Mathematics - 2019-10-17 (page 3 of 3)

Jacksoja

23:00

or x = i

so Z[x] become Z[i] and x^2+1 becomes 0

MatheinBoulomenos

that's the idea yeah. If you want to make this rigorous, you need the homomorphism theorem

Leaky Nun

in a certain sense that is what Z[i] means

Jacksoja

in finding a homomorphism

I struggle sometimes to find the image

Will

Can someone please explain the "base" in a logarithm

Jacksoja

I can work this out

Leaky Nun

23:02

@Will $\log_b y$ is the answer to "what is $x$ such that $b^x = y$?"

$b$ is the base because it's the base of the exponent $b^x$

so e.g. $3^2 = 9$, so $2$ is $x$ such that $3^x = 9$, so $\log_3 9 = 2$

oh the what

and that generalizes to tensor product?

MatheinBoulomenos

wait, is that right? hmm, no I should be more careful

We have $k[G \times H]=k[G] \otimes_k k[H]$. To give a $A \otimes_R B$-module for two $R$-algebras $A$ and $B$ is the same as giving a $R$-module $M$ and two ring homomorphisms $A,B \to \mathrm{End}_R(M)$ whose images commute pointwise. I don't know of a concise categorical description for this

Jacksoja

@MatheinBoulomenos am sorry but, how did you choose x^2+1 , and how did you know that that is what we need for ideal to make it work out , to be isomprhic to Z[i] ?

MatheinBoulomenos

@Jacksoja well, can you think of a surjective ring homomorphism Z[x]->Z

@LeakyNun at least you can say that $k[G * H]$ is a pullback of categories, lol

Jacksoja

@MatheinBoulomenos evaluation at zero works right?

Leaky Nun

@MatheinBoulomenos oh btw a pset had me prove that there are only finitely many extensions of Qp of degree n

MatheinBoulomenos

23:13

@Jacksoja oops

Leaky Nun

there were like 5 subquestions as guide

and then the next question had me show that there are infinitely many extensions of Fp((x)) of degree p

and then the next question had me pinpoint why the proof fails for Fp((x))

MatheinBoulomenos

@LeakyNun didn't I already tell you that you reduce it to Kummer theory?

Leaky Nun

I was staring at it for a long time

and then I was so excited when I finally discovered it

MatheinBoulomenos

@Jacksoja I meant a surjective homomorphism Z[x]->Z[i]

Jacksoja

@MatheinBoulomenos okay let me see

Leaky Nun

23:14

I even worked out the Artin--Schreier extensions of F2((x))

and found that the corresponding Eisenstein polynomials are y^2 - t^k y - t = 0

which converges to y^2 - t = 0 !

ÍgjøgnumMeg

y^2 - t = 1?

Leaky Nun

I already included a space -,-

ÍgjøgnumMeg

hehehehehe

MatheinBoulomenos

@LeakyNun there are some substeps, but it does't seem that hard.
1. Reduce to the case that the base field has sufficient roots of unity
2. Reduce to the Galois case
3. Use the fact that local Galois groups are solvalble, reducing it to the cyclic case
4. Kummer theory
5. profit

ÍgjøgnumMeg

@Mathein you forgot the intermediate step "???????????"

MatheinBoulomenos

23:17

but I gave all the relevant steps

Leaky Nun

@MatheinBoulomenos then for this proof the Fp((t)) fails at step 1 :P

ÍgjøgnumMeg

ite

MatheinBoulomenos

@LeakyNun I think the precise relationship between k[GxH] and k[G] and k[H] likely involved 2-category theory. It's probably a 2-limit of categories or something

but I'm not sure about the details

Leaky Nun

oh no

cries in infinity category

MatheinBoulomenos

the reason is that the universal property of GxH that is relevant here is not that of a product. the UMP of products is completely useless here since we are interested in morphisms out of GxH. But luckily GxH has another universal property: it's a 2-colimit! (This holds for general semidirect products)

Leaky Nun

23:21

maybe inflation restriction will help

or not

MatheinBoulomenos

rather not

Leaky Nun

but I like inflation restriction

MatheinBoulomenos

so do I

Leaky Nun

what is Galois representation?

MatheinBoulomenos

a representation of a Galois group

But in elementary terms the universal property of GxH as a 2-colimit is just $\mathrm{Hom}_{Grp}(G\times H,K) = \{(\phi,\psi) \in \mathrm{Hom}_{Grp}(G,K) \times \mathrm{Hom}_{Grp}(H,K) \mid \forall g \in G, h \in H: [\phi(g),\psi(h)]=1\}$. That's totally reasonable, despite the scary-sounding description "2-colimit"

Leaky Nun

23:25

what are some interesting theorems / definitions?

Jacksoja

@MatheinBoulomenos is it evaluation at i ?

Leaky Nun

@Jacksoja precies

MatheinBoulomenos

@Jacksoja yes

@LeakyNun oh boy

Jacksoja

What I fail to see is how from Z[i

Leaky Nun

@MatheinBoulomenos what on earth is that

Jacksoja

23:26

you jumped to Z x

Leaky Nun

Z[x] must not be viewed as just any old ring @Jacksoja

how it interacts with other objects is what makes it interesting

(ok it is also interesting as a ring unto itself)

Jacksoja

okay intresting

and if we want to show that this map is surjective

Leaky Nun

for any ring $A$, any $x \in A$ gives us an evaluation homomorphism $ev_x: \Bbb Z[x] \to A$

Jacksoja

if I pick an element as a+bi in Z[x]

what is the polynomial that when evaluates at i gives a+bi

MatheinBoulomenos

@LeakyNun there's a plethora of interesting and difficult theorems on Galois representations. But maybe a more reasonable question for the beginning is how do you get a Galois representation? If you want a representation of, say the absolute Galois group of $\Bbb Q$, it's not exactly straightforward to just "write down" a representation since $\mathrm{Gal}(\overline{\Bbb Q}/\Bbb Q)$ is pretty hard

Leaky Nun

23:29

well you reduce it to a finite extension because it has finite image anyway

MatheinBoulomenos

if you want Artin representation sure

but a continuous representation to $\mathrm{GL}_n(\Bbb Q_p)$ won't necessarily have finite image

Leaky Nun

oh no

what do we do then

Jacksoja

ok I see how it is surjective

MatheinBoulomenos

well, there are a lot of objects that actually give rise to a Galois representation. One of the easiest example is the Tate module of an abelian variety

Leaky Nun

wow

Jacksoja

23:33

@MatheinBoulomenos so the strategy is to construct a surjective ring hom in such questions?

going back to my problem

MatheinBoulomenos

@Jacksoja yes, then compute the kernel

Jacksoja

Z[i] / (2+i)

I want to find a surjective hom from Z[x] to Z[i]

with kernel (2+i)

Sha Vuklia

hey

anyone willing to give a hint for this:

MatheinBoulomenos

@LeakyNun so if you have an abelian variety (e.g. an elliptic curve) $A$ over $\Bbb Q$, then $A(\overline{\Bbb Q})$ is an abelian group. This consists of the closed points that are cut out in some projective space by homogenous polynomials equations with coefficients in $\Bbb Q$ and the group operation is also given by polynomials with coefficients in $\Bbb Q$. This meas that $G_{\Bbb Q}$ acts on $A(\overline{\Bbb Q})$ and the action is compatible with the group operation

MrCauchy

Hey, can anyone tell me; If i have a field homomorphism acting between two Q-subspaces over C can I automatically assume that it is a Q-linear map or is scalar multiplication not guaranteed?

Sha Vuklia

23:36

lol sorry that was wrong

MatheinBoulomenos

@MrCauchy it's automatically $\Bbb Q$-linear

Leaky Nun

@MrCauchy any field homomorphism must preserve Q (why?)

also "field homomorphism between two Q-subspaces over C" doesn't really make sense because subspaces don't need to be fields

MrCauchy

so i originally had two subfields with a field homomorphism, i have shown they are Q-subspaces. I want to show the the image of the homomorphism restricted to Q is exactly Q itself, i figured that if i could show it was a linear map then f(q) = qf(1) for any q in Q

Leaky Nun

@MatheinBoulomenos then we take out the p-torsion part and make it Z_(p)?

@MrCauchy that makes the approach circular :P

MatheinBoulomenos

So what you can do now is fix some prime $p$. Then for all $n$, look at the $p^n$-torsion. $A(\overline{\Bbb Q})[p^n]$. This is a finite abelian group, which is also a module over $\Bbb Z/p^n \Bbb Z$. Now take the inverse limit $\varprojlim A(\overline{\Bbb Q})[p^n]$, this will be a module over $\Bbb Z_p$.
The transition maps are equivariant with respect to the action of the Galois group (they are just given by multiplication with some suitable power of $p$, i.e. adding an element to itself a bunch of times), so $G_{\Bbb Q}$ acts on this inverse limit

Jacksoja

23:40

@MatheinBoulomenos if the kernel is 2+i , what will the map be in this case?

MrCauchy

@LeakyNun that is what i feared, wrong approach then?

Jacksoja

@MatheinBoulomenos am sorry but am failing to see this, i want to understand it before i go to bed

MatheinBoulomenos

now we can take the tensor product $\Bbb Q_p \otimes_{\Bbb Z_p} \varprojlim A(\overline{\Bbb Q})[p^n]$ and voilà, there's a $p$-adic Galois representation

Leaky Nun

@MrCauchy show that f(q) = q for any q in Q using the fact that f is a field homomorphism

@MatheinBoulomenos cool

and how do I know it isn't 0?

@Jacksoja maybe you should follow Ted's approach

Jacksoja

@LeakyNun I like this aproch of finding a map better, but I dont understand it compeltely

seems if we already know a surjective map and all

we already know too much

MatheinBoulomenos

23:43

@LeakyNun we know pretty well what these torsion points look like. $A(\overline{\Bbb Q}[p^n]) \cong (\Bbb Z/p^n\Bbb Z)^{2d}$ where $d$ is the dimension of $A$

but the Galois action is more mysterious

Leaky Nun

hmm

MatheinBoulomenos

so the inverse limit will be $\Bbb Z_p^{2d}$

Jacksoja

so 5 is in (2+i)

Since intesection of ideals is an ideal

Z intersect (2+i) can either be Z or 5Z

@MatheinBoulomenos @LeakyNun does this make sense?

MatheinBoulomenos

yes

I want to sleep now

should I tell you an algebraic solution before I go?

Jacksoja

@MatheinBoulomenos okay good night and thanks!

If you have time yes id love to

MatheinBoulomenos

23:48

@Jacksoja Z[i]/(2+i)=Z[x]/(x^2+1,2+x)=Z[x]/(2+x,5)=Z/5

Jacksoja

I will solve other problems

MatheinBoulomenos

that's really short

Leaky Nun

@Jacksoja and you should really work out the elementary approach Ted gave

namely draw out the lattice

Jacksoja

okay thanks ! I will try to work on these until tomorrow

that was very fast

No idea how I would come up with such solution

Leaky Nun

Si triste d'avoir appris par mon père le #mathématicien Jean-Pierre Serre, la disparition aujourd'hui de son ami le mathématicien #américain John Tate, #prix Abel, longtemps professeur à @Harvard @abel_prize @SocMathFr @USEmbassyFrance @INSMI_CNRS @cdf1530

schn

23:55

When switching bounds on a definite integral, one puts a minus sign in front of the integral. However, the same would be achieved by switching bounds and then putting a minus sign on the new upper limit, correct?

Leaky Nun

"We are sad to have learnt from my father the #mathematician Jean-Pierre Serre that his #American friend, the mathematician John Tate, Abel #prize winner, professor emeritus at Harvard, has passed away"

@MatheinBoulomenos

MatheinBoulomenos

@LeakyNun yeah I heard this

it's quite sad

Leaky Nun

rip

Galois in the Field

rip tate

MatheinBoulomenos

rip

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$Mathematics$ Mathematics