Consider the multispace $$M:=\Bbb P^{n_1}\times \Bbb P^{n_2}\times ...\times \Bbb P^{n_r}\times \Bbb A^m.$$Let $$f\in k[x_{11},...,x_{1n_1},x_{21},...,x_{2n_2},...,x_{r1},...,x_{rn_r},y_1,...,y_m]=k[X_1,...,X_r,Y]$$ be a polynomial such that $f$ is homogeneous in each set of variables that correspondence to a projective space $\Bbb P^{n_i}$, but there is no restriction on those corresponding to $\Bbb A^m$. Then we say $f$ is a multiform.
Let $S$ be a collection of multiforms and $$V(S):=\{(x_1,...,x_{r},y)\in M:f(x_1,...,x_{r},y)=0,\forall f\in S\}.$$
Suppose $V\subseteq M$ such that $V=V(S)$ for some collection $S$ of multiforms, then we say $V$ is algebraic. An algebraic set which can not be written union of two proper algebraic subsets is called irreducible algebraic set.
Let $V$ be an irreducible algebraic set and $$I(V):=\{f\in k[X_1,...,X_r,Y]:f(x_1, ...,x_r,y)=0,\forall (x_1,...,x_r,y)\in V\}.$$ My problem is that, how to show $I(V)$ is prime? Also, is the Converse true? I don't know how do I start with this problem.
It's the same proof as in the affine/projective cases.
Let $R$ be the coordinate ring for $M$. Let $f,g\in R$ with $fg\in I(V)$. Then if both $f$ and $g$ are nonzero on $V$, $V(I(V),f)\subsetneq V$ and $V(I(V),g)\subsetneq V$, but $V=V(I(V),f)\cup V(I(V),g)$. Thus if $V$ is irreducible, $I(V)$ is prime.
Conversely if $I(V)$ is prime, and $V=V_1\cup V_2$, with $V_1=V(I)$, $V_2=V(J)$, then if $I\subset I(V)$, $V \subset V_1$, so $V_1=V$ in this case. Thus we can assume $I\not\subset I(V)$, so choose $i\in I$ with $i\not\in I(V)$. Then for all $j\in J$, $ij \in I(V)$, since for any point $x$ in $V$, either $x\in V_1$, and $i(x)=0$, or $x\in V_2$ and $j(x)=0$. Thus since $i\not\in I(V)$ and $I(V)$ is prime, $j\in I(V)$. Thus $J\subseteq I(V)$, so $V\subseteq V_2$. Thus when $V=V_1\cup V_2$, we have shown that either $V\subseteq V_1$ or $V\subseteq V_2$. Thus $V$ is irreducible.