I'm trying to learn some basic abstract algebra from Pinter's A Book of Abstract Algebra and I find myself puzzled by the following simple question about ring homomorphisms:
Let $A$ and $B$ be rings. If $f : A \to B$ is a homomorphism from $A$ onto $B$ with kernel $K$, and $J$ is an ideal of $A$ such that $K \subseteq J$, then $f(J)$ is an ideal of $B$.
I'm clearly missing the obvious, but I don't see where the requirement $K \subseteq J$ comes into the proof. Since $f$ is also a homomorphism of additive groups, the image $f(J)$ must be closed under addition and negatives (correct?). Then for $f(J)$ to be an ideal we have to show that it is closed under multiplication by an arbitrary element $b \in B$. Since $f$ is onto, there is some $a \in A$ such that $b = f(a)$. Let $j'$ be any element of $f(J)$, so $j' = f(j)$ for some $j \in J$. Then $bj' = f(a)f(j) = f(aj)$, and $aj \in J$ since $J$ is an ideal. Then it seems that $f(J)$ is closed under multiplication by $B$.
What mistaken assumption am I making?