Must an ideal contain the kernel for its image to be an ideal?

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?

You are right, it is not required. However maybe later he wants to point out the following fact: There is a bijection of ideals of $A$ containing $K$ and ideals of $B$. You give the one direction: If $J$ is an ideal of $A$ then $f(J)$ is an ideal of $B$. The inverse of this map is given by: If $I$ is an ideal of $B$ then $f^{-1}(I)$ is an ideal of $A$ that contains $K$.

Edit: Using this we can answer to Jon O's question in the comment below: If $B$ is a field and $I$ is an ideal of $A$ that contains $\ker f$ properly, then $f(I)$ is an ideal of $B$ that contains $0$ properly. Since every non zero element of $B$ is a unit $f(I)$ must be the whole of $B$. By the above we have $I=f^{-1}(f(I))=f^{-1}(B)=A$ so every ideal properly containing $\ker f$ is the whole of $A$, i.e. $\ker f$ is maximal.