Let $F$ be a field and $R=F[X_1,X_2,\ldots,X_n]$ be the polynomial ring in $n$ variables over $F$ and $P$ be a prime ideal in $R$, I'm trying to prove that$$\operatorname{ht}P+\dim R/P=\dim R$$where $\dim$ is the Krull dimension of a ring.
Any ideas, thank you for your help.
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Martin Sleziak
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i.a.m
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$\mathrm{ht}(P) + \dim(R/P) \leq \dim(R)$ is trivial, but the converse is not so easy to prove directly. You can find a proof in several books on algebraic geometry, for example Qing Liu's Algebraic geometry and arithmetic curves, Proposition 2.5.23. Another reference is the CRing Project, Theorem §10.3.22. It holds more generally for every domain which is a finitely generated $F$-algebra (this generalization is important for the induction step).
For a more detailed discussion about the equation, see math.SE/49136.
Martin Brandenburg
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Thank you for your help Martin. I'll look for the book and find the proof, thank you again. – i.a.m Mar 20 '13 at 01:37
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This is also Theorem 15.12 in the book $\textit{A Term of Commutative Algebra}$ by Altman&Kleiman, after the observation that we always have $\mathrm{ht}(P)=\dim(R_P)$, where $R_P$ is the localization at $P$. – Marco Flores Jan 07 '18 at 22:00
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Is $\operatorname{ht}I+\dim R/I=\dim R$ true for all ideals instead of just prime ideals? – reggie Mar 09 '21 at 14:05