Dimension of polynomial algebras - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T07:11:14Z http://mathoverflow.net/feeds/question/115403 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/115403/dimension-of-polynomial-algebras Dimension of polynomial algebras Fred Rohrer 2012-12-04T15:01:30Z 2012-12-05T08:28:06Z <p>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.</p> <p>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).</p> <p>Are there other interesting classes of rings with the property that the Krull dimensions of their polynomial algebras are minimal in the above sense?</p> http://mathoverflow.net/questions/115403/dimension-of-polynomial-algebras/115411#115411 Answer by J.C. Ottem for Dimension of polynomial algebras J.C. Ottem 2012-12-04T15:41:42Z 2012-12-04T15:41:42Z <p>This class of rings is studied by Arnold and Gilmer in the paper </p> <p>Jimmy T. Arnold and Robert Gilmer '<a href="http://www.jstor.org/stable/2373549" rel="nofollow">The Dimension Sequence of a Commutative Ring</a>' American Journal of Mathematics , Vol. 96, No. 3 (1974), pp. 385-408.</p> http://mathoverflow.net/questions/115403/dimension-of-polynomial-algebras/115481#115481 Answer by François Brunault for Dimension of polynomial algebras François Brunault 2012-12-05T08:28:06Z 2012-12-05T08:28:06Z <p>The following reference should be of interest :</p> <p>Brewer, Montgomery, Rutter, Heinzer, <em>Krull dimension of polynomial rings</em>.</p> <p>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.</p> <p>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.</p> <p>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. </p>