# How to prove that $\dim \mathrm{Spec}~A = \dim \mathrm{Spec}~ A_\mathfrak{p} + \dim \mathrm{Spec}~A/\mathfrak{p}$.

$\DeclareMathOperator{Spec}{Spec}$ $\newcommand{p}{\mathfrak{p}}$ I was trying to solve the following exercise.

Let $A$ be a finitely generated $k$-algebra over an infinite field $k$. Assume that $A$ is a domain, thus $\Spec A$ is irreducible. Let $\p \subset A$ be a prime ideal and let $A_\p$ be the localization of $A$ at $\p$. Show that $\dim \Spec A = \dim \Spec A_\p + \dim \Spec A/\p$.

I'm not very sure how to proceed. Is it possible to proceed somehow by using Noether Normalization Theorem and that in this setting we have $\dim \Spec A = \mathrm{tr.deg}_k A$? To be clear, this has been asked in a somewhat different form previously, but if possible, I wonder if one can furnish a proof using the transcendence degree and the normalization lemma?

• @Arthur I don't think so. 2 reasons: (1) the linked question does not have a complete answer nor does it provide a proof for the question at hand. (2) I am interested to see if it is possible to get a proof that only relies on the Normalization Lemma + transcendence degree. Both are not addressed in the linked question. – Drew Brady Feb 22 '18 at 10:05
• Fair enough. Let's degrade that to a related question, then, at least. – Arthur Feb 22 '18 at 10:06
• @Arthur, yes, it is related for sure! I took a look at the reference in Eisenbud as well, but it uses more sophisticated ideas in dimension theory with which I am less familiar. – Drew Brady Feb 22 '18 at 10:09

Proposition: Let $A$ be a finitely generated $k$-algebra and suppose $A$ is a domain. Suppose $\mathfrak p_0\subsetneqq \mathfrak p_1\subsetneqq \dots \subsetneqq \mathfrak p_r$ is a chain of prime ideals and denote by $d$ the transcendence degree of $K=\mathrm{Frac}(A)$. Then $r\leq d$ with equality if and only if the chain is a maximal one.
Proof: By Noether Normalization, $A$ is module finite over $P=k[t_1,\dots,t_\nu]$ such that $\mathfrak p_i\cap P=\langle t_1,\dots,t_{h_i}\rangle$ for suitable $h_i$. If $L$ is the fraction field of $P$, then $\nu=\mathrm{tr.deg}_k(L)$. But $P\hookrightarrow A$ is an integral extension and module finite, hence $K/L$ is algebraic and therefore $\nu=d$. By the incomparability property of integral extensions we find $h_i<h_{i+1}$ for every $i$, hence $r\leq h_r$. But $h_r\leq \nu=d$, hence $r\leq d$. If $r=d$, then $r$ is maximal as we just showed that no chain can be longer. Conversely, assume $r$ maximal. Then $\mathfrak p_0=\langle 0\rangle$, so $h_0=0$. Furthermore $\mathfrak p_r$ is maximal, hence $\mathfrak p_r\cap P$ must be maximal too. Hence $h_r=\nu$. Now suppose there is $i$ such that $h_i+1<h_{i+1}$. Then $(\mathfrak p_i\cap P)\subsetneqq\langle t_1,\dots,t_{h_i+1}\rangle \subsetneqq (\mathfrak p_{i+1}\cap P)$. But $P/(\mathfrak p_i\cap P)$ is equal to $k[t_{h_i+1},\dots,t_\nu]$, which is a normal ring. As furthermore the extension $P/(\mathfrak p_i\cap P)\hookrightarrow A/{\mathfrak p_i}$ must be integral as $P\subset A$ is, Going-down yields a prime $\mathfrak p$ between $\mathfrak p_i$ and $\mathfrak p_{i+1}$ such that $\mathfrak p\cap P=\langle t_1,\dots,t_{h_i+1}\rangle$. Then $\mathfrak p_i\subsetneqq \mathfrak p\subsetneqq \mathfrak p_{i+1}$ which contradicts the maximality assumption of $r$. Hence $h_i+1=h_{i+1}$ for all $i$, which implies $r=h_r=\nu$.
Now you can prove your formula quite easily: Pick a maximal chain $\mathfrak p_0\subsetneqq \dots\subsetneqq \mathfrak p\subsetneqq\dots\subsetneqq \mathfrak p_n$ in $A$ (which is possible by the above proposition and has length $\dim (A)$), then $\mathfrak p_0A_{\mathfrak p}\subsetneqq \dots\subsetneqq \mathfrak pA_{\mathfrak p}$ is a maximal chain in $A_{\mathfrak p}$ and $\mathfrak p/\mathfrak p\subsetneqq \dots\subsetneqq \mathfrak p_n/\mathfrak p$ is a maximal chain in $A/\mathfrak p$ and since every such chain in $A$ can be constructed from two chains in $A_{\mathfrak p}$ and $A/\mathfrak p$, respectively, we see that the lengths of both chains must be the dimension of both rings.
• why is the statement that $h_i < h_{i + 1}$ a consequence of incomparability? – Drew Brady Feb 22 '18 at 17:08
• If $h_i=h_{i+1}$, then $\mathfrak p_i$ and $\mathfrak p_{i+1}$ both lie over $\langle t_1,\dots,t_{h_i}\rangle$, which would imply $\mathfrak p_i=\mathfrak p_{i+1}$ by incomparability. But this we excluded, hence $h_i<h_{i+1}$ (as $h_i\leq h_{i+1}$ holds by construction). – asdq Feb 22 '18 at 17:38