Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up.
Sign up to join this community
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.
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 $$ \dim(A)= \dim(A/xA)+1 \quad (***)$$
Here is a solution, but unfortunately it invokes several deep results of dimension theory.
Theorem. (Krull's Hauptidealsatz)
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 $1$.
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. $\qquad \blacksquare$.
