6
$\begingroup$

Let $R$ be a commutative ring of Krull dimension $d$, let $n\in\mathbb{N}$, and let $R[X_1,\ldots,X_n]$ denote the polynomial algebra in $n$ indeterminates over $R$. One can show that then we have $\dim(R)+n\leq\dim(R[X_1,\ldots,X_n])$. So, it is natural to wonder about the class of rings $R$ for which this inequality is an equality.

I know of only two subclasses of this class: Namely, the class of noetherian rings (Krull 1951) and the class of Prüfer rings (Seidenberg 1954).

Are there other interesting classes of rings with the property that the Krull dimensions of their polynomial algebras are minimal in the above sense?

$\endgroup$
3
  • 1
    $\begingroup$ The following reference should be of interest : Brewer, Montgomery, Rutter, Heinzer, Krull dimension of polynomial rings. For example, they prove, see Corollary 2 p. 30, that any semi-hereditary ring (all finitely generated ideals are projective) satisfies the dimension formula above. This generalizes Seidenberg's result since a Prüfer ring is a semi-hereditary integral domain. $\endgroup$ Commented Dec 4, 2012 at 16:28
  • $\begingroup$ Dear @François, this is a very interesting reference! May I ask you to add it as a proper answer? $\endgroup$ Commented Dec 5, 2012 at 7:37
  • 1
    $\begingroup$ Dear Fred, done -- and slightly expanded the answer. $\endgroup$ Commented Dec 5, 2012 at 8:28

3 Answers 3

2
$\begingroup$

The following reference should be of interest :

Brewer, Montgomery, Rutter, Heinzer, Krull dimension of polynomial rings.

For example, they prove, see Corollary 2 p. 30, that any semi-hereditary ring (all finitely generated ideals are projective) satisfies the dimension formula above. This generalizes Seidenberg's result since a Prüfer ring is a semi-hereditary integral domain.

It might be interesting to study the class of rings $R$ satisfying the following condition : for every prime ideal $P$ of $R$ and every $n \geq 1$, we have $\textrm{height}(P[X_1,\ldots,X_n]) = \textrm{height}(P)$. This condition implies $\dim R[X_1,\ldots,X_n] = \dim R +n$ for every $n$ (this can be deduced from Thm 1 of this paper), but I don't know about the converse.

The authors also discuss the class of strong $S$-rings introduced by Kaplansky (see the paper for the definition). This class contains the Noetherian rings and the Prüfer rings, and is stable by localizations and quotients. Kaplansky proved that a strong $S$-ring $R$ satisfies $\mathrm{height}(P[X]) = \textrm{height}(P)$ for every prime ideal $P$ of $R$, and thus $\textrm{dim}(R[X])=\textrm{dim}(R)+1$. But a strong $S$-ring doesn't necessarily satisfy the dimension formula for every $n$. In the other direction, the authors give an example of a ring which satisfies the height formula $\textrm{height}(P[X_1,\ldots,X_n]) = \textrm{height}(P)$ for every prime ideal $P$, but which is not a strong $S$-ring.

$\endgroup$
2
  • 1
    $\begingroup$ The converse of the implication you mention in the second paragraph does not hold; see p. 406 in the paper by Arnold and Gilmer mentioned by J.C. Ottem in his answer. $\endgroup$ Commented Dec 5, 2012 at 9:43
  • $\begingroup$ The class of rings satisfying the "height formula" $h(P[X_1,\ldots,X_n])=h(P)$ for every $P$ and $n$ is denoted by $\mathcal{R}_\infty$ in Arnold-Gilmer's paper. $\endgroup$ Commented Dec 5, 2012 at 14:49
3
$\begingroup$

This class of rings is studied by Arnold and Gilmer in the paper

Jimmy T. Arnold and Robert Gilmer 'The Dimension Sequence of a Commutative Ring' American Journal of Mathematics , Vol. 96, No. 3 (1974), pp. 385-408.

$\endgroup$
1
  • $\begingroup$ This - of course very interesting paper - contains a lot of information that is related to my question. But does it indeed concretely describe classes of rings with the desired properties? $\endgroup$ Commented Dec 4, 2012 at 16:02
2
$\begingroup$

The finite-dimensional integral rings satisfying the property in question are the so-called Jaffard domains. They are extensively studied in:

D. F. Anderson, A. Bouvier, D. E. Dobbs, M. Fontana, S. Kabbaj, On Jaffard domains, Expo.Math. 6 (1988). 145--175.

(A scan of this is available on S. Kabbaj's homepage.)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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