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Let $R=k[x_1,\dots,x_r]$ be the polynomial ring over the field $k$. Denote by $R_1$ the vector space of linear forms, i.e. all the degree-$1$ elements of $R$. Let $M \neq 0$ be a finitely generated graded $R$-module and suppose $H_m^0(M)=0$, i.e. no non-zero element of $M$ is annihilated by a power of $m=(x_1,\dots,x_r)$.

Question: $R_1$ can be viewed as the affine space $\mathbb{A}^r$, which is irreducible in the Zariski topology. Can we show that there exists a Zariski-open subset of $R_1$ consisting of regular elements of $M$?

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Yes: The set $\text{Reg}_R(M)$ of homogeneous $M$-regular elements is the complement of the union of the homogeneous associated primes of $M$, i.e. those ${\mathfrak p}$ for which a $R/{\mathfrak p}\langle k\rangle\hookrightarrow M$ for some $k$. That is, $$\quad\quad\ \ \ \ \text{Reg}_R(M) = R \setminus\bigcup\limits_{{\mathfrak p}\in\text{Ass}_R(M)} {\mathfrak p},\\(\ddagger)\quad\quad\text{Reg}_R(M)_1 = R_1 \setminus\bigcup_{{\mathfrak p}\in\text{Ass}_R(M)} {\mathfrak p}_1.$$ Since the ${\mathfrak p}\in\text{Ass}_R(M)$ ocurring on the right hand side of $(\ddagger)$ are ideals and in particular a ${\mathbb k}$-subspaces of $R$, their degree-$1$ components ${\mathfrak p}_1$ are ${\mathbb k}$-subspaces of $R_1\cong {\mathbb A}^r$. Moreover, they are proper subspaces of $R_1$ as otherwise $(x_1,...,x_n)={\mathfrak m}\in\text{Ass}_R(M)$ which contradicts the assumption that $\text{H}^0_{\mathfrak m}(R;M)=0$. Hence, $(\ddagger)$ expresses $\text{Reg}_R(M)_1\subset{\mathbb A}^r$ as the complement of a finite ($\text{Ass}_R(M)$ is finite for finitely generated $M$) union of proper subspaces of ${\mathbb A}^r$, which is hence a Zariski open set and non-empty if ${\mathbb k}$ is infinite.

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  • $\begingroup$ Dear Manos, I tried to clarify now. $\endgroup$ – Hanno Dec 28 '14 at 7:11
  • $\begingroup$ Excellent. Thank you. $\endgroup$ – Manos Dec 28 '14 at 7:19
  • $\begingroup$ You're welcome! :) $\endgroup$ – Hanno Dec 28 '14 at 8:01

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