# Krull dimension of a finitely generated integral domain over $k$ is equal to the transcendence degree.

This theorem is from Matsumura (p.34)

Let $$k$$ be a field and $$A$$ an integral domain which is finitely generated over $$k$$. Then $$\dim A = \operatorname{trdeg}_k A$$ (where $$\operatorname{trdeg}_k A$$ is the transcendence degree of the field of fractions of $$A$$ over $$k$$).

I've been trying to read this proof but there are so many things that are confusing me, so I would appreciate any kind of help.

Proof

Let $$A = k[X_1, ... , X_n]/P$$, and set $$r = \operatorname{trdeg}_k A$$

Question 1: From what I understand $$A = k[a_1, ... , a_n]$$ for some $$a_1, ... , a_n$$ since it is finitely generated over $$k$$, and we have $$f: k[X_1, ... ,X_n] \rightarrow k[a_1, ... , a_n]$$ sending $$X_i$$ to $$a_i$$. So the first isomorphism theorem gives us $$k[X_1, ... , X_n]/\ker f \simeq k[\alpha_1, ... ,\alpha_n]$$. Since $$A$$ is an integral domain, $$\ker f$$ must be prime and we just call it $$P$$, right?

Question 2: Since $$A$$ is a module over $$k$$, the dimension of $$A$$ is defined to be the Krull dimension of $$k/\operatorname{ann}(A)$$, right?

To prove that $$r \geq \dim A$$ it is enough to show that if $$P$$ and $$Q$$ are prime ideals of $$k[X] = k[X_1, ... ,X_n]$$ with $$Q \supset P$$ and $$Q \not= P$$, then $$\operatorname{trdeg}_k k[X]/Q < \operatorname{trdeg}_k k[X]/P.$$

Question 3: Why would it be enough to show that? How does that say anything about the Krull dimension of $$k/\operatorname{ann}(A)$$?

2. False: $\operatorname{ann}_k(A)=0$. The considered Krull dimension is that of $A$ as a commutative ring.
3. Let's say $n=\dim A$. Then there is a chain of prime ideals $(0)=p_0\subset\cdots\subset p_n$ in $A$. Now note that $p_i=P_i/\ker f$, where $P_i$ is a prime ideal in $k[X]$. Moreover, $A/p_i=k[X]/P_i$. In particular, $P_0=\ker f$ and $A=k[X]/P_0$. From the inequality mentioned by the proof we get $$\operatorname{trdeg}_k k[X]/P_n+n\le \operatorname{trdeg}_k k[X]/P_0=\operatorname{trdeg}_kA.$$ Now use that $\operatorname{trdeg}_k k[X]/P_n=0$ (why?).
• @user26857 $tr \deg_k k[X]/P_n = 0$ because $P_n$ is maximal, meaning that $P_n = (X)$...so we have $tr \deg_k k = 0$. But I have one more question (if you don't mind). The inequality tells us that $tr \deg k[X]/P_n \leq tr \deg k[X]/P_0$, right? It does not tell us that $tr \deg k[X]/P_n + n \leq tr \deg_k k[X]/P_0$. – Artus Oct 18 '14 at 18:41
• @Artos In your question it is written $<$, not $\le$, right? Or if $a,b$ are integers and $a<b$ then $a+1\le b$. (Btw, $P_n$ is maximal, but isn't necessarily $(X)$.) – user26857 Oct 18 '14 at 18:44