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Let $S = \sum_{n\ge 0} S_n$ be a graded commutative ring. Let $f$ be a homogeneous element of $S$ of degree $> 0$. Let $S_{(f)}$ be the degree $0$ part of $S_f$. If $S$ is Noetherian, $S_{(f)}$ is also Noetherian?

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This is a special case of the following result: Let $R$ be a $\mathbb{Z}$-graded ring. Then $R$ is Noetherian iff $R_0$ is Noetherian and $R$ is a finitely generated $R_0$ algebra. (This result can be found, for instance, in Bruns and Herzog, Theorem 1.5.5.) – user26857 Nov 8 '12 at 16:38
up vote 3 down vote accepted

Yes. If $I$ is an ideal of $S_{(f)}$, then $IS_f$ is a finitely generated ideal of $S_f$ because $S_f$ is noetherian. We can find a set of generators $a_1, \dots, a_r\in I$ of $IS_f$.

Let us show $a_1, \dots, a_r$ generate $I$. Let $a\in I$. Then $$a=a_1 b_1 + \cdots + a_r b_r, \quad b_i\in S_f.$$ Decomposing the $b_i$ into sums of homogeneous elements of $S_f$ and identifying elements of degree $0$ in the above equality, we find $a\in a_1S_{(f)}+\cdots a_rS_{(f)}$.

We used the fact that $S_f$ is a graded algebra over $S_{(f)}$: $$ S_f=\oplus_{k\in \mathbb Z} (S_f)_k$$ where $(S_f)_k$ denotes the elements of the form $b/f^N$ with $b\in S$ homogeneous of degree $N+k$.

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Let $A\subset B$ be a (commutative) ring extension such that $A$ is a direct summand of $B$ (considered as an $A$-module). If $B$ is noetherian then $A$ is noetherian.

Hint. For any ideal $\mathfrak a$ of $A$ we have $\mathfrak a=\mathfrak aB\cap A$.

In order to solve the proposed question set $B=S_f$ and $A=S_{(f)}$. If $S$ is noetherian, then $B$ is noetherian, and therefore $A$ is noetherian.

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Hint for the Hint: show that an increasing sequence of ideals of $A$ is stationary (I'm writing this comment because I first thought one could find finitely many generators for $\mathfrak a$ using the Hint but this got me nowhere). +1 for this proof and the interesting result on which it relies. – Georges Elencwajg Dec 19 '12 at 12:08
+1 for the same reason as Georges'. – Makoto Kato Dec 19 '12 at 14:47

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