Why is the homogenization of a radical ideal radical? Let $I\subseteq k[X_1,\ldots,X_n]$ be a radical ideal and $I^*\subseteq k[X_1,\ldots,X_{n+1}]$ be its homogenization which by definition is generated by the set $\{F^*:F\in I\}$ where $F^*$ is the homogenization of the polynomial $F$ with respect to $X_{n+1}$ (e.g. $(X_1^3+X_1X_2+1)^*=X_1^3+X_1X_2X_3+X_3^3$).
Why is $I^*$ a radical ideal? I know that it being a homogeneous ideal it's enough to show that if a power of a homogeneous polynomial belongs to the ideal then the polynomial itself belongs but even this turns out to be difficult to me.

Then you dehomogenize it and arrive at $(G_{*})^N \in I$, and use that $I$ is radical. So $G_{*}\in I$, and then $(G_{*})^* \in I^*$ gives $x_{n+1}^r\cdot (G_{*})^* \in I^*$, and therefore
$G \in I^* $ ( Prop.5 of Alg. Curves by Fulton may be helpful).
