$\dim(R/x) = \dim(R)-1$ for Noetherian integral domains?

Question

Let $R$ be a Noetherian integral domain of finite Krull-dimension and $0 \neq x \in R$ a non-unit. Do we have $\dim(R/x) = \dim(R) -1$ in general?

If this is wrong, does it change something if we further assume that $R$ is positively graded, finitely generated by homogeneous elements of degree one over $R_0$, which is Artinian local, and $x$ is homogeneous of degree one?

Context: This dimension formula is used here.

You should also require $x$ to be non-unit, otherwise $R / (x) \cong { 0 }$. — Zhen Lin, Aug 08 '14 at 17:46
@Hamou: Do you mean Lemma 5.13? The other results on those pages are about local rings. — Dune, Aug 08 '14 at 18:00
The additional assumption that we want is for $R$ to be catenary. Then this follows from Krull's principal ideal theorem. — Andrew Dudzik, Aug 08 '14 at 18:20
@you-sir-33433: catenary is not quite enough - see the example in the linked answer in my post. However, it does suffice that $x$ is in the Jacobson radical of $R$ — zcn, Aug 08 '14 at 18:35
@zcn You are right. I believe I was thinking of equidimensionality. — Andrew Dudzik, Aug 08 '14 at 18:55

score 2 · Accepted Answer · edited Apr 13 '17 at 12:21

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In general $\dim R/(x)$ can be $< \dim R - 1$ - see here. However it does hold if $R$ is (in addition) local. Thus in the specific setting, where $R$ is a standard graded Noetherian domain (so that $R_0$ is a field, not just Artinian), then if $\DeclareMathOperator{\m}{\mathfrak{m}}$$\m$ is the homogeneous maximal ideal (which is maximal), we have $\dim R - 1 = \dim R_m - 1 = \dim R_m/xR_m = \dim (R/x)_m = \dim R/(x)$ (since $\dim R = {}^*\!\dim R$ if $R$ is positively graded with $R_0$ local).

edited Apr 13 '17 at 12:21

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answered Aug 08 '14 at 18:08

zcn

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Thank you very much! I just don't understand why you want to assume that $R_0$ is a field. Doesn't your argumentation hold as well for $R_0$ Artinian local, since then we also have a unique homogeneous maximal ideal? – Dune Aug 08 '14 at 18:27
It's true that there is a unique homogeneous maximal ideal if $R_0$ is only assumed Artinian local. But if $R$ is a domain, then $R_0 \subseteq R$ is also a domain, hence will be a field – zcn Aug 08 '14 at 18:29
Of course you are right. Thanks again! – Dune Aug 08 '14 at 18:31

$\dim(R/x) = \dim(R)-1$ for Noetherian integral domains?

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