# Can the zero-degree part of $M_f \otimes_{S_f} N_f$ be identified with $M_{(f)} \otimes_{S_{(f)}} N_{(f)}$?

The isomorphism ${(M \otimes _ {S} N)} _ {f} = M _ {f} \otimes _ {S _ {f}} N _ {f}$ is well-known. Here, $S$ is a graded ring, and $M,N$ are graded $S$ modules.

Now, let $f$ be any homogeneous element of $S$ of degree 1. I want to have ${(M \otimes _ {S} N)} _ {(f)} = M _ {(f)} \otimes _ {S _ {(f)}} N _ {(f)}$. Here, $L _ {(f)}$ means the zero-degree part of $L _ {f}$ for any graded $S$-module $L$.

By using the first formula, the problem reduces to the following assertion: Can the zero-degree part of $M _ {f} \otimes _ {S _ {f}} N _ {f}$ be identified with $M _ {(f)} \otimes _ {S _ {(f)}} N _ {(f)}$ ?

I see that there is a canonical homomorphism $M _ {(f)} \otimes _ {S _ {(f)}} N _ {(f)} \to M _ {f} \otimes _ {S _ {f}} N _ {f}$, and I also see that the image of this homomorphism equals to the zero-degree part of $M _ {f} \otimes _ {S _ {f}} N _ {f}$.

However, I cannot prove that this homomorphism is injective.

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Also, you may consult therisingsea.org/notes/ModulesOverProjectiveSchemes.pdf – Martin Brandenburg Aug 5 '11 at 10:30
Thank you for your kindness. I will study this PDF file. – Hiro Aug 5 '11 at 10:49