I’m wondering if the following statement is correct:
$k$ is algebraically closed,Given finitely many $g_i\in k[y_1,...,y_t]$ and a $f\in k[x_1,...,x_s,y_1,...,y_t]$, there exists several $h_l\in k[x_1,...,x_s]$ such that $(\forall i(g_i(y)=0) \Rightarrow f(x,y)=0)\Leftrightarrow (\forall l(h_l(x)=0))$,$y\in A^t_k$.where $A_k^t$ is the affine space of dim $t$
I met this question when trying to prove the product of two varieties is irreducible:Given $X,Y$ finite type,integral over algebraically closed k,to prove their product being irreducible,I reduced to the case where $X,Y$ is affine.
If $X\times_kY =F_1\cup F_2$ as a union of two closed sets,considering $X_i=\lbrace x\in X\mid \lbrace x \rbrace\times Y\subset F_i\rbrace$.now I have to show $X_i$ is closed.Here is one way to do this.
the other way to prove this I thought might work may be(take $X_1$ as example): embed $X,Y$ into affine space as $X=\text{Spec } k[x_1,...,x_s]/I$,$Y=\text{Spec }k[y_1,...,y_t]/J$,$J$ is generated by some $g_i$'s.Without loss of generality,set $F_1$ generated by a single equation $f(x,y)=0$.now we have the following
$\lbrace x \rbrace\times Y\subset F_1$ is equivalent to $(\forall i(g_i(y)=0) \Rightarrow f(x,y)=0)$
so I want to use the statement above to show the last condition $(\forall i(g_i(y)=0) \Rightarrow f(x,y)=0)$ is a condition on $x$ satisfied exactly by points on a variety in $A^s_k$ given by several equations $h_l(x)=0$.
