# Nullstellensatz-type result for polynomial rings over infinite fields, which are not algebraically closed?

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?

$$V(J)$$ 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 $$\Phi\in J$$.