# If $I=I(V)$ is an ideal of $F[x_1,\cdots,x_n]$ obtained from an algebraic set then can be factorized into prime ideals?

Let $I=I(V)$ where $$I(V)=\{f\in F[x_1,\cdots,x_n]\mid(\forall x\in V), f(x)=0\},$$ and where $V\subseteq F^n$ and $F$ is a field. Then there exist prime ideals $P_1,\cdots,P_m$ of $F[x_1\cdots,x_n]$ such that $I=P_1\cdots P_m$.

Is this proposition true? If not, how could I find a counterexample?

No, it's not. (Instead it's true that $I$ is a finite intersection of prime ideals since it is a radical ideal.)
Set $V=(\{0\}\times\{0\}\times F)\cup(\{0\}\times F\times\{0\})\subset F^3$. Then $I(V)=(X,Y)\cap(X,Z)$. (I assume that $F$ is an infinite field.) Set $Q_1=(X,Y)$ and $Q_2=(X,Z)$. If $I(V)=P_1\cdots P_m$ with $P_i$ prime ideals, then $Q_1\cap Q_2\subseteq P_i$ for all $i$, so each $P_i$ contains $Q_1$ or $Q_2$. On the other side, $Q_1$ contains some $P_i$, say $P_1\subseteq Q_1$. If $Q_2\subseteq P_1$, then $Q_2\subseteq Q_1$, false. So $Q_1=P_1$. Similarly we get $Q_2=P_2$, so $Q_1\cap Q_2=Q_1Q_2\cdots\subseteq Q_1Q_2$. Thus we have $Q_1\cap Q_2=Q_1Q_2$, false. ($X\notin Q_1Q_2$, but $X\in Q_1\cap Q_2$.)