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$A$ 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.

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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)

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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
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