Suppose $k = \mathbb{C}$ (or any algebrically colsed field) and I call $I^h = \{ f^h | f \in I\}$ where $f^h(x_0, \dots , x_n) = x_0^{deg(f)}f(\dfrac{x_1}{x_0},\dots, \dfrac{x_n}{x_0}) $ is the homogenization of $f$ with respect to $x_0$ and $I$ is an ideal.
Now, I think is quite simple to prove that "$I^h$ prime ideal implies $I$ prime ideal". Indeed: if $a, b \in k[x_1, \dots , x_n]$; $ab \in I$ and $b \notin I$ then $(ab)^h = a^hb^h \in I^h$. But $b^h \notin I^h$ since $b \notin I$ and so $a^h \in I$, hence $a \in I$.
Problems arise when I try to prove the other implication. Any hint? In particular I can't use the fact that $I$ is prime and $b \notin I^h$.
For Example:
if $ab \in I^h, b \notin I^h$ then there exist $f \in I$ such that $ab=F=f^h$. Hence $a(1, \dots x_n)b(1, \dots, x_n) = f(x_1, \dots , x_n ) \in I$ but now I'm stuck. Any hint?