# Dimension of polynomial algebras

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?

-
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. –  François Brunault Dec 4 '12 at 16:28
Dear @François, this is a very interesting reference! May I ask you to add it as a proper answer? –  Fred Rohrer Dec 5 '12 at 7:37
Dear Fred, done -- and slightly expanded the answer. –  François Brunault Dec 5 '12 at 8:28

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.

-
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. –  Fred Rohrer Dec 5 '12 at 9:43
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. –  François Brunault Dec 5 '12 at 14:49