Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

$A$ and $B$ are commutative noetherian local rings with maximal ideals $\mathfrak{m}$ and $\mathfrak{n}$ respectively.

If $f\colon A \to B$ is a local ring homomorphism, how do I prove the inequality $$\dim(B) \leq \dim(A) + \dim(B/f(\mathfrak{m})B),$$ where $\dim$ denotes the Krull dimension?

I know that dim equals the minimal number of generators of an $\mathfrak{m}$-primary ideal in the noetherian local case, but so far I cannot prove this statement.

share|cite|improve this question
up vote 1 down vote accepted

Prologue: As you probably know, a system of parameters of $A$ is a set of $\dim(A)$ elements of $\mathfrak m$ generating an $\mathfrak m$-primary ideal.

Sketch of proof: Take a system of parameters $a_1,...,a_d$ of $A$ and a system of parameters $\bar b_1,...\bar b_e$ of $B/(a_1,...,a_d)B$.
The sequence $a_1,...,a_d; b_1,... b_e$ of elements of $\mathfrak n$ generates an $\mathfrak n$-primary ideal of $B$ so that $\dim(B)\leq d+e$ according to the definition of dimension you mentioned.
(You might want to use that, since $(a_1,...,a_d)$ is $\mathfrak m$-primary, you have $(a_1,...,a_d)\supset \mathfrak m^r$ for $r$ big enough. )

Although there are some details to fill in, the proof is, as you see, pretty natural ( well, if you take into account that commutative algebra is as a rule more austere than detective fiction)

share|cite|improve this answer
Excellent, thank you. I would give +1 just for the "austere"-comment :) but do you really mean $B/(a_1,...,a_d)B$ in the beginning of the sketch? – Joni Jan 22 '12 at 19:30
Dear @Joni: yes, that is what I meant. But you can also replace $B/(a_1,...,a_d)B$ by $B/\mathfrak mB$ if you prefer and the exact same proof works. This freedom is due to the inclusions $\mathfrak mB \supset (a_1,...,a_d)B \supset \mathfrak m^rB$ – Georges Elencwajg Jan 22 '12 at 20:24

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.