Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
1  
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
    
@navigetor23 Thanks for the info. –  Makoto Kato Nov 9 '12 at 1:04

2 Answers 2

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 $N+k$.

share|improve this answer

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.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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