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There's a small remark I don't quite follow in some reading I've been doing.

Let $A$ be integrally closed in its quotient field $K$, and let $B$ be its integral closure in a finite Galois extension $L$, with group $G$. Let $\mathfrak{p}$ be maximal in $A$, and $\mathfrak{P}$ maximal in $B$ lying above $\mathfrak{p}$. Denote by $G_\mathfrak{P}$ the subgroup of $G$ of automorphisms such that $\sigma\mathfrak{P}=\mathfrak{P}$. Denote by $L^\text{dec}$ the fixed field of $G_\mathfrak{P}$, and let $B^\text{dec}$ be the integral closure of $A$ in $L^\text{dec}$, and let $\mathfrak{Q}=\mathfrak{P}\cap B^\text{dec}$.

It's remarked that $\mathfrak{P}$ is the only prime of $B$ lying above $\mathfrak{Q}$. Why is this?

I know that if $A$ is an integral domain integrally closed in its quotient field $K$, and $L$ is a finite Galois extension of $K$, then the Galois group acts transitively on the set of prime ideals in the integral closure of $A$ in $L$ which lie above some maximal ideal $\mathfrak{p}$ of $A$. Does this fact somehow immediately imply that $\mathfrak{P}$ is the unique prime of $B$ lying above $\mathfrak{Q}$? Maybe it's clear but I'm strugging, unfortunately. Thank you.

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The Galois group of $L$ over $L^{\mathrm{dec}}$ is precisely $G_{\mathfrak{P}}$, by the Fundamental Theorem of Galois Theory. We also know that the action of the Galois group is transitive on the primes lying above $\mathfrak{Q}$. But for every $\tau$ in the Galois group of $L/L^{\mathrm{dec}}$, you have $\tau(\mathfrak{P})=\mathfrak{P}$ by construction; so the only prime lying above $\mathfrak{Q}$ is $\mathfrak{P}$, since it is the only image of $\mathfrak{P}$ under the action of the Galois group.

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Thanks Arturo. In order use the fact that the Galois group is transitive on primes lying above $\mathfrak{Q}$, do we need to know if $\mathfrak{Q}$ is maximal in $A$? It seems $\mathfrak{Q}=\mathfrak{P}\cap B^\text{dec}\supseteq\mathfrak{P}\cap A=\mathfrak{p}$. –  yunone Jan 21 '12 at 22:47
    
@yunone: It is maximal by the Going Up Theorem; if $\mathfrak{Q}\subseteq \mathfrak{M}\subseteq B^{\mathrm dec}$ for some maximal ideal $\mathfrak{M}$, then there is an ideal of $B$ lying over $\mathfrak{M}$ and which contains $\mathfrak{P}$; but $\mathfrak{P}$ is maximal, so $\mathfrak{M}=\mathfrak{Q}$. And, yes, $\mathfrak{Q}$ is a prime of $B^{\mathrm{dec}}$ that lies above $\mathfrak{p}$. –  Arturo Magidin Jan 21 '12 at 22:50
    
I see, thanks again. –  yunone Jan 21 '12 at 22:56
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