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Let $f(x)$ be a polynomial over an algebraic number field $K$ with coefficients in $\mathcal{O}_K$ such that its reduction splits completely in $\mathcal{O}_K/\mathfrak{p}\mathcal{O}_K$. Let $L$ be the splitting field of $f$ over $K$ and $\mathfrak{q}$ be a prime in $\mathcal{O}_L$ such that $\mathfrak{q}|\mathfrak{p}$. Then I want to show the decomposition group of $\mathfrak{q}$ is trivial.

A polynomial splits completely by definition is that it splits as linear factors. Could someone explain how to show $D(\mathfrak{q})$ is trivial?

The motivation is this question.

Thanks.

  • On an elliptic curve $p$ is inseparable because the Frobenius, purely inseparable of degree $p$, has a dual isogeny such that $\pi^* \pi=p$. The curve is supersingular iff $\pi^$ is inseparable too, in which case it is $= -\pi$ and $tr(\pi)=\pi+\pi^=0$. – reuns Dec 14 '20 at 16:53

1 Answers1

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Assume $f$ is monic, and let $\mathfrak{q}|\mathfrak{p}$ where $\mathfrak{q}$ prime ideal in $L$. Then $f(X)=\prod_i{(X-a_i)}$ for some $a_i \in \mathcal{O}_L$. Reducing mod $\mathfrak{q}$, we find that $\overline{f} = \prod_i{(X-\overline{a_i})}$. As $\overline{f}$ splits in $\mathcal{O}_K/(\mathfrak{p})$, we find the $\overline{a_i}$ belong to $\mathcal{O}_K/\mathfrak{p}$, and they're distinct because $\overline{f}$ splits completely.

Now, let $\sigma$ be in the decomposition group of $\mathfrak{q}$, ie $\sigma \in Gal(L/K)$ with $\sigma(\mathfrak{q}) \subset \mathfrak{q}$. Then $\sigma$ acts on the roots of $f$, and since their reductions mod $\mathfrak{q}$ are in $\mathcal{O}_K/(\mathfrak{p})$, we know that $\sigma(a_i) = a_i$ mod $\mathfrak{q}$. But by the above, the $a_i$ mod $\mathfrak{q}$ are distinct, so $\sigma(a_i)=a_i$. $L$ is the splitting field of $f$ so is generated by $K$ and the $a_i$ so $\sigma$ is the identity and the decomposition group is trivial.

Aphelli
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  • Dear @Mindlack, I'd like to ask that, split completely means splits as product of different linear factors? So a polynomial likes $f(x)=(x-a)^2(x-b)$ doesn't split completely? –  Dec 08 '20 at 16:42
  • Another question, what do you mean by "their reductions mod $\mathfrak{q}$ are in $\mathcal{O}_K/(\mathfrak{p})$"? And if $\sigma(a_i) = a_i$ mod $\mathfrak{q}$ and $a_i$ mod $\mathfrak{q}$ are different, I can only see that $\sigma(a_i)$ are different, but why $\sigma(a_i)=a_i$? –  Dec 08 '20 at 16:47
  • To be honest, I've had that class a while ago so I'm not 100 percent sure for the terminology. What I know is that when a field extension splits completely, this means that the extension is unramified, and in particular that the reduction of the polynomial is separable. – Aphelli Dec 08 '20 at 16:48
  • You have a natural field inclusion $\mathcal{O}_K/\mathfrak{p} \subset \mathcal{O}_L/\mathfrak{q}$ and I mean that each $a_i \text{ mod }\mathfrak{q}$ is in the image of this inclusion. – Aphelli Dec 08 '20 at 16:49
  • For the last question, $\sigma(a_i)$ is some $a_k$ (it's a root of $f$) which is congruent to $a_i$ mod $\mathfrak{q}$. So as the $a_j \text{ mod }\mathfrak{q}$ are different, $a_k$ must be $a_i$. – Aphelli Dec 08 '20 at 16:50
  • No, just the uniqueness of factorization, since $f$ has simple roots in $\mathcal{O}_K/\mathfrak{p}$ according to your problem statement. – Aphelli Dec 08 '20 at 17:01
  • Thank you very much! –  Dec 08 '20 at 17:02