Mathematics - 2021-02-22 (page 4 of 4)

but I think the standard way is to take $(\sigma_\Bbb R x, \sigma_\Bbb C x)$ where $\sigma_\Bbb R$ is the real embedding and $\sigma_\Bbb C$ is the complex embedding

so it's a lattice inside $\Bbb R \times \Bbb C$

Semiclassical

should be $2^{1/3}1^{2/3}$ (or just $2^{1/3}\omega$)

Balarka Sen

9:08 PM

Yeah, C was actually correct

Also yeah I meant 2^1/3 * w

Lol this is a mess

Fix it, someone

Semiclassical

lol

isomorphic to Eisenstein is fine with me

Balarka Sen

$\Bbb Z[2^{1/3}]$ is the correct ring, what I did was not the correct lattice

$2^{2/3}$ is fine, but you should do the natural roots, yeah

Semiclassical

oh, hey

Cubic reciprocity

Cubic reciprocity is a collection of theorems in elementary and algebraic number theory that state conditions under which the congruence x3 ≡ p (mod q) is solvable; the word "reciprocity" comes from the form of the main theorem, which states that if p and q are primary numbers in the ring of Eisenstein integers, both coprime to 3, the congruence x3 ≡ p (mod q) is solvable if and only if x3 ≡ q (mod p) is solvable. == History == Sometime before 1748 Euler made the first conjectures about the cubic residuacity of small integers, but they were not published until 1849, after his death.Gaus...

so this is a thing

Balarka Sen

Yeah, yeah, we know

Let's just compute the correct lattice first

Semiclassical

i didn't

Balarka Sen

9:10 PM

$2^{1/3} \Bbb Z \oplus 2^{1/3}\omega \Bbb Z \oplus 2^{1/3}\omega^2 \Bbb Z$ in $\Bbb R \oplus \Bbb C$

OK, so the lattice basis is $(2^{1/3}, 0, 0)$, $(0, 2^{1/3}/2, 2^{1/3} 3^{1/2}/2)$ and $(0, 2^{1/3}/2, -2^{1/3} 3^{1/2}/2)$

$\Bbb R \oplus \Bbb C = \Bbb R \oplus \Bbb R \oplus i \Bbb R$ mind you

Semiclassical

i guess i shouldn't be surprised that most of this stuff can be labelled as "crazy stuff that Eisenstein worked out almost 200 years ago"

@BalarkaSen computing the volume of that seems straightforward but tedious

Balarka Sen

It should be $-2^2 \cdot 3^3$

Like we want it to be. -108

But you get the procedure, right?

Semiclassical

yeah, i guess it's not that bad

Balarka Sen

Just to be out of this mess

OK, thanks to @Astyx and @Leaky

Semiclassical

$$2^{1/3}(-2^{1/3}\cdot 2^{1/3}3^{1/2}/2-2^{1/3}3^{1/2}/2\cdot 2^{1/3}/2)$$

hmm, i don't think that gives me enough powers of 3

Astyx

9:15 PM

yeah mb, I confused everyone by saying R^3

Wait, no I didn't

Semiclassical

i think that only gives $-3^{1/2}$

Astyx

But I was about to

Semiclassical

i'd be lying if i could be arsed to work this out though

Astyx

So the general way of finding the lattice is what Leaky said, look at embeddings into R and into C

Balarka Sen

yeah, something

i give up

Semiclassical

9:16 PM

yeah, and i guess there's no way to avoid having such lattices

which is a good thing

Balarka Sen

I can tell you the covolume for the lattice for $\Bbb Q(\zeta_n)$

It's

$$\frac{n^{\phi(n)}}{\prod_{p|n} p^{\frac{\phi(n)}{p-1}}}$$

Lol

Semiclassical

gross

Balarka Sen

The prime divisors of the covolume say interesting things about arithmetic in your field

Astyx

Say you have $\sigma_1,\dots,\sigma_r$ embeddings into $\Bbb R$ and $\tau_1,\dots, \tau_s$ embeddings into $\Bbb C$, then $i:K\to \Bbb R^r\times \Bbb C^s = V$, $i(x) = (\sigma_1(x), \dots, \sigma_r(x),\tau_1(x),\dots,\tau_s(x))$ projects $O_K$ to a lattice of $V$

Semiclassical

i'd say i'd stick to crystallographic lattices but i can't say i actually work with those

Balarka Sen

9:19 PM

This relates to cubic reciprocity, or cyclotomic reciprocity

Semiclassical

for context, this was in the vein of an undergrad abstract algebra project i had

which was to figure out roots of x^2+x+1=0 in various settings

Balarka Sen

Someone should do algebraic number theory just in terms of lattices

I dont have the energy for it

By doing geometric manipulation with lattices figure out how primes ramify

Astyx

People do

Balarka Sen

ref?

Semiclassical

i think back in the day i'd noticed the pattern that x^2+x+1=0 had solutions mod prime p if p=1 mod 6

but had no explanation for it

Astyx

9:25 PM

The idea is that class group elements are represented by ideals with a small norm (less that Minkowski's bound)

Balarka Sen

I know Minkowski bound, more or less

I was just saying you should do all of ANT in terms of lattices

Astyx

${n!\over n^n}\left({4\over \pi}\right)^{r_2}\sqrt{d}$

I'm not sure what you mean by that

Balarka Sen

If a prime p divides covol of the lattice, why does it ramify?

Give a geometric explanation

Stuff like that

You should mod the lattice by p, then you get some lattice in F_p^n and try to see there

I don't have the energy for it but its possible

Astyx

hmm

Balarka Sen

@Semiclassical I am confused. x = 1 is a solution mod 3, and 3 is not 1 mod 6

You did not mean iff?

Semiclassical

9:30 PM

sufficient but not necessary yeah

Balarka Sen

I mean I guess 1 mod 6 primes are not inert in Z[(1 + sqrt(-3))/2]

That's an interesting statement, yeah

Astyx

I'm not even sure what ramification looks like in terms of lattice

Balarka Sen

Ramification means you have e, so you are getting nilpotent factors in the fiber algebra $\mathcal{O}_L \otimes \kappa(\mathfrak{p})$... what does this mean in terms of lattices?

Must mean something...

Ted Shifrin

Saying $1$ mod $3$ suffices to get $3$ together with $1$ mod $6$. :P

Balarka Sen

? 3 is not 1 mod 3. You mean 1 mod 2.

But that's a lot more than 3 and 1 mod 6.

Ted Shifrin

9:40 PM

No, I was an idiot. But I did mean to say primes $1$ mod $3$.

We still don't include $3$.

I was never meant to discuss number theory. Never.

Balarka Sen

If arithmetic mistakes is sufficient condition for being an idiot imagine my state

Ted Shifrin

@Balarka This bothers me a lot. How can this person say $E^2=1$?

Balarka Sen

I never did any arithmetic correctly.

Ted Shifrin

I can usually do arithmetic (e.g., in a calculus course).

Balarka Sen

Did they not mean $E^2 = -1$?

@TedShifrin Of course, I don't even compare

Mathphile

9:43 PM

If we know $a \mod b$ then is there any easy method of finding $a \mod b^n$?

Ted Shifrin

They wrote standard generator of $\overline{\Bbb CP^2}$. I don't know what they meant.

Thorgott

that's not a well-posed question

Ted Shifrin

If all you know is $a\pmod b$, then you only have a list of options $\mod b^n$.

Thorgott

yeah, this map's going the wrong way

Balarka Sen

@TedShifrin Yeah, $E^2 = 1$ is wrong. $E^2 = -1$ is correct, probably a typo.

Upvoted comment

Ted Shifrin

9:47 PM

Of course, I don't think about blow-ups in terms of connected sums. I know that this is the way topologists do it; I've known that since Kirby days.

BigSocks

like Kirby moves Kirby?

Balarka Sen

It makes computing cohomology easier

Cuz you can Mayer-Vietoris it out

Ted Shifrin

Yes, Socks.

I can Mayer-Vietoris just fine thinking about a tubular neighborhood of the exceptional divisor.

Balarka Sen

Yeah also true

The answer to this guy's question is yes, right?

BigSocks

@TedShifrin man you met all the coolest guys

Balarka Sen

9:50 PM

The antiholomorphic map given by conjugation takes $H$ to $E$, and $E$ to $H$

Ted Shifrin

I actually didn't read much further.

@BigSocks: I was lucky to have two of the best as PhD advisers.

BigSocks

Chern and Kirby? wow

Ted Shifrin

No, not Kirby. Nooooo.

Griffiths.

Balarka Sen

@TedShifrin lol

Ted Shifrin

I have never remotely resembled a topologist.

BigSocks

9:52 PM

oooh. I only know the E&M book author Griffiths I think

Ted Shifrin

One of the most famous complex algebraic geometers and differential geometers of the 20th century. Phillip.

Mathphile

I'll try improving my previous question

BigSocks

k lemme see

Mathphile

Using Fermat's little theorem we know that $a^{(p-1)} \equiv 1 \mod p$

Is there any good way of finding $x$ in $a^{(p-1)} \equiv x \mod p^n$?

BigSocks

ok yeah man looks like a big deal

Thorgott

9:54 PM

the issue with the question remains the same

a residue class mod p does not determine a residue class mod p^n

Mathphile

alright

Ted Shifrin

@BigSocks: Arguably one of the most famous (maybe not most respected) complex algebraic geometry books is by Griffiths and Joe Harris (who was his student). Harris has since written zillions of other books.

Mathphile

Is there any efficient way in general to find $a^{(p-1)} \mod p^n$ where $p$ is prime?

Mike Miller

I always forget that's Joe and not Michael.

Ted Shifrin

I don't think I know Michael.

Mike Miller

9:58 PM

Colleague here.

Ted Shifrin

Never met him.

I generally have had nonnegative experiences with most mathematicians, but your former geometry prof at UCLA was one of the exceptions.

We've talked about it before.

Too bad $p$ and $p$ aren't relatively prime.

Thorgott

p=1

Balarka Sen

$p$ is called a Wieferich prime if $2^{p-1} = 1 \pmod{p^2}$

Mathphile

@TedShifrin lol yes otherwise we could use Chinese remainder theorem

Balarka Sen

It was known for a long time that FLT is true for all such primes

It is unknown if there are more than 2 such primes

Ted Shifrin

10:03 PM

Yup @Mathphile :)

Mike Miller

@BalarkaSen Hahaha

Ted Shifrin

I only have found one so far.

Mathphile

Is there something like CRT for cases in which our mod's are not relatively prime?

Ted Shifrin

Nope.

Balarka Sen

@TedShifrin You have probably found 11.

The other one is 1093 :)

Ted Shifrin

10:05 PM

Um, no.

I screwed up my easy arithmetic yet again. I should resign.

Mathphile

@TedShifrin :(

Balarka Sen

Hahah

Welcome to the Other Side

Mike Miller

I looked at the wikipedia page. It's an immense amount of information about two numbers

Ted Shifrin

Balarka is still a repository of unbelievable amounts of esoterica.

Balarka Sen

I just love that joke

Mathphile

10:07 PM

I'm trying to find an efficient way of calculating $a^b \mod p^n$

Balarka Sen

Wieferich himself proved FLT for all his primes

Mike Miller

2^10 = 1024 = 1024-968 = 56, it's not 11

Ted Shifrin

Interesting.

Oh oh.

Thorgott

lol

Mike Miller

1093 and 3511 according to wiki

Balarka Sen

10:07 PM

Oh lol we both screwed up

Yeah, 3511. Knew it had to end with 11

Thorgott

elementary arithmetic is too hard

Ted Shifrin

Balarka was trying to make me feel better for screwing up $2$. :D

Balarka Sen

Knee-bump, Ted

Ted Shifrin

LOL

Mike Miller

Ahahahahah

It's not known that there are infinitely primes not of this form

Balarka Sen

10:10 PM

lol

Massive troll Wieferich

Mike Miller

abc implies infinitely many. I just checked that twin primes implies infinitely many

Balarka Sen

Your next research project is to prove there are at least 3 Wieferich primes, Mike

Mike Miller

I will collaborate with you. You will tell me your ideas and I will say I am confused to force you to be more precise

Ted Shifrin

Some collaborations like that have made for some powerful results.

Mike Miller

It's the only way I know how

Balarka Sen

10:13 PM

You mean to say proving existence of a new Wieferich prime is not a powerful result?

Ted Shifrin

@BalarkaSen Where did that inference come from?

Balarka Sen

Oh, ok, you did not mean that. I am just really protective of new Wieferich prime facts.

Sorry to be insecure.

Ted Shifrin

LOL

I had never heard of these before. Hence my allusion to esoterica.

Balarka Sen

They are esoterica. I was bringing it up as a joke, and $a^{p-1} \pmod{p^n}$, mostly.

Thorgott

Balarka gonna become a leading expert on Wieferich primes in the future

life goals

Ted Shifrin

10:15 PM

Primes in the future? But not so prime in the past?

Balarka Sen

lol

I'm at my inert prime.

Alessandro Codenotti

@TedShifrin What do you mean, you're homeomorphic to many topologists

Ted Shifrin

You might think so, @Alessandro, but it's not been verified.

Leaky Nun

11:01 PM

So if we use $(\sigma_1 x, \sigma_2 x)$, which is really $(\sigma_1 x, \operatorname{Re}(\sigma_2 x), \operatorname{Im}(\sigma_2 x))$, then under a complex change of basis which scales by $2$ ($(a,b,c) \mapsto (a,b+ic,b-ic)$), you would obtain $(\sigma_1 x, \sigma_2 x, \sigma_3 x)$; the covolume would then be $\det \sigma_i b_j$ where $B = \{1, 2^{1/3}, 2^{2/3}\}$ is a basis. The trick here is that $(\sigma_i b_j)^T (\sigma_i b_j) = (\sum_j \sigma_j (b_i b_k)) = (Tr(b_i b_k))$, and

the trace is easier to compute using this basis, since $2^{1/3}$ and $2^{2/3}$ have zero trace

Balarka Sen

if you have power basis youre through right

i mean its just vandermonde

Leaky Nun

so we obtain $(Tr(2^{(i+j-2)/3})) = 3 \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 2 \\ 0 & 2 & 0 \end{pmatrix}$

why is this negative...???

Balarka Sen

dunno, i know disc(x^3 - 2) = -27 * (-2)^2 tho

i should compute covol by hand but dont have energy

Leaky Nun

that's exactly the determinant of the matrix I just wrote

Balarka Sen

you have signs for vandermonde

maybe thats where you screw up

who gives a shit dude you did the right computation upto signs

Leaky Nun

11:06 PM

what's the Jacobian of my transformation?

$(a,b,c) \mapsto (a,b+ic,b-ic)$

user2103480

@TedShifrin Codenotti's conjecture

Balarka Sen

Cody's conjecture

"Gabagool over there"

Leaky Nun

$(dx+idy) \land (dx-idy) = (-i-i) dx \land dy = -2i dx \land dy$?

yeah the $i$ is why the square is negative

so the covolume is $6\sqrt3$?

@BalarkaSen I mean, the $(\sigma_i b_j)^T (\sigma_i b_j) = (\sum_j \sigma_j (b_i b_k)) = (Tr(b_i b_k))$ trick is really nice

Balarka Sen

thats what i got last time lol

which makes sense, square root of shit

Leaky Nun

just keep in mind that the measure on $\Bbb C$ is 2 times the normal one

since $dz \land d\overline{z} = 2 dx \land dy$

I mean I just computed it to be $-2idx \land dy$

where am I pulling this out of

who cares, $-i$ is a unit

Balarka Sen

11:10 PM

lol

measure upto units

Ted should see this

Ok I gotta sleep

Thanks!

Leaky Nun

np

for $p > 3$, $x^2+x+1=0$ has root mod $p$ iff $\left(\dfrac{3}{p}\right) = 1$ iff $\left(\dfrac{p}{3}\right) = 1$ iff $p \equiv 1 \pmod 3$ right

Semiclassical

@LeakyNun yeah. but primes aren't divisible by 2, so that can be improved to p=1 mod 6

Leaky Nun

right

it has no solution mod $2$ and it has a solution mod $3$

Semiclassical

(i mean i'm taking for granted that you're using quadratic reciprocity correctly, but it's not as if i can contradict you)

Leaky Nun

so an equivalent condition is p = 1, 3 mod 6

Semiclassical

11:18 PM

right. and it's just that there's not a lot of primes = 3 mod 6

Leaky Nun

actually I messed up the quadratic reciprocity

Semiclassical

i wonder what the simplest elementary proof would be

without appealing to reciprocity

I think that's what I ran aground with when I had this problem as an undergrad. I could tell that quadratic reciprocity would do it (even if i couldn't understand how to use it) but didn't know how to give an elementary proof.

Leaky Nun

for $p > 3$, $x^2+x+1=0$ has root mod $p$ iff $\left(\dfrac{3}{p}\right) = 1$.
For $p = 4k+1$, $\left(\dfrac{3}{p}\right) = \left(\dfrac{p}{3}\right)$, so it is equivalent to $p \equiv 1 \pmod 3$;
for $p = 4k+3$, $\left(\dfrac{3}{p}\right) = -\left(\dfrac{p}{3}\right)$, so it is equivalent to $p \equiv 2 \pmod 3$

so the combined condition is $p \equiv 1, 11 \pmod{12}$

i.e. $p \equiv \pm 1 \pmod{12}$

and also $p = 3$

wait, but this contradicts your statement

at $p=7$

indeed $3$ is not a square modulo $7$

so you shouldn't be able to find a solution mod $7$

Ted Shifrin

@Leaky Only you would use \land for \wedge !

Leaky Nun

$\wedge$

oh I was afraid it would go the other way