The following is the problem 11 of Chaper 8 Section 4 of Ideals, Varieties, and Algorithms by Cox, Little and O'Shea.
A homogeneous ideal is said to be prime if it is prime as an ideal in $k[x_0,\cdots,x_n]$. Show that a homogeneous ideal $I \subset k[x_0,\cdots,x_n]$ is prime if and only if whenever the product of two homogeneous polynomials $F,G$ satisfies $F \cdot G \in I$, then $F \in I$ or $G \in I$.
So, I have to prove that for $f,g \in k[x_0,\cdots,x_n]$ if $f \cdot g \in I$ then $f\in I$ or $g\in I$ given the assumption that it is true if $f,g$ are homogeneous polynomials.
I've been struggling this for two days without any progress. Could someone help me?
Suppose it's not true, so there exists $f$ and $g$ in $k\left[ x_0,\ldots,x_n\right]$ such that $f\cdot g \in I$ but $f\notin I$ and $g\notin I$. Let $f$ and $g$ be such a pair such that $\deg(f) + \deg(g)$ is least possible ( ie. there is no such pair $p$ and $q$ with $\deg(p) + \deg(q)$ lower ). Since $I$ is graded, each graded piece of $f \cdot g$ must be in $I$. In particular, the highest degree term is in $I$. The highest degree term is just the product of the leading terms of $f$ and $g$, which we'll call $a_f$ and $a_g$. So $a_f \cdot a_g \in I$, and since they are homogeneous, we see that either $a_f$ or $a_g$ is in $I$. Without loss of generality, let's say it's $a_f$. In that case $\left( f - a_f \right) \cdot g \in I$, but neither $\left( f - a_f \right)$ nor $g$ are in $I$, and this violates the minimality of the pair $f$ and $g$.