In Eisenbud's book Commutative Algebra with a View Toward Algebraic Geometry, in the prove of Proposition 1.4, the auther seems to use the following fact.

Let $R$ be a Noetherian ring, $M$ is a finitely generated $R$-module, $N$ is a submodule of $M$, and $f_1\in M$. Then the author says:

Since $Rf_1\subset M$ is generated by one element, its submodule $N\cap Rf_1$ is finitely generated.

My question is, I don't think that if $K$ is finitely generated (even with one element), then any submodule of $K$ is also finitely generated. For example, any Non-Noetherian ring $R$ is finitely generated by ONE element, say 1, but there exists an ideal which is not finitely generated.

Why can Eisenbud deduce such conclusion?

  • 3
    $\begingroup$ because $R$ is Noetherian, then your $K$ is a Noetherian $R$-module, meaning that any $R$-submodule is finitely generated. $\endgroup$ – user27126 Nov 10 '13 at 22:26
  • $\begingroup$ Submodules of finitely generated modules over Noetherian rings are finitely generated. $\endgroup$ – user40276 Nov 10 '13 at 22:26
  • $\begingroup$ @Sanchez, Oh I see, thanks $\endgroup$ – hxhxhx88 Nov 10 '13 at 23:10

Specifically for the case, Let $R_N=\{r \in R/ rf_1 \in N\}$. It's easy to see that $R_N$ is an ideal of $R$. Since $R$ is Noetherian, the ideal $R_N$ of $R$ is finitely generated.Let $r1,..,r_s$ générators of $R_N$, then it's easye to sea that $r_1f_1, \cdots, r_sf_1$ are generators of $N \cap Rf$


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