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Let $f(x)$ be a monic irreducible polynomial with integer coefficient. Let $K$ be the splitting field of $f$ and $\alpha$ one of its roots. Let $p$ a prime number such that $p$ does not divide $disc(f)$ and suppose that $f$ has a root mod $p$. Then there exists a prime $P$ of $K$ over $p$ such that the decomposition group $G(P)\leq G(K|Q)$ is contained in $G(K|Q(\alpha))$. Why this is true?

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  • $\begingroup$ Since $K$ is a field the only (prime) ideals are the zero ideal and $K$ itself. Maybe you meant $P$ to be a prime ideal of the ring of integers of $K$, or a prime ideal of the ring of integers of $\Bbb{Q}(\alpha)$? $\endgroup$ – Marc Bogaerts Sep 19 '14 at 8:59
  • $\begingroup$ @Nimda, I may be wrong, but I believe that it is "standard" abuse of language to refer to non-zero prime ideals of the ring of integers of $K$ as "primes of $K$". What else could it mean? $\endgroup$ – Jyrki Lahtonen Sep 19 '14 at 10:09
  • $\begingroup$ I think that this is because the existence of that zero of $f$ modulo $p$ implies that at the level of $\Bbb{Q}(\alpha)$ there will be (at least) one prime ideal $\mathfrak{p}$ above $p$ such that $f(\mathfrak{p}|p)=1$. Then any prime of $K$ lying above $\mathfrak{p}$ should have this property, because $f$ is multiplicative in a tower of extensions. $\endgroup$ – Jyrki Lahtonen Sep 19 '14 at 10:15
  • $\begingroup$ I am confused (I'm new to this subject) because of the following example. Take $\alpha$ a root of $f=x^3-3$. Then $(5)$ is the product of $(5, \alpha+3)$ and $(5, \alpha^2+2\alpha-1)$ with norms $5$ and $25$ respectively. $\endgroup$ – Marc Bogaerts Sep 19 '14 at 11:08
  • $\begingroup$ @Marbor: If there are five prime ideals, each with decomposition group of order two, then $\Bbb{Q}(\alpha)$ cannot be the splitting field, because it is a degree five extension (assuming that $f$ is irreducible). $\endgroup$ – Jyrki Lahtonen Sep 19 '14 at 13:20
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$F$ is a separable extension of $Q$. Take $p$ relatively prime with the conductor of $Z[\alpha]$, that is the ideal $\{y \in O_F \ | \ y O_F\subseteq Z[\alpha]\}$. Notice that this ideal measure how far $Z[\alpha]$ is from $O_F$ (that is not monogenic in general). Then by Neukirch, Jürgen, Algebraic Number Theory, pag. 47, we have that, since $f$ has a root $\bmod p$ and then a linear factor, there is a prime $\mathcal{P}$ over $p$ in $F$ such that ${\rm f}(\mathcal{P}|p)=1$. Thus, for any $\mathfrak{p}$ over $\mathcal{P}$ in $K$, we have $f(\mathfrak{p}|p)=f(\mathfrak{p}|\mathcal{P})$.

If $p$ does not divide the discriminant of $K$, then $p$ doesn't ramify, so that $e(\mathfrak{p}|p)=1$. Then $e(\mathfrak{p}|\mathcal{P})=1$.

Now, since $K$ is Galois both over $F$ and $Q$, the order of the decomposition group $$\# D_{K|F}(\mathfrak{p})=e(\mathfrak{p}|\mathcal{P})f(\mathfrak{p}|\mathcal{P})=f(\mathfrak{p}|\mathcal{P})=f(\mathfrak{p}|p)=e(\mathfrak{p}|p)f(\mathfrak{p}|p) =\# D_{K|Q}(\mathfrak{p})$$ and then $$ D_{K|Q}(\mathfrak{p})=D_{K|F}(\mathfrak{p}) \leq G(K|F).$$

The only problem is now to prove that if $p$ does not divide $disc(f)$ then it is relatively prime with the conductor of $Z[\alpha]$ and does not divide the discriminant of $K$.

For this see Lang, Algebraic Number Theory, III.3.

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  • $\begingroup$ @JyrkiLahtonen thanks for the hint. Here is my answer. Is it correct? $\endgroup$ – Marbor Sep 23 '14 at 15:25

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