I am studying commutative algebra at the moment, so all rings are assumed commutative (and unital).

Does there exist a Jacobson local ring $\newcommand{\mfm}{\mathfrak{m}}(A, \mfm)$ that is not an Artinian ring?

For the context I've been studying in, Jacobson rings are defined as rings where all prime ideals are intersections of maximal ideals. A local ring is a ring containing exactly one maximal ideal.

I have come across the result that a ring is Artinian if and only if it is Noetherian and every prime ideal is maximal. Since $A$ is Jacobson, we can deduce that $\mfm$ is the unique prime ideal of $A$, so in particular every prime ideal of $A$ is maximal. My question should then be equivalent to finding a Jacobson local ring that is not Noetherian.

  • $\begingroup$ Is your question "is there a local ring that is not Artinian" (as in the body of the question) or "is there a Jacobson local ring that is not Artinian" (as in the title)? $\endgroup$ May 14, 2018 at 2:17
  • $\begingroup$ Sorry, missed the word in the body. Fixed. $\endgroup$
    – Mike Ho
    May 14, 2018 at 2:24
  • $\begingroup$ DaRT query for commutative nonartinian local jacobson rings. $\endgroup$
    – rschwieb
    May 3, 2022 at 15:35

1 Answer 1


Let $I = (x_ix_j ~|~ i,j \in \mathbb N) \subset \mathbb Q[x_0,x_1,x_2, \dotsc] =: R$. Then $R/I$ is Jacobson and local because every prime ideal contains $x_i^2$ thus contains $x_i$, i.e. is equal to the maximal ideal $(x_0,x_1, x_2, \dotsc)$.

But it is not noetherian, because the maximal ideal is not finitely generated. Hence it is also not artinian.

  • $\begingroup$ Amazing! This is the first time I have seen a local ring that did not come out of localising at a prime. For sake of generalisation, this means that if a ring $A$ has maximal ideal $\mfm$, then for every $k \in \mathbb{N}$ we should have that $A/\mfm^k$ is Jacobson and local (with unique prime/maximal ideal $\mfm/\mfm^k$), so as long as $\mfm$ is not finitely generated as an ideal then we have a counterexample. $\endgroup$
    – Mike Ho
    May 14, 2018 at 6:36
  • $\begingroup$ Yes, but note that in general there are cases, where $A/\mathfrak m^k$ is noetherian, although $A$ was not. $\endgroup$
    – MooS
    May 14, 2018 at 6:51

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