# How to show the intersection of a prime ideal and a subring is a prime ideal

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.

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You need to show that $Q \cap R$ is an ideal in $R$, not in $S$! Thus, $a,b$ are automatically in $R$. – N. S. Jan 18 '13 at 22:37
Since you're trying to prove $Q\cap R$ is prime in $R,$ shouldn't you start that $a,b$ are in $R$? – user17794 Jan 18 '13 at 22:37
@N.S. You're right. I can show that it's an ideal. (Sometimes I'm hopeless.) Thanks very much. – TheBirdistheWord Jan 18 '13 at 22:42
@TimDuff Thanks very much. Embarrassed. – TheBirdistheWord Jan 18 '13 at 22:44

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.