0
$\begingroup$

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.

$\endgroup$
2
  • 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

1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.