Let $R$ be a commutative unitary ring of finite Krull dimension and let $x \in R$. Is it true that $\dim R/(x) \ge \dim R 1$ ?
1 Answers
This is not true, even if $R$ is Noetherian! It is possible to have a Noetherian ring $R$ and $x \in R$ a nonunit, such that $\dim R/(x) = \dim R  2 < \infty$. The easiest counterexample I know is the following: if $S$ is a Noetherian local ring, $f \in S$ is a nonnilpotent nonunit, then $\dim S_f \le \dim S  1$ (since the only maximal ideal of $S$ is no longer prime in $S_f$). But $S_f = S[\frac{1}{f}] \cong S[t]/(tf  1)$, and $\dim S[t] = \dim S + 1$ (since $S$ is Noetherian), so taking $R = S[t]$, $x = tf  1$ gives a Noetherian ring with $\dim R/(x) \le \dim R  2$ (and equality holds if $f$ avoids all minimal primes of $S$).
As an explicit example: the ring $R = \mathbb{Z}_{(2)}[t]$ is Noetherian of dimension $2$, and the element $x = 2t  1$ is not a unit, but $R/(x) = \mathbb{Z}_{(2)}[t]/(2t1) \cong \mathbb{Z}_{(2)}[\frac{1}{2}] \cong \mathbb{Q}$ has dimension $0$.
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4This is a fantastic answer: not only is the idea excellent but it is writtenup in an extremely clear, didactical way. I encourage all users to frenetically upvote this post! And I encourage zcn to post under his real name! – Georges Elencwajg May 18 '14 at 07:59

1@zcn: Thank you very much for this great example. That helps me a lot. – user120513 May 18 '14 at 08:19

3However it is true that $\dim R/(x) \ge \dim R 1$ for noetherian rings if one knows that $x$ is in the Jacobson radical of $R$. – user26857 May 18 '14 at 10:33

2Another example: $R=K[[X]][Y]$ and $x=XY1$; for details see [this answer](http://math.stackexchange.com/a/99372) of @GeorgesElencwajg ;) – user26857 May 18 '14 at 11:27

@user26857: Concerning your first comment: In the noetherian case I think it's even enough to require that $x$ is in a maximal ideal of maximal height. But it's really appalling to see what can happen if $x$ is not. – user120513 May 18 '14 at 12:16

@zcn It appears to me that your example has a geometric analogue: take $k$ a field, $R = (k [s]_{(s)}) [t]$ and $x = s t  1$. Then the picture one thinks of is the hyperbola $\{ s t = 1 \} \subset \mathbb{A}^2_k$ (which is 1dimensional) and the germ of $\mathbb{A}^2_k$ around $\{ s = 0 \}$ (which is 2dimensional), and the intersection is unsurprisingly 0dimensional. – Zhen Lin Aug 08 '14 at 20:26