I am trying to understand the proposition 5 in page 99 of 'Commutative Algebra' by NS. Gopalakrishnan.
Let $R$ be an integrally closed domain with quotient field $K$ and $S$ a normal extension of $R$ with Galois group $G=Gal(L/K)$. Then two prime ideals $P'$ and $Q'$ of $S$ lie over the same prime ideal in $R$ if and only if there exists a $\sigma\in G$ such that $\sigma(P')=Q'$.
I am having some doubts in the proof.
- While proving for the case when G is infinite, a set $\Sigma$ is constructed in order to apply Zorn's lemma. I am not able to show that $\Sigma$ is non empty.
- If $P'$ is a prime ideal in $S$ and $L_\alpha$ is a normal extension of $K$, how is $P'\cap L_\alpha$ defined? Also how can I show that $P'\cap L_\alpha$ is a prime ideal in $S\cap L_\alpha$? Need help. Thank you.
Addressing part of your second question first, $P' \cap L_\alpha$ is defined as the usual intersection of sets. $P'$ is a prime ideal of $S$ which has field of fractions $L$, so $P' \subset S \subseteq L$. $L_\alpha$ is a subextension of $L/K$, so we have $R \subseteq K \subseteq L_\alpha \subseteq L$.
From what I can see looking in the book, $\Sigma$ is the set of normal subextensions $L_\alpha$ of $L/K$ such that there is a $\sigma_\alpha \in \operatorname{Gal}(L_\alpha/K)$ with $\sigma_\alpha(P' \cap L_\alpha) = Q' \cap L_\alpha$. If we presume that $P'$ and $Q'$ lie over the same prime ideal $P$ in $R$ then we can take a finite subextension $E$ and $P' \cap E$ and $Q' \cap E$ will lie over $P$. The result for the finite case thus implies there is a $\sigma$ in $\operatorname{Gal}(E/K)$ with $\sigma(P' \cap E) = Q' \cap E$. Therefore $\Sigma$ is non-empty.
Your second question is a consequence of a general property of rings: if $f \colon A \to B$ is a ring homomorphism then the contraction of a prime ideal in $B$ is a prime ideal in $A$. In this case we are taking the inclusion map $i \colon S \cap L_\alpha \to S$, where the contraction of $P'$ in $S \cap L_\alpha$ is $P' \cap S \cap L_\alpha = P' \cap L_\alpha$.