Suppose $f\colon B\to A$ is a ring homomorphism, and $I\subseteq B$ is an ideal. What's the geometric interpretation for the extension $f(I)A$ of the ideal $I$? Especially, I'm interested in the case that $I$ is prime or maximal.
Let's consider a special case. If $k$ is an algebraically closed field, and $X,Y$ are isomorphic to affine lines $\mathbb A_k^1$. Let's consider the morphism $\varphi\colon X\to Y,x\mapsto y=x^2$, which induces the pullback $k$-algebra homomorphism $f\colon B=k[Y]\to A=k[X]$. If $I=(Y-a^2)$ is a maximal ideal in $B$, then $f(I)A=(X^2-a^2)$. If $a\neq0$, $f(I)A$ naturally corresponds to the fiber $\varphi^{-1}(a)$. However, if $a=0$, the case is complicated. It's related to $\varphi^{-1}(0)$, but contains an infinitesimal neighborhood.
I need some general precise statement for this. Any idea? Thanks!
The ideal $f(I)A$ is the ideal of the scheme-theoretic fiber $\text{Spec}(A) \times_f \text{Spec}(B/I)$, that is, the restriction of the morphism $\text{Spec}(A) \to \text{Spec}(B)$ to the subscheme $\text{Spec}(B/I)$.
So algebraically, $f(I)A$ is the kernel of the map of $B$-algebras
$$A \to A \otimes_B B/I \cong A/f(I)A.$$
So it is the (ideal of the) preimage, as you said, but it has the correct scheme structure.