In the proof that an algebraic set $V$ is irreducible iff $I(V)$ is prime. Let $k$ be a field, $\Bbb{A}^n(k)$ be affine $n$-space over $k$, and $R = k[X_1, \dots, X_n]$.  

If $V \subset \Bbb{A}^n(k)$ is algebraic, then it's irreducible iff $I(V)$ is prime.

The proof in the first direction goes:

If $I(V )$ is not prime, suppose $F_1F_2 ∈ I(V )$, $F_i \notin I(V ), i=1,2$. Then $V = (V \cap V (F_1)) \cup (V \cap V (F_2))$, and $V \cap V (F_i) \subsetneq V$ , so $V$ is reducible.

I'm not seeing how they get that equality involving $V$.  Also, what lemma says we can pick such $F_1, F_2$ to begin with?

I know that $V(F_1 F_2) = V(F_1) \cup V(F_2)$.
 A: If $\mathcal I(V)$ is not prime, then by definition, there is $F_1,F_2$ both not in $\mathcal I(V)$ such that $F_1F_2\in \mathcal I(V)$. The notation here is confusing. By $\mathcal{V}(F_i)$ they mean the variety of the ideal $(F_1)$, not $V$ times $(F_i)$ (which doesn't even make sense).
I will use $\mathcal V$ for variety. We claim
$$V=(V\cap \mathcal V(F_1))\cup(V\cap \mathcal V(F_2)).$$
Clearly $(V\cap \mathcal V(F_1))\cup(V\cap \mathcal V(F_2))\subseteq V$. On the other hand, any element of $x\in V$ must be a root of either $F_1$ or $F_2$ since $x$ is a root of $F_1F_2$. Thus $x$ is in either $\mathcal V(F_1)$ or $\mathcal V(F_2)$, so $V\subseteq (V\cap \mathcal V(F_1))\cup(V\cap \mathcal V(F_2))$. Thus $V$ is reducible.
Conversely, suppose that $V$ is reducible. Then $V$ is a finite union of irreducible algebraic sets, which we can write as $\mathcal V(J_k)$ for prime ideals $J_k$. Then
$$\mathcal I(V)= \mathcal I\left(\bigcup_{k=1}^n\mathcal{V}(J_k)\right) = \bigcap_{k=1}^n\mathcal I(\mathcal V(J_k))=\bigcap_{k=1}^n\sqrt{J_k}=\bigcap_{k=1}^n J_k$$
by the Nullstellensatz. Since these ideals are distinct by definition, $\mathcal I(V)$ is not prime.
