Let $G$ be an abelian group and $R$ be a $G$-graded ring.
Is there a map $\phi:\mathbb{N}\rightarrow\mathbb{N}$ such that for every $n\in \mathbb{N}$ and any homogeneous ideal $I$ of $R$ generated by $n$ elements, $I$ can be generated by $\phi(n)$ homogeneous elements ?
If we denote by $\mu_R(I)$ the minimal number of homogeneous generators of $I$, this is equivalent to the following:
Is for every $n\in \mathbb{N}$, $Sup_I \ \ \mu_R(I) < \infty$, where $I$ runs over the set of homogeneous ideal of $R$ generated by $n$ elements ?
Thank you very much.
No, not in general. For instance, consider the $\mathbb{Z}$-graded ring $R$ with $R_n=\mathbb{Z}/(n)$ for each $n$ where all products of homogenenous elements of nonzero degree are $0$. Given any pairwise coprime integers $n_1,\dots,n_k$, consider the element $x$ which is $1$ in degrees $n_1,\dots,n_k$ and $0$ in all other degrees. Then $x$ generates a homogeneous ideal, since each homogeneous part of $x$ can be written as $mx$ for some $m\in\mathbb{Z}$ (choose $m$ which is $1$ mod $n_i$ and $0$ mod $n_j$ for all $j\neq i$). But clearly $(x)$ cannot be generated by fewer than $k$ homogeneous elements. So, there are principal homogeneous ideals in $R$ which require arbitrarily large numbers of homogeneous generators.