Mathematics

5:38 AM

@anon Suppose a finite group $G$ acts transtively on a set $X$. Given $x,y \in X$ Is there a way to count the number of $g \in G$ so that $gx = y$?

Sanchez

same as stabilizer of $x$?

BenjaLim

@Sanchez why is that so?

Sanchez

If both $g,g'$ sends $x$ to $y$, then $g^{-1}g'$ fixes x

BenjaLim

ok yes that is true.

so $g^{-1}g' \in stab(x)$

@Sanchez Ah I got it.

Sanchez

@BenjaLim, ?

BenjaLim

5:45 AM

@Sanchez My actual question is why if $L/K$ a Galois extension and $Q,Q'$ primes lying over $P$ then the number of $\sigma$ such that $\sigma(Q) = Q$ is the same as the number of $\sigma$ such that $\sigma(Q) = Q'$.

Sanchez

@BenjaLim, ah I see.

BenjaLim

The proof is actually quite simple.

Sanchez

That's because $gStab(x)$ are exactly the elements that send $x$ to $y$.

BenjaLim

@Sanchez well that was not exactly the proof I had in mind but I think it is the same

Sanchez

probably.

BenjaLim

5:46 AM

for example since $G$ the galois group acts transitively on the primes lying over $P$ there is $\sigma$ so that $\sigma(Q) = Q'$.

Now define a map $f : \{\sigma : \sigma(Q) = Q'\} = B \to stab(x)$

that sends $\tau$ to $\sigma^{-1}\tau$

Sanchez

Yeah, that's the same.

BenjaLim

yea.

the map is injective

and it's also surjective because we can write any $\tau \in stab(x)$ as $\tau = \sigma^{-1}(\sigma\tau)$ @Sanchez

Sanchez

That's right.

Samuel Handwich

is 2z(|z|^2 - 1) the zbar derivative of (|z|^2-1)^2?

BenjaLim

yeah and that's a bijective map between two finite sets

so they must have the same cardinality @Sanchez

Sanchez

5:49 AM

@Samuel, yes.

@BenjaLim, yes.

BenjaLim

@Sanchez hmmm I'm wondering if we can prove properties about the relative norm by localisation.

Sanchez

@BenjaLim, yes.

What properties are you thinking about?

BenjaLim

@Sanchez Well I have a definition of relative norm as:

$I$ an ideal of $\mathcal{O}_L$

$N(I) = \mathcal{O}_K \cap \prod_{\sigma \in G} \sigma(I)$

$G = \text{Gal}(L/K)$

Sanchez

This is not how I learnt it.

Another way to define norm is this

if you have a number field $K/\mathbb{Q}$, then for $I$ an ideal of $\mathcal{O}_K$, the norm can be defined as $|\mathcal{O}_K/I|$.

BenjaLim

yea @Sanchez ?

that's the absolute norm.

Sanchez

5:56 AM

In general, you can do the same thing.

BenjaLim

?

Sanchez

But you would need to measure how much $\mathcal{O}_K$ and $I$ differ, as we want to produce an ideal in the ground field.

BenjaLim

ok yes.

Sanchez

There is a notion of module index (see Cassels Frohlich chapter 1 for example), where such notion can be defined, by multiplying local things up.

BenjaLim

@Sanchez but what do you mean by <<in general, you can do the same thing>> ?

Sanchez

5:58 AM

In general = relative situation.

BenjaLim

ok.

Sanchez

The essence is this. If $I = (a)$ is principal, then $N(a)$ can be viewed as the determinant of the $\mathbb{Z}$-linear transformation that sends $\mathcal{O}_K$ to $(a)$.

Now of course, a general ideal may not be principal, so we cannot do this. But for a Dedekind domain whenever we localize we get a DVR, which means that locally we can always do this.

BenjaLim

which norm are you talking about now? The absolute norm yes?

Sanchez

Yes.

I'm trying to give you another interpretation of the absolute norm, that can be generalized to the relative situation, elaborating what i said just now.

The claim is that, the product of all the local modules indices can be defined (i.e. most of them are 1), and that it would agree with this situation.

brb by the way.

BenjaLim

ok.

@Sanchez I will look at cassels and frolich.

Sanchez

6:12 AM

Sure.

Mariano Suárez-Alvarez

6:30 AM

we have a great candidate for the Silliest Question/Answer Pair Award on MO...

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$Mathematics$ Mathematics