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.

I was looking for an example of a non-Noetherian complete local commutative ring with $1$. I would appreciate if anyone can point to a reference.

share|improve this question
add comment

2 Answers

up vote 4 down vote accepted

I think that you should look at the ring $k[[X_1,X_2,...]]$ of power series with an infinity of variables. This ring is complete (as the completion of the ring $k[X_1,X_2,...]$ of polynomials with an infinity of variables) and its unique maximal ideal is the one consisting of power series without constant term. It is clearly not noetherian, considering the chain of ideals generated by $X_1$, $\{X_1,X_2\}$, $\{X_1,X_2,X_3\}$, etc.

share|improve this answer
add comment

Take an algebraic closure of $\mathbb{Q}_p$, the field of $p$-adic numbers, and complete it to get a complete field $\mathbb{C}_p$. It turns out that $\mathbb{C}_p$ is still algebraically closed. The valuation ring $R$ of $\mathbb{C}_p$ is a complete local ring whose maximal ideal $\mathfrak{m}$ satisfies $\mathfrak{m}^2=\mathfrak{m}$. It follows from Nakyama's lemma and the fact that $\mathfrak{m}\neq 0$ that $R$ is not Noetherian.

EDIT: I was thinking about this example recently and realized that it is not an example of the type requested by the OP, because the valuation ring $R$, while complete with respect to its valuation topology, is not $\mathfrak{m}$-adically complete due to the relation $\mathfrak{m}^2=\mathfrak{m}\neq 0$, which implies $\mathfrak{m}^n=\mathfrak{m}$ for all $n\geq 1$. This relation shows that $R$ is not $\mathfrak{m}$-adically separated, which is part of the definition of $\mathfrak{m}$-adically complete. Its $\mathfrak{m}$-adic completion is the residue field $R/\mathfrak{m}$, which is an algebraic closure of $\mathbf{F}_p$.

share|improve this answer
    
Very interesting! Do you have a reference? –  Alex Becker Apr 6 '12 at 23:32
1  
There is a more direct way to see the maximal ideal of $R$ is not finitely generated: for any finite set of elements $x_1,\dots,x_k$ in the maximal ideal their absolute values all satisfy $|x_i| \leq 1-\varepsilon$ for some small $\varepsilon$. Then the ideal in $R$ generated by the $x_i$'s is a subset of $\{x \in R : |x| \leq 1-\varepsilon\}$, which is a proper subset of the maximal ideal. Hence the maximal ideal is not finitely generated. –  KCd Apr 6 '12 at 23:59
add comment

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.