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Let $I$ be a homogeneous ideal in a graded local commutative ring $R$, $S$ be its minimal homogeneous system of generators. So, we know that the cardinality of $S$ is unique as the dimension of the vector space $I/\mathfrak{m}I$, where $\mathfrak{m}$ is the maximal graded ideal of $R$.
My question is the following: Is the degree of each element of $S$ uniquely determined by $I$?
Thank for reading my question.
Fix $i\ge 1$ and let $S_i=S\cap I_i$ be the set of elements of degree $i$ in $S$. Let us prove that $S_i$ has $\dim_k (I/\mathfrak m I)_i$ elements, where $k$ is the residue field of $R$.
Let $f\in I_i$. Then $$f=r_1 s_1 + \dots +r_n s_n$$ where $r_j\in R$ are homogeneous and $S=\{ s_1, \dots, s_n\}$. We can just keep the homogeneous components in $r_j$ of the right degree so that $\deg r_j+\deg s_j=i$ for all $j\le n$. If $\deg r_j>0$, then $r_j\in \mathfrak m$ and $r_js_j\in \mathfrak m I$. This means that modulo $\mathfrak m I$, $I_i$ is generated by the elements of $S_i$.
Now suppose $s_1, s_2, \dots, s_m$ are the elements of $S_i$. If $m$ is bigger than the dimension of $I/\mathfrak m I$ as $k$-vector space, then, up to renumbering, we have $$s_1\in t_2s_2+\dots + t_{m} s_{m} + \mathfrak m I, \quad t_i\in R$$ Expanding $I$ with $s_1, \dots, s_n$, we see that $s_1\in s_2 R + \dots + s_n R$ and $S$ would not be minimal. Contradiction.
If $R$ is a graded $K$-algebra, then the answer is yes (in the folowing sense): exactly $\dim_K (I/\mathfrak{m}I)_i$ among the elements of $S$ have degree $i$.
