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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.

Jake Voigt
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3 Answers3

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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.

bzc
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Louis La Brocante
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  • If by $k[[X_1,X_2,\ldots]]$ you mean the completion of $k[X_1,X_2,\ldots]$ at the maximal ideal $(X_1,X_2,\ldots)$, the resulting local ring with maximal ideal $\mathfrak{m}$ is not $\mathfrak{m}$-adically complete. See https://stacks.math.columbia.edu/tag/05JA. Maybe if $k[[X_1,X_2,\ldots]]$ instead means formal sums of monomials $\prod X_i^{a_i}$ with $a_i \geq 0$ and $\sum a_i < \infty$ (e.g. including $X_1 + X_2 + \cdots $) then maybe this could work...but I didn't check this. – 654897419 Jul 11 '22 at 00:05
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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 $\mathbf{F}_p$.

Keenan Kidwell
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  • Very interesting! Do you have a reference? – Alex Becker Apr 06 '12 at 23:32
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    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 06 '12 at 23:59
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As another example, you can consider the commutative ring of series with shuffle product. Let $\mathbb F$ be a field (say the rational numbers $\mathbb F = \mathbb Q$), $\Sigma$ a finite set of symbols, and consider the set of series $\mathbb F\langle\!\langle\Sigma \rangle\!\rangle := \Sigma^* \to \mathbb F$ with componentwise sum "$+$" and shuffle product "⧢". For instance, $aab ⧢ c = caab + acab + aacb + aabc$. The resulting structure of shuffle series $$(\mathbb F\langle\!\langle\Sigma \rangle\!\rangle, +, ⧢, 0, 1)$$ is a commutative ring with unit. I mentioned in this answer that the the shuffle series ring is not Noetherian.

This ring $R$ has a valuation $v : R \to \mathbb N \cup \{\infty\}$ mapping a nonzero series to the shortest length of a word in its support. For instance $v(ba^2 + a^5 + \cdots) = 3$. $R$ is also a local ring with maximal ideal $\mathfrak m$ consisting of all series $f \in R$ with $v(f) \geq 1$, since all series outside $\mathfrak m$ are units of $R$.

Finally, this ring is complete w.r.t. the topology induced by the valuation. We have thus presented a complete local (commutative) ring which is not Noetherian.

Lorenzo
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