This is part (a) of exercise 1.8.4 in Geck's Introduction to Algebraic Geometry and Algebraic Groups
Consider the polynomial ring $R=k[X_1,\dots,X_n,Y_1,\dots,Y_m]$, where $k$ is an infinite field. Let $g_1,\dots,g_n\in k[Y_1,\dots,Y_m]\subset R$, and consider the ideal $J=(X_1-g_1,\dots,X_n-g_n)\subseteq R$.
Show that every $f\in R$ is of the form $f=g+h$ where $g\in J$, $h\in k[Y_1,\dots,Y_m]$. Deduce from this that $R/J\simeq k[Y_1,\dots,Y_m]$, and $I(V(J))=J$. (Here $I$ and $V$ are the usual operations on a Zariski topology.)
I noted that for any monomial with a power $X_i^{a_i}$ of some $X_i$ can be expanded as $X_i^{a_i}=((X_i-g_i)+g_i)^{a_i}\in J+k[Y_1,\dots,Y_m]$, via the binomial theorem. It follows quickly that every $f\in R$ has form $f=g+h$ as above. The projection map $g+h\mapsto h$ is surjective with kernel $J$, giving the ring isomorphism $R/J\simeq k[Y_1,\dots,Y_m]$. In particular, this shows $J$ is prime in $R$.
My problem is with $I(V(J))=J$. Of course $J\subseteq I(V(J))$. If $k$ were algebraically closed, the nullstellensatz would give $I(V(J))=\sqrt{J}=J$, since $J$ is prime. But if $k$ is merely infinite, is there a salvage?
Best Answer
$V(J)$ is the set of points of the form $$(g_1(b_1,\ldots,b_m),\ldots,g_m(b_1,\ldots,b_m),b_1,\ldots,b_m)$$ with the $b_j\in k$. Suppose $\Phi$ is a polynomial vanishing on $V(J)$. We can certainly write $\Phi$ as an element of $J$ plus a polynomial $\Psi$ in $k[Y_1,\ldots,Y_n]$. Then $\Psi$ vanishes on $V(J)$ which means that $$\Psi(b_1,\ldots,b_m)=0$$ for all $b_1,\ldots,b_m\in k$. But $k$ is an infnite field, so $$\Psi(X_1,\ldots,X_m)=0$$ as a polynomial. Thus $\Phi\in J$.