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.