@MatheiBoulomenos let's say $f$ has exactly one real root and is irreducible and is of degree $5$. Then, The splitting field of $f$ over $\Bbb R$ has an automorphism given by conjugate which transposes $2$ pairs of roots. Therefore, the splitting field of $f$ over $\Bbb Q$ has an element of cycle type 2+2. What is this theorem?
@TedShifrin I did. Can't say I understand your paper fully though.
22:08
Well, I can't say I do either now. ... But the various notions of blow-ups of Gauss maps are interesting.
@LeakyNun I'd say that's hardly a theorem. That's just considering complex conjugation
@MatheiBoulomenos what's the justification for that?
@TedShifrin Yeah I saw the Nash blowup mentioned
I think I encountered that in Joe Harris's book first.
@LeakyNun as I said, just consider complex conjugation, restricted to the splitting field, this is an element of the Galois group
@MatheiBoulomenos what if I insist to consider $\Bbb R$?
22:10
Really $\mathbb{R}$ does not have much to with it
$\mathbb R$ is not algebraic over $\mathbb Q$
@TedShifrin @LeakyNun Is the following justification sufficient?

We want to show that the set $\{ 1+a+ \dots+a^k+ \dots+ a^n | n \in \mathbb{N}\}$ is bounded.

We have that $\{ 1+a+ \dots+a^k+ \dots+ a^n | n \in \mathbb{N}\}=\{ \frac{1-a^{n+1}}{1-a} \mid n \in \mathbb{N}\}$ .

Since $|a|<1$, we have that $a^{n+1}>a$. So $1-a^{n+1}<1-a$.

Thus $0<\frac{1-a^{n+1}}{1-a}<1, \forall n$, so we deduce that the set $\{ 1+a+ \dots+a^k+ \dots+ a^n | n \in \mathbb{N}\}$ is bounded. The minimum upper bound is $1$.
so it's not really relevant as a field for Galois theory purposes
I think I accidentally misdiscovered the Nash blowup once while thinking about blowups of cusps like $y^2 = x^3$ and $y^2 = x^5$.
22:11
Huh? @evinda $a^{n+1}>a$??
@TedShifrin lol you're faster than me
Also, remember that you could have $a$ negative.
I thought so since $|a|<1$ . @TedShifrin
Like, classically you have to blowup $y^2 = x^3$ once to desingularize it and $y^2 = x^5$ twice.
There's no apparent pattern behind that that I know of
What's cool, @Balarka, is that you see scheme structure rearing its ugly head. There's an embedded component in the blow-up.
22:13
But the number of times you have to Nash blowup to desingularize is the same as the smallest order at which the limiting tangents at (0, 0) differ from each other in the two branches of the curve
So, I guess 2 for the first and 4 for the second
@Balarka: I think there's a clear pattern.
So doesn't this hold? @TedShifrin
@TedShifrin Yeah
Try examples, @Evinda.
@TedShifrin Really? That being?
22:15
Start with the discrepancy between the two exponents.
$y^m = x^n$, so assuming $m<n$, we set $y=xu$ and we end up with $u^m = x^{n-m}$.
Now we have to decide which is greater, $m$ or $n-m$, etc.
Yes, y^2 = x^5 blows up to y^2 = x^3 (and another component corresponding to the exceptional divisor)
Yeah, we look only at proper transforms.
So now you've reduced the previous case. :)
But, eh, that doesn't explain the picture to me :) Locally y^m = x^n at the origin is like the cone of the (m, n) torus knot to me.
Why does #\barCP^2 reduce that to cone of the (m, n-m) torus knot?
22:18
Oh, you're trying to see the topological picture.
Milnor would have an answer immediately. That might even be in his Isolated Singular Points of Complex Hypersurfaces where he does Milnor numbers.
Ah. Maybe I'll check it out.
This is remindful of the gcd Euclidean algorithm game.
I gave away that book, so I can't check.
Let me note down the name of the book.
Ah, yikes, it's about 4. I should sleep now, have to wake up early tomorrow.
22:21
Night, @Balarka.
G'night, @Ted. Thanks for the nice conversation.
For once we actually talked about topology and algebra both, instead of the rest of the crew in this chat.
LOL ... g'night :)
It is a nightmare to compute Galois group for arbitrary quintic polynomials.
Math is a nightmare.
@Jasper no it isn't
Galois theory is dank
22:24
Dank is your new word, lol.
@LeakyNun @TedShifrin Oh no, it does not hold
It's a short-lived infatuation, I'm sure, @Jasper.
@TedShifrin why so?
0
A: Galois group of splitting field

JoanpemoYou can read about this nice stuff either in Milne's "Fields and Galois Theory", or in this nice paper by K. Conrad. Anyway, in a nutshell: Your polynomial's resolvent is $\;x^3-4x-1\;$ , which is irreducible (all the time over the rationals) as it has no roots there (Eisentein's), so its Galois...

exactly at which prime is $x^3-4x-1$ Eisenstein at?
Hi everybody
rip latex
22:28
@MatheiBoulomenos: Don't post that ... let other people work on it.
I am looking for a result in a French paper
@TedShifrin sorry
Any help will be appreciated.
LOL, kein Problem, @Mathei.
Wow, du kannst ja Deutsch
22:30
@MatheiBoulomenos du bist deutsch :o
Ja, ich hab' deutsch auf der Universität studiert.
I was thinking way too long about the factorizations
Very nice :D
@Mambo: You've hardly posed a question!
rip @Mambo
22:36
@TedShifrin I guess I mostly worry about my English while posting! I may phrase it so bad that people don't usually answer :(
@Mambo: But all you said is "I am looking for a result in a French paper." A little vague?
@MatheiBoulomenos wow the method of modulo p is really useful
@Ted
@LeakyNun I also can't see why this polynomial is Eisenstein. But it's easy to see that it is irreducible by reducing modulo $3$
but that's just when you're trying to establish lower bounds of group
@MatheiBoulomenos I've left a comment
do you have any insight on the Z5 question?
rip finding such a polynomial
22:37
I haven't thought much about it
@Leaky: Throw away "rip" while you're at it. You're overdoing that one, too.
inverse Galois theory is hard, you know :P
@TedShifrin rip rip
@MatheiBoulomenos I know
OK, what about them?
22:42
This has a construction of sequence $b_n$ positive tending to infinity such that $\sum b_n c_n$ is finite whenever $sum c_n(>0)$ is finite.
Oh that question again :)
Still stuck at this one :)(
If it's in Borel's lectures, why are you stuck?
I don't understand where exactly it is!
Leaky, he's trying to do this for an arbitrary convergent series $\sum c_n$ ...
But someone told you it's in there, @Mambo?
So I take it you don't read French?
22:44
I thought Fatou's phd thesis had it, but he referrred to this lecture notes, which I realized just now
Interesting.
yes, I don't read French
By skimming you should still see something that looks relevant :P
There seem to be so many relevances
I guess page 17 is the guy
@LeakyNun I'm sure you can find a polynomial of degree 5 with cyclic Galois group in this paper: ams.org/journals/mcom/1991-57-195/S0025-5718-1991-1079014-X/…
I give up
22:47
@MatheiBoulomenos lol alright
It looks like he's doing something with divergent on pp. 13-14 and making it diverge more slowly.
I haven't got to p. 17 yet. :)
p. 14 ... we're going to construct a series that converges more slowly ...
$x^5-110x^3-55x^2+2310x+979$ @MatheiBoulomenos the third example
Eisenstein at 11
yeah, I would've never come up with that
lol
Dummit is a genius
I find it a bit hard reading the scanned book, @Mambo. Some of the letters/symbols are very hard to decipher.
@Leaky: I bet you they used a computer, just as I did in writing some of my problems.
22:51
@Ted I will search for a more clearer copy
@TedShifrin still he's a genius
Anyhow, I think it's pages 12-14, @Mambo.
Have a look on this one
https://ia800302.us.archive.org/19/items/leonssurlessrie00boregoog/leonssurlessrie00boregoog.pdf
@Mambo: Try googling du Bois Reymond and see what you find. It seems to be his theorems.
@LeakyNun if you just want to know that such a polynomial exists, then you could argue that $\mathbb{Z}/(5)$ is solvable, hence by a result due to Shafarevich, it can be realized as a Galois group over $\mathbb Q$ :P
22:54
Oh, that's a much better scan.
@MatheiBoulomenos hmm
Hope :)
He gives a reference for du Bois Reymond, too.
Crelle, v. 76. I bet you can find stuff on-line if you google his name.
But it is all in those few pages.
That was more of a joke, by the way. That result by Shafarevich is out of reach for us
@MatheiBoulomenos I see
@MatheiBoulomenos let's just play with $\Bbb Q(X)$
$\operatorname{Gal}(\Bbb Q(X)/\Bbb Q)$ itself
wait I don't think it is normal lol
22:58
Well, you typically only write $\operatorname{Gal}$ if the extension is algebraic
@TedShifrin Thank you very much. I am doing that search right on.
@MatheiBoulomenos what about $\operatorname{Gal}(\overline{\Bbb Q}/\Bbb Q)$?
OK ... Hardy seems to refer to it in his book, too.
@LeakyNun that group is a mystery which many number theorists seek to understand

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