TITLE: How to show the intersection of a prime ideal and a subring is a prime ideal
QUESTION [6 upvotes]: I am a self-studier. This is a problem (1.5b) from Andrew Baker's excellent notes (freely available for download) on Galois Theory.
$R \subseteq S$ are rings containing $1$. $Q$ is a prime ideal of $S$. Show $Q \cap R$ is a prime ideal of $R$.
I have seen this can be proved knowing that the inverse of a ring homomorphism (where $1$ maps to $1$) of a prime ideal is a prime ideal. And considering $\phi: R \rightarrow S$ as an inclusion map and taking the inverse of $Q$, one can get the desired result.
I was trying to prove this more "directly" and it is here I would appreciate help.
The approach I am taking is to show that if $ab \in Q \cap R$, then WLOG $a \in Q \cap R$. Since $Q$ is prime, $a \in Q$.
I would appreciate help as to how to show $a \in R$. Is it valid to say that since $1 \in R$ then $a1 \in R$?
As always thanks for you help.

REPLY [9 votes]: You are making it harder than it needs to be.  To show $Q \cap R$ is a prime ideal, we need only consider a pair of elements $a,b \in R$ such that $ab \in Q \cap R$.  Then since this also happens in $Q$, we know one of $a$ or $b$ is in $Q$ (and both already are in $R$).  The desired result follows, one of them is in $Q \cap R$ and thus the latter is a prime ideal.