Let $A=\bigoplus_{n\geq 0}A_n$, $B=\bigoplus_{n\geq 0}B_n$ be two graded Noetherian rings, where $A_n\subset B_n$ and $A_0=B_0$ is a local ring. Suppose $B$ is a finitely generated $A$ module. Let $I$ be an ideal in $A_0$. Define $M=\bigoplus_{n\geq 0}M_n$ such that $M_0=I$ and $M_i=0$ for $i\geq 1$, with the property $am=0$ for all $a\in A_n$ with $n\geq 1$ and $m\in I$. Then is the following statement true

Thm: $M$ and $\bigoplus(\frac{B_n}{A_n}\otimes I)$ are finitely generated $A$ modules.

I think this result is true but I want to be sure. Any suggestion would be really helpful. Thanks in advance.

  • 1
    $\begingroup$ The expression $B_n/A_n \otimes I$ doesn't make sense, since $B_n$ isn't an $A$-module, for any $n$. This follows from the fact that $A_1B_n \subseteq B_{n+1}$, for example. Do you mean $B/A \otimes M$ as an $A$-module? $\endgroup$
    – Exit path
    Jul 24, 2017 at 15:34
  • $\begingroup$ Yes. I wrote direct sum. $\endgroup$ Jul 24, 2017 at 16:53

1 Answer 1


To expand on my comment, if you're asking whether the modules $M$ and $B/A \otimes_A M$ are finitely generated, the answer is yes. Since $A$ is noetherian, so is $A_0$. In particular, $I$ is a finitely generated ideal in $A_0$, so it's easy to see that $M$ is a finitely generated module over $A$. Lastly, since $B/A$ is the quotient of a finitely generated module, it too is a finitely generated module, and so $B/A \otimes M$ is finitely generated as well.


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