Do surjective ring homomorphisms commute with intersection of ideals?

Question

Let $f:A\longrightarrow B$ be a surjective ring homomorphism. Is it true that for any intersection of ideals, the image of the intersection is equal to the intersection of the images of the ideals?

One containment is trivial.

Edit: In particular the image of the intersection is contained in every image of the ideals individually and hence in the intersection of the images.

Answer 1

Let $k$ be a field, let $A=k[x,y]$, let $B=k[z]$, and let $f:A\to B$ be defined by $f(x)=f(y)=z$.

Consider the ideals $I=(x)$ and $J=(y)$ of $A$. Then $I\cap J=(xy)$ so $f(I\cap J)=(z\cdot z)=(z^2)$, whereas $f(I)\cap f(J)=(z)\cap (z)=(z)$. So in this case we have $$f(I\cap J)\subsetneq f(I)\cap f(J)$$