Krull dimension of mixed extensions
Let \(R\) be a commutative ring. Then it is well known that \(\dim R+1\leq \dim RX\leq 2\dim R+1\). What happens if we replace \(R[X]\) by \(RX\)? Certainly \(\dim RX\geq \dim R+1\) and we have equality if \(R\) is Noetherian. Recall that \(R\) is an \textit{SFT ring} if for each ideal \(I\) of \(R\), there exists a finitely generated ideal \(B\subseteq I\) and a positive integer \(k\) depending on \(I\) such that \(a^{k}\in B\) for each \(a\in I\). \textit{J. T. Arnold} [Trans. Am. Math. Soc. 177, 299--304 (1973; Zbl 0262.13007)] has shown that if \(R\) is non SFT, then \(\dim RX=\infty \) and \(\dim D[[X_{1},\dots,X_{n}]]=n\dim D+1\) if \(D\) is an SFT Prüfer domain [\textit{J. T. Arnold}, Indiana Univ. Math. J. 31, 897--911 (1982; Zbl 0475.13014)]. \textit{J. Coykendall} [J. Algebra 256, No. 1, 85--96 (2002; Zbl 1069.13011)] gave an example of a finite dimension SFT ring with \(\dim RX=\infty \) and raised the question (also raised by Gilmer) of whether \(\dim RX<\infty \) implies that \(\dim RX\leq 2\dim R+1\). This paper shows that the answer is no. This paper considers ``mixed extensions'' \(R[X_{1}]]\cdots [ X_{n}]]\) where \([X]]\) means that \([X]]\) could either be \([X]\) or \(X\). Generalizing Arnold's result they show that for \(D\) a finite dimensional SFT Prüfer domain (e.g., a finite dimensional discrete valuation domain) \(\dim D[X_{1}]]\cdots [ X_{n}]]=n\dim D+1\) provided at least one of the extensions is a power series extension (of course, for \(D\) Prüfer we have \( \dim D[X_{1},\dots,X_{n}]=\dim D+n\)). In particular, if \(R=D[X_{1},\dots,X_{n-1}]\) and \(\dim D=d>2(n-1)/(n-2)\) (where \(n>2\)), then \(\dim RX=dn+1>2\dim R+1.\)
scientific article; zbMATH DE number 5566280
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