# Krull dimension of local Noetherian ring (2)

Let $(A,\mathfrak{m})$ be a local Noetherian ring and $x \in \mathfrak{m}$.

Prove that $\dim(A/xA) \geq \dim(A)-1$, with equality if $x$ is $A$-regular (i.e. multiplication with $x,$ as a map $A\rightarrow A$ is injective).

The dimensions are Krull dimensions.

It may have something to do with this question Dimension inequality for homomorphisms between noetherian local rings, but I simply can't figure it out.

Thank you.

• This is basically the statement of Krull's Hauptidealsatz. I don't think this is supposed to be easy... – Zhen Lin Jun 4 '12 at 13:33
• Well, the statement that we proved at the course was: Let $(A,\mathfrak{m})$ local Noetherian, $f\in A$ regular and $\mathfrak{p} \in Min((f))$. Then $ht(\mathfrak{p})=1$. So I suppose that now we have to prove their equivalence. – Adrian Manea Jun 4 '12 at 13:39
• your statement is the equality case above. I'd guess in the proof of it, if you take out where you use regularity, you get the inequality – uncookedfalcon Jun 4 '12 at 13:57
• Actually, we proved Hauptidealsatz in the context of the Krull-Chevalley-Samuel theorem and primary decompositions, so the connection doesn't look obvious to me. I could present the proof, if it would serve our purpose. – Adrian Manea Jun 4 '12 at 14:04
• I'm very sorry, I seem to have been thinking of a different inequality. This seems even harder than I thought! – Zhen Lin Jun 4 '12 at 15:24

## 2 Answers

Prologue
For any commutative ring $R$ and any ideal $I\subsetneq R$ we have $$\dim(R/I)+ht(I)\leq \dim (R) \quad (*)$$ This does not assume $R$ noetherian, nor local, nor... but just follows from the definitions.

Inequality
Suppose now that $(A,\mathfrak m)$ is local noetherian.
The trick is to use that $\dim(A)$ is the smallest number of elements in $\mathfrak m$ generating an $\mathfrak m$-primary ideal (cf. Atiyah-Macdonald Theorem 11.14). Let's do that for $A/xA$:
If $\bar x_1,...,\bar x_k\in \mathfrak m/xA$ generate an $\mathfrak m/xA$-primary ideal, then $x, x_1,..., x_k$ generate an $\mathfrak m$-primary ideal and this immediately yields the required inequality $$\dim(A)\leq \dim(A/xA)+1 \quad (**)$$
Equality
The Prologue implies that equality in $(**)$ will hold if $ht(xA)=1$.
The principal ideal theorem says that we always have $ht(xA)\leq 1$.
Now to say that $ht(xA)=0$ means that $x\in \mathfrak p$ for some minimal ideal $\mathfrak p$.
But it is well known that minimal ideals consist of zero divisors (= non-regular elements).
Hence if $x$ is regular we have $ht(xA)=1$ (since we don't have $ht(xA)=0$ !) and the required equality follows $$\dim(A)= \dim(A/xA)+1 \quad (***)$$

• The solution looks comprehensive and more appropriate for my current level of knowledge. I will read it more carefully tomorrow and come back if there is anything left. Thanks again for your time! You, Mr. Georges, have style of exposition that I'm very fond of. Your help is, again, much appreciated, like Mr. Zhen's. – Adrian Manea Jun 4 '12 at 18:53
• Thank you Adrian, it is very pleasant for a teacher to read that. By all means come back tomorrow if something is not clear. – Georges Elencwajg Jun 4 '12 at 19:05
• This must have been the intended solution! I guess I was too focused on trying to make the geometric approach work (and also unfamiliar with primary ideals...) – Zhen Lin Jun 4 '12 at 22:56
• Dear @Zhen, I'm happy we have your solution : the more alternatives, the better! – Georges Elencwajg Jun 4 '12 at 23:31

Here is a solution, but unfortunately it invokes several deep results of dimension theory.

Theorem. (Krull's Hauptidealsatz)

1. Let $A$ be a ring, and let $a \in A$. If $a$ is not invertible, then every prime of $A$ minimal over $(a)$ has height at most $1$.

2. Let $A$ be a ring, and let $a_1, \ldots, a_r$ be elements of $A$. If $(a_1, \ldots, a_r) \ne A$, then every prime of $A$ minimal over $(a)$ has height at most $r$.

Proposition. A noetherian ring $R$ has finitely many minimal primes.

Proof. Let $\Sigma$ be the set of all ideals $I$ of $R$ such that there are infinitely many primes minimal over $I$. (Recall that every ring has at least one minimal prime, so there is always at least one prime minimal over $I$.) Partially order $\Sigma$ by inclusion; suppose, for a contradiction, that $\Sigma$ is non-empty – then $\Sigma$ has a maximal element $I$, because $R$ is noetherian. Obviously, $I$ is not prime, so there are $f$ and $g$ in $R$ such that $f g \in I$ with $f \notin I$ and $g \notin I$. Yet, any prime minimal over $I$ must also be minimal over $I + (f)$ or $I + (g)$, so $I$ has only finitely many minimal primes – a contradiction. $\qquad \blacksquare$

Corollary. Let $R$ be a noetherian ring, and let $\mathfrak{n}_1, \ldots, \mathfrak{n}_c$ be the minimal primes of $R$. If $\dim R < \infty$, then $$\dim R = \max_i \dim R / \mathfrak{n}_i$$

Proof. If $\dim R < \infty$, every maximal chain of primes of $R$ must start at a minimal prime and end at a maximal ideal. $\qquad \blacksquare$

Geometrically, what we're saying is that $\operatorname{Spec} R$ can be decomposed into finitely many irreducible components, and the dimension of $\operatorname{Spec} R$ is the maximum of the dimensions of those irreducible components. Thus, when doing dimension theory, we can sometimes get away with assuming that $R$ is an integral domain.

Fact. If $A$ is a local ring with maximal ideal $\mathfrak{m}$ and $\hat{A}$ is its $\mathfrak{m}$-adic completion, then $\dim A = \dim \hat{A}$.

Fact. A complete noetherian local ring is universally catenary. In particular, in a complete noetherian local domain, every saturated chain of primes has the same length.

Proposition. Let $A$ be a noetherian local ring with maximal ideal $\mathfrak{m}$ and residue field $k$. If $a \in \mathfrak{m}$, then $\dim A/(a) \ge \dim A - 1$.

Proof. We start by reducing to more tractable cases. First, observe that the completion of $A / (a)$ as a $A$-module is the same as the completion of $A / (a)$ as a local ring, so we will assume that $A$ is a complete local ring. Notice that it is enough to show that, for each minimal prime $\mathfrak{n}$ of $A$, there exists a prime $\mathfrak{c}$ of $A$ with $\mathfrak{n} \subseteq \mathfrak{n} + (a) \subseteq \mathfrak{c}$ and $\dim A/\mathfrak{n} - \dim A/\mathfrak{c} \le 1$. Moreover, it is enough to do this for a minimal prime $\mathfrak{n}$ such that $\dim A / \mathfrak{n} = \dim A$. Thus we may assume without loss of generality that $A$ is complete noetherian local domain.

Let $\mathfrak{c}_1, \ldots, \mathfrak{c}_n$ be the primes of $A$ minimal over $(a)$. The case $a = 0$ is uninteresting, so we assume $a \ne 0$. Then, our hypotheses together with Krull's Hauptidealsatz implies the height of each $\mathfrak{c}_i$ is exactly $1$. So $\dim A / (a) + 1 \le \dim A$ – therefore we are looking to prove that $\dim A / (a) = \dim A - 1$ exactly. Any maximal chain of primes containing $(a)$ extended to the left by $(0)$ yields a saturated chain of primes, but $A$ is catenary, so this is also a maximal chain of primes. Therefore $\dim A / (a) = \dim A - 1$, as required. $\qquad \blacksquare$.

• Wow... It exceeds my knowledge, at least for this moment. That's why I won't be accepting it officially. If there are not other easier answers in the next few days, I will, though, and call it a close. Thank you very much for your time. – Adrian Manea Jun 4 '12 at 17:19