Find the inverse image of the algebraic variety

Question

Let $h: K^n \rightarrow K^m$ be a polynomial function and $Y= V_k(I)$ in $K^m$ the algebraic variety of an Ideal in $K[y_1,...,y_m]$.

What would the inverse image $X=h^{-1}(Y)$ described as an ideal be? Is there any example which illustrates this?

Thanks for any help

Answer 1

Your map $h$ is induced (via contraction) by a $K$-algebras homomorphism $\bar h:K[y_1,\dots,y_m]\to K[x_1,\dots, x_n]$. For any ideal $I\subseteq K[y_1,\dots, y_m]$, it is known that $h^{-1} (V(I))=V(J)$, where $J$ is the ideal generated in $K[x_1,\dots ,x_n]$ by $\bar h(I)$.

(This is usually proved in order to show that a homomorphism of finitely generated $K$-algebras induces a continuous map between affine varieties, if $K=\bar K$).

For example, $\pi:K^2\to K$ is induced by $\iota:K[x]\to K[x,y]$. If you take the closed set $V(f)\subseteq K$, for some $f\in K[x]$, you can see that $\pi^{-1}(V(f))=V(f)\times K= V(\iota (f))$.