Why are Artinian rings of Krull dimension 0?

As in the example of $\mathbb{Z}/(6)$, the ideal $\mathbb{Z}/(2)$ is prime, I think. So, Artinian rings may contain prime ideals. But why does not the primes ideal contain other prime ideals properly?

Another question is that, when a ring is of Krull dimension 0, is it necessarily Artinian?



2 Answers 2


The following proof that any artinian ring $R$ has $\text{dim}(R)=0$ is from Theorem 8.1 of Atiyah Macdonald:

Let $\frak{p}$ be a prime ideal of $R$, so that $S=R/\frak{p}$ is an artinian domain (this is because any infinite descending chain of ideals in $S$ could be lifted to an infinite descending chain of ideals in $R$, so because $R$ is artinian, $S$ must be artinian too). For any non-zero $x\in S$, we must have that $(x^n)=(x^{n+1})$ for some $n$ (this is because $S$ is artinian, so we can't have an infinite descending chain $S\supset (x)\supset (x^2)\supset\cdots$), hence $x^n=x^{n+1}y$ for some $y\in S$, but because $S$ is a domain, we can cancel to get $1=xy$, hence $x$ is invertible. Any non-zero element of $S$ is invertible, hence $S=R/\frak{p}$ is a field, hence $\frak{p}$ is maximal. Because any prime ideal of $R$ is maximal, we must have that $\text{dim}(R)=0$.

Theorem 8.5 of Atiyah Macdonald says that $R$ is artinian $\iff$ $R$ is noetherian and $\text{dim}(R)=0$.

So here is an example of a ring $R$ with $\text{dim}(R)=0$ but $R$ not noetherian, and hence not artinian:

$$R=k[x_1,x_2,\ldots]/(x_1,x_2,\ldots)^2\cong k[\epsilon_1,\epsilon_2,\ldots]$$ which is a field $k$ with infinitely many nilpotent elements $\epsilon_i$ added in. There's only one prime ideal of $k$, namely the zero ideal, and nilpotents won't change that, so the ideal $(\epsilon_1,\epsilon_2,\ldots)$ is the only prime ideal of $R$, but $R$ is certainly not noetherian - the chain of ideals $$(0)\subset (\epsilon_1)\subset(\epsilon_1,\epsilon_2)\subset\cdots$$ is an infinite ascending chain.

  • $\begingroup$ My algebra professor proved 8.5 in class just the other day. Funny I should run into it so soon. $\endgroup$ Mar 16, 2011 at 6:22
  • $\begingroup$ Shouldn't the intersection of all prime ideals in $R$ be $\mathfrak{p}:=(\epsilon_1,\epsilon_2,...)$? I.e., all nilpotent elements? The ring still has krull dimension $0$ because any domain that occurs as the quotient must be by an ideal containing all of these elements and this is ideal is also maximal (quotienting gives $k$). $\endgroup$
    – Eoin
    Jun 25, 2015 at 17:47
  • $\begingroup$ @Eoin: You're quite right, I'm not sure how I got so mixed up. I'll edit my answer. $\endgroup$ Jul 2, 2015 at 20:42

HINT $\ $ By factoring out a prime ideal it reduces to showing that an Artinian domain is a field, which follows immediately by DCC.

This method of factoring out by a prime ideal to reduce from rings to domains is a ubiquitous algebraic problem solving technique. Indeed, the eminent algebraist Irving Kaplansky explicitly pauses to mention this method (in a less trivial context) in his classic textbook Commutative Rings. Follow the above link for an excerpt. Not only was Kap a great algebraist but also a great expositor - a rare combination. I wholeheartedly recommend his expositions, where I recall learning many beautiful mathematical ideas.

  • 3
    $\begingroup$ Certainly your answer isn't wrong, but uh... it does seem a bit redundant, no? $\endgroup$ Mar 15, 2011 at 16:16
  • 6
    $\begingroup$ I don't agree. The proof lifted from A&M doesn't give any hint that it is a special case of the ubiquitous algebraic technique of factoring out a prime. Pedagogically it is essential to make explicit such ubiquitous problem-solving techniques so that they can be consciously reified by students. If the student knew this technique he would not be asking this question. $\endgroup$ Mar 15, 2011 at 16:37

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