I'm unsure about a basic property of regular maps. Can someone please clarify?
Let $\varphi\colon X\to Y$ a regular map between quasi-affine sets. If $k$ algebraically closed, and $J$ an ideal in the coordinate ring $k[X]$, then why do we have $\overline{\varphi(Z(J))}=Z((\varphi^{*})^{-1}(J))$? Here I'm denoting $\varphi^*$ as the pushforward $\varphi^*\colon k[Y]\to k[X]$.
I tried to show $\varphi(Z(J))\subseteq Z((\varphi^*)^{-1}(J))$ to at least show the containment of the closure in the closed set $Z((\varphi^*)^{-1}(J))$, by taking $\varphi(f)\in \varphi(Z(J))$, and $f\in Z(J)$. I then take $g\in(\varphi^*)^{-1}(J)$ so $\varphi^*g=g\circ\varphi\in J$. Then I think $g(\varphi(f))=0$ since $g\circ\varphi\in J$, and $f\in Z(J)$, and then conclude $\varphi(Z(J))\subseteq Z((\varphi^*)^{-1}(J))$. Thus $\overline{\varphi(Z(J))}\subseteq Z((\varphi^{*})^{-1}(J))$.
However, I can't verify for myself the other containment. What's the best way to see that? Thanks.
For an ideal $J$, I will denote by $\mathfrak{r}(J)$ its radical. Then \begin{align} y\in \overline{\varphi(Z(J))}&\Leftrightarrow \left(g(y)=0\ \forall g\in K[Y]/\ (g(\varphi(x))=0\ \forall x\in Z(J))\right)\\ &\Leftrightarrow\left( g(y)=0\ \forall g\in K[Y]/\ (\varphi^*(g)(x)=0\ \forall x\in Z(J))\right)\\ &\Leftrightarrow \left(g(y)=0\ \forall g\in K[Y]/\varphi^*(g)\in\mathfrak{r}(J)\right)\\ &\Leftrightarrow \left(g(y)=0\ \forall g\in (\varphi^*)^{-1}(\mathfrak{r}(J))=\mathfrak{r}((\varphi^*)^{-1}(J))\right)\\ &\Leftrightarrow y\in Z((\varphi^*)^{-1}(J)) \end{align}