Let $I$ be an ideal. If $I^{h}$ is the homogenization of that ideal, defined by $I^{h}:= \langle f^h: f \in I\rangle$ that is the ideal generated by the homogenized elements of I, then why is $\sqrt{I^h}=\sqrt{I}^h$?
I know, that for one single element $(f^h)^n=(f^n)^h$, but I have trouble representing an arbitrary element in $I^h$, because they do not need to be homogeneous anymore.
I use the notation $I^*$ for the homogenization of $I$.
Take $F\in\sqrt{I^*}$ homogeneous. Then $F^N\in I^*$ for some $N$. But then $F^N_*\in I$ which means $F_*\in\sqrt I$. So that $(F_*)^*\in(\sqrt I)^*$. But then $F=X_{n+1}^r(F_*)^*\in (\sqrt I)^*$. For a general $F=\sum F_i$, each homogeneous component $F_i$ is in $\sqrt{I^*}$ since the radical of a homogeneous ideal is still homogeneous (verify).
Now take $F\in(\sqrt{I})^*$. Then $F=\sum{A_iG_i^*}$ where $G_i\in\sqrt{I}$. So there is an $N$ large enough so that each $G_i^N\in{I}$. And so there is an $M$ large enough so that $F^M\in I^*$ or $F\in\sqrt{I^*}$.