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Theorem 2.5 in Conca and Herzog, Castelnuovo-Mumford regularity of products of ideals http://arxiv.org/abs/math/0210065, says that if $R$ is a polynomial ring over a field $k$, $I$ a homogeneous ideal of $R$ such that $\dim R /I \le 1$ and $M$ is a finitely generated, graded $R$-module, then $reg(IM) \le reg(M) + reg(I)$, where $reg(\cdot)$ denotes Castelnuovo-Mumford regularity.

Definition: An element $x \in R$ is called almost regular on $M$, if it is not a zero divisor of $M / H_m^0(M)$.

The proof begins by the statement "let $x \in R_1$ be almost regular on $M, M/IM, R/I$". Observe that if $x \in R_1$ is almost regular on $M/IM$, then it will be almost regular on $M, R/I$ as well. So $x \in R_1$ is almost regular on $M, M/IM, R/I$ if and only if it is almost regular on $M / IM$.

TheThe proof does not seem to treat the case where no such $x$ exists, i.e. no element of $R_1$ is almost regular on $M/IM$. I suspect this is true because the statement of the theorem holds true in that case.

Question: Suppose that no element of $R_1$ is almost regular on $M /IM$. Then why is it true that $reg(IM) \le reg(M) + reg(I)$?

Theorem 2.5 in Conca and Herzog, Castelnuovo-Mumford regularity of products of ideals http://arxiv.org/abs/math/0210065, says that if $R$ is a polynomial ring over a field $k$, $I$ a homogeneous ideal of $R$ such that $\dim R /I \le 1$ and $M$ is a finitely generated, graded $R$-module, then $reg(IM) \le reg(M) + reg(I)$, where $reg(\cdot)$ denotes Castelnuovo-Mumford regularity.

Definition: An element $x \in R$ is called almost regular on $M$, if it is not a zero divisor of $M / H_m^0(M)$.

The proof begins by the statement "let $x \in R_1$ be almost regular on $M, M/IM, R/I$". Observe that if $x \in R_1$ is almost regular on $M/IM$, then it will be almost regular on $M, R/I$ as well. So $x \in R_1$ is almost regular on $M, M/IM, R/I$ if and only if it is almost regular on $M / IM$.

The proof does not seem to treat the case where no such $x$ exists, i.e. no element of $R_1$ is almost regular on $M/IM$. I suspect this is true because the statement of the theorem holds true in that case.

Question: Suppose that no element of $R_1$ is almost regular on $M /IM$. Then why is it true that $reg(IM) \le reg(M) + reg(I)$?

Theorem 2.5 in Conca and Herzog, Castelnuovo-Mumford regularity of products of ideals http://arxiv.org/abs/math/0210065, says that if $R$ is a polynomial ring over a field $k$, $I$ a homogeneous ideal of $R$ such that $\dim R /I \le 1$ and $M$ is a finitely generated, graded $R$-module, then $reg(IM) \le reg(M) + reg(I)$, where $reg(\cdot)$ denotes Castelnuovo-Mumford regularity.

Definition: An element $x \in R$ is called almost regular on $M$, if it is not a zero divisor of $M / H_m^0(M)$.

The proof begins by the statement "let $x \in R_1$ be almost regular on $M, M/IM, R/I$". The proof does not seem to treat the case where no such $x$ exists, i.e. no element of $R_1$ is almost regular on $M/IM$. I suspect this is true because the statement of the theorem holds true in that case.

Question: Suppose that no element of $R_1$ is almost regular on $M /IM$. Then why is it true that $reg(IM) \le reg(M) + reg(I)$?

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Manos
Manos
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Theorem 2.5 in "Castelnuovo-Mumford regularity of products of ideals" by Conca & Herzog

Theorem 2.5 in Conca and Herzog, Castelnuovo-Mumford regularity of products of ideals http://arxiv.org/abs/math/0210065, says that if $R$ is a polynomial ring over a field $k$, $I$ a homogeneous ideal of $R$ such that $\dim R /I \le 1$ and $M$ is a finitely generated, graded $R$-module, then $reg(IM) \le reg(M) + reg(I)$, where $reg(\cdot)$ denotes Castelnuovo-Mumford regularity.

Definition: An element $x \in R$ is called almost regular on $M$, if it is not a zero divisor of $M / H_m^0(M)$.

The proof begins by the statement "let $x \in R_1$ be almost regular on $M, M/IM, R/I$". Observe that if $x \in R_1$ is almost regular on $M/IM$, then it will be almost regular on $M, R/I$ as well. So $x \in R_1$ is almost regular on $M, M/IM, R/I$ if and only if it is almost regular on $M / IM$.

The proof does not seem to treat the case where no such $x$ exists, i.e. no element of $R_1$ is almost regular on $M/IM$. I suspect this is true because the statement of the theorem holds true in that case.

Question: Suppose that no element of $R_1$ is almost regular on $M /IM$. Then why is it true that $reg(IM) \le reg(M) + reg(I)$?