If $S$ is a Noetherian graded ring, $S_{(f)}$ is also Noetherian? 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?
 A: 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 $N+k$. 
A: 
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.
