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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.

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    It should reduce to $(F^*)^k = (F^k)^*$ – 2017-01-24
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    @user1952009 Could you please elaborate because I don't see the reduction? If I assume $G^N\in I^*$ then the connection with $F^*$, $F\in I$ is that $G^N$ is a finite sum of those, each multiplied by an element of $k[X_1,\ldots,X_{n+1}]$. I don't really know where to go from there. – 2017-01-24

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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).