Let $R$ be a Noetherian $\mathbb{N}$-graded ring where $R_0=K$ is an infinite field. I explicitly do not want to assume that $R$ is standard graded (i.e. $R\neq K[R_1]$).
If $I$ is a homogeneous ideal of $R$, we know that $I$ is generated by homogeneous elements of $R$. We know $I$ also has some minimal generating set as an ideal of $R$. However, I am interested in finding a minimal generating set of $I$ which consists only of homogeneous elements. For example, in $R=K[x,y]$, we can take $I=(x,y)=(x+y^2,y)$, so not every minimal generating set needs to be homogeneous. If we write $\mu(I)$ to be the cardinality of a minimal generating set (which need not be homogeneous) and $\mu_h(I)$ to be the be the cardinality of a minimal generating set which is homogeneous, it is clear that $\mu_h(I) \ge \mu(I)$.
Is it possible that no minimal generating set of $I$ needs to be homogeneous? That is, it seems possible that $\mu_h(I)>\mu(I)$. I see this answer (https://math.stackexchange.com/a/3578204/98077) which notes that it is possible for an $\mathbb{Z}$-graded ring to have a principal ideal ($\mu(I)=1$) where $\mu_h(I)$ is arbitrarily large, but I do not immediately see how to use that answer to create a similar Noetherian $\mathbb{N}$-graded example where the degree zero part is a field.
In case more restrictive hypotheses on $I$ helps the issue, my primary case of interest is that $I$ is a homogeneous ideal which is a reduction of $R^+$, so that for some $N\in \mathbb{N}$, we have $I(R^+)^N=(R^+)^{N+1}$. By standard tricks in local algebra, $\mu(I)\ge \dim(R)$ and that there is a (homogeneous) ideal $J\subset I$ such that $\mu(J)=\dim(R)$. I worry however that we may have $\mu_h(J)>\dim(R)$ for every minimal reduction $J$ of $R^+$. In other words, is it possible that $R$ has no homogeneous system of parameters which also form a reduction of $R^+$?
This is not a worry if $R$ is standard graded, since a theorem from Matsumura (Commutative Ring Theory, Theorem 14.14) seems to indicate that minimal reductions of $R^+$ can be generated by linear forms, given by generic linear combinations of $K$-vector space generators of $R_1$. However, in the rings I am thinking about, $R_1=0$ is a common occurrence.
An example with $\mu_h(J)>\dim(R)$ or a reference to a paper where $\mu_h(J)=\dim(R)$ is shown under some hypotheses would be appreciated!
Best Answer
After some research, I think I can answer this question myself.
If $R$ is $\mathbb{N}$-graded over a field $K$ with homogeneous maximal ideal $R^+$ and $I$ is a homogeneous ideal, it will always have a minimal homogeneous generating set. The idea is that $I/(R^+)I$ is a graded $K$-vector space, and has a homogeneous basis consisting of $\mu_h(I)$ elements.
But, if we consider instead the local ring $(R_{R^+},(R^+)R_{R^+},K)=(S,\mathfrak{m},K)$ and let $J=IS$, $J/\mathfrak{m}J$ is canonically isomorphic to $I/(R^+)I$ as $K$-vector spaces, since $I$ and $R^+I$ are inside $R^+$. Then, $\mu(J)=\mu(I)$ and $\mu(J)=\dim_K(J/\mathfrak{m}J) = \dim_K(I/(R^+)I)=\mu_h(I)$. From this, we can even get the number of minimal generators of $I$ of degree $m$ as the dimension of the $K$-vector space $[I/(R^+)I]_m$. This is a standard argument about graded Betti numbers as well, see this MathOverflow answer.
The reason this proof doesn't apply to the example in the answer to this previously linked question is that the $\mathbb{Z}$-graded ring mentioned does not contain a field in its degree 0 part.
This also settles the existence of homogeneous reductions of the maximal ideal. In fact, a quick graded adaptation of the proof of the existence of minimal reductions in the excellent paper "Reductions of Ideals in Local Rings" by D. Northcott and D. Rees will guarantee homogeneous minimal reductions also exist. It does seem possible that the reduction number measured using only homogeneous minimal reductions may be larger than the reduction number measured using all minimal reductions, but I am not sure about this.