I am reading Fulton's Algebraic Curves, and there is a remark which I do not fully understand. In Section 2 of Chapter 4, he defines irreducible projective algebraic set as follows:
An algebraic set $V\subset \mathbb{P}^n$ is irreducible if it is not the union of two smaller algebraic sets.
He then says
(1) The same proof as in the affine case, but using Problem 4.4 below, shows that $V$ is irreducible if and only if $I(V)$ is prime.
Problem 4.4 asks you to prove the following:
(2) A homogeneous ideal $I\subset k[X_1,\cdots, X_{n+1}]$ is prime if and only if for any forms $F,G$ such that $FG\in I$, we have $F\in I$ or $G\in I$.
What I am having trouble understanding is that I do not know why the result (2) is necessary to prove (1). I record below my attempted proof of (1) (,which I essentially copied from the affine case) which does not use (2). Can someone tell me what is wrong with my argument? I feel very stupid to ask this question, but I honestly do not see where the argument fails... Thanks in advance.
(Attempted Proof of (1))
If $V$ is reducible, say $V=V_1\cup V_2$, where $V_j\subsetneq V$ are proper algebraic sets, then we have $I(V)\supsetneq I(V_j)$ and hence there are $F_j\in I(V_j)\setminus I(V)$. We then have $F_1F_2\in I(V)$ but $F_j\not\in I(V)$, so $I(V)$ is not prime. Conversely, if $I(V)$ is not prime, then we can find polynomials $F,G\not \in I(V)$ such that $FG\in I(V)$. We then have $$\begin{align}V&=V(I(V)\cup {FG})\\ &=V(I(V))\cup (V(F)\cap V(G))\\ &=(V\cap(V(F)))\cup(V\cap V(G)). \end{align}$$ Now since $F\in I(V\cap V(F))\setminus I(V)$, we have $V\cap V(F)\subsetneq V$. For the same reason, $V\cap V(G)\subsetneq V$. Hence $V$ is a union of two proper algebraic sets, i.e., it is reducible. $\blacksquare$
The problem is that $V(F)$ is meaningless if $F$ is not homogeneous. You can only talk about the vanishing sets in projective space of homogeneous polynomials, not arbitrary polynomials. If you could choose $F$ and $G$ to be homogeneous then your argument would work, and that's exactly what Problem 4.4 gives you: it says that if $I(V)$ is not prime, then there are homogeneous polynomials (or "forms") $F,G\not\in I(V)$ such that $FG\in I(V)$.