Let $f:R\to S$ be a surjective homomorphism of commutative rings. If $J$ is a prime ideal in $S$, and $I=\{r \in R: f(r) \in J\}$, prove that $I$ is a prime ideal in $R$.
I am perfectly fine with proving that $I$ is prime, however, I am struggling to prove that $I$ is an ideal in $R$ in the first place.
Clearly $I \neq \emptyset$, since if $j \in J\subseteq S$, by surjectivity of $f$, there exists an $r \in R$ s.t. $f(r)=j$, i.e. $r \in I$.
Now let $a,b \in I$, then $f(a), f(b) \in J$. Then $f(ab)=f(a)f(b) \in J$, i.e. $ab \in I$.
Also notice that $f(a-b)=f(a)-f(b) \in J$ (since $J$ is an ideal, and hence a subring, of $S$), and so $a-b \in I$.
We thus have that $I$ is a subring of $R$. In order to prove that $I$ is an ideal in $R$, we require the following
If $a \in I$ and $r \in R$, then $ar \in I$. Since $I$ is commutative (since $R$ is commutative), this will be enough to ensure that $ra=ar \in I$ is also satisfied.
This is the part I am struggling with:
Let $a \in I$, then $f(a) \in J$.Now consider any $r \in R$, then clearly $f(r) \in S$ and so $f(ar)=f(a)f(r) \in S$. The problem, however, is I require $f(ar) \in J$ to have that $ar \in I$.
How can I do this?
It's simple. Since $J$ is an ideal, for any $f(r) \in J$ and $f(a)\in S$ $$ f(ar)=f(a)f(r) \in J $$