Suppose that $F$ is a field, and $\overline F$ is it's algebraic closure. Suppose further, that $I \subseteq \overline F[x_0, \ldots, x_d]$ is a homogenous ideal, that is, it is generated by homogenous polynomials.
Suppose now that $I$ can be generated by elements of $F[x_0, \ldots, x_d]$. Does it follow that $I$ can be generated by homogenous elements of $F[x_0, \ldots, x_d]$?
Yes. An alternative characterization of a homogeneous ideal in a graded ring is as an ideal that contains the homogeneous component of any element of itself.
So if $I\subseteq \overline{F}[x_0,\ldots,x_d]$ is a homogeneous ideal generated by the subset $S\subseteq F[x_0,\ldots,x_d]$, then $I$ is also generated by the set of homogeneous components of elements of $S$.
Here's a proof of the alternative characterization of homogeneity: Suppose $I$ is an ideal in the graded ring $R=\bigoplus_{n\in\mathbb{Z}}R_n$. Let us denote homogeneous components of elements $x\in R$ by $x_n\in R_n$. On the one hand if $I$ has the property that it contains the homogeneous component of any element of itself, then it is clearly generated by the set of homogeneous components of its elements and is therefore homogeneous. Conversely assume that $I$ is homogeneous, generated by $S\subseteq \bigcup_{n\in \mathbb{Z}} R_n$. Assume that $x\in I$ and write $x$ as $\sum_{s\in S}r_ss$ where all but finitely many of the $r_s\in R$ are zero. Then for each $n\in\mathbb{Z}$, $x_n$ is $\sum_{s\in {S\cap R_n}}r_ss$ which is clearly in $I$.