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In a lecture I was attending, I was presented with the following "fact" (that word was used since the result was not proven):

If $k$ is a field and $I \subseteq k[x_1,...,x_n]$ an ideal with Rad($I$) a prime ideal, and $m \subseteq R:= k[x_1,...,x_n]/I$ is a maximal ideal, then dim $R$ = dim $R_m$ (localization of $R$ at $m$).

I am looking for a source for this statement because I would like to see the proof but I can't find anything. Please help if you know where this might be from or if you know the proof.

Joe
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why
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1 Answers1

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This is a fairly difficult result. You can find a proof in, say, Qing Liu's Algebraic Geometry and Arithmetic Curves (Proposition 2.5.23 on page 74 of the 2006 edition). Among other things, the proof relies upon the characterisation of dimension of varieties using transcendence degree, which in turn relies upon on nontrivial theorems from commutative algebra such as Krull's Hauptidealsatz.

As proven in Liu, if $k$ is a field and $A$ is a finitely generated algebra over $k$ which is an integral domain, then for every prime ideal $\mathfrak p\subset A$, we have $\operatorname{ht}(\mathfrak p)+\dim(A/\mathfrak p)=\dim(A)$. By taking $\mathfrak p\subset A$ to be a maximal ideal in this theorem, we see that $\dim(A_\mathfrak p)=\operatorname{ht}(\mathfrak p)=\dim(A)$.

Let $X$ be an irreducible, but not necesesarily reduced, affine variety over $k$. Geometrically, the first assertion of the preceding paragraph corresponds to the fact that if $Y\subseteq X$ is a closed subvariety, then $\operatorname{codim}(Y,X)+\dim(Y)=\dim(X)$. The second assertion corresponds to the fact that if $x\in X$ is a closed point, then $\dim(X)=\dim(\mathcal O_{X,x})$.

You can find more details in this thread. Note that these results are very much special to varieties (as opposed to more general schemes). Even if $X$ is an irreducible Noetherian affine scheme, it is not necessarily the case that $\operatorname{codim}(Y,X)+\dim(Y)=\dim(X)$ for closed subschemes $Y\subseteq X$. See Chapter 5 of Görtz and Wedhorn's Algebraic Geometry I: Schemes (in particular Remark 5.29 and Exercise 5.7).

Joe
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  • Maybe a stupid question, but I've seen someone in lecture notes refer with this to Eisenbud's Commutative Algera (page 286, Theorem A) where it says the following: "If $R$ is an affine domain over a field $k$, then dim $R =$ trdeg$_{k}R$, and this number is the length of every maximal chain of prime ideals in $R$." The person that referred to this did so without further explanation so I'm assuming one can deduce the result I asked about from that very easily. I simply don't see how exactly right now. Any help? – why May 23 '25 at 18:55
  • @why: So, to be clear, are you asking whether the fact that $\operatorname{ht}(\mathfrak p)+\dim(A/\mathfrak p)=\dim(A)$ (for affine domains $A$ over $k$) can be easily be deduced from the equality $\dim(A)=\operatorname{trdeg}_k(\operatorname{Frac}(A))$? – Joe May 23 '25 at 19:13
  • Yes. Since there was no further explanation there should be a fairly easy way right? I simply can't see that right now. – why May 23 '25 at 19:39
  • @why: I just checked the reference I provided in my answer again. It doesn't seem easy to me. The proof involves more than just the equality $\dim(A)=\operatorname{trdeg}_k(\operatorname{Frac}(A))$; it also requires nontrivial theorems from commutative algebra such as Noether Normalization, and so forth. Unfortunately I need to go off now, so I don't have the time to comment any further on what ingredients are needed in the proof. Perhaps you could try consulting the reference yourself (I'm sure you can find a PDF somewhere...). – Joe May 23 '25 at 19:51
  • Thank you still for your effort. I thought I was overlooking something obvious and that made me curious but maybe someone just made a mistake by referring to Eisenbud. – why May 23 '25 at 20:29