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 $k/\operatorname{ann}(A)$?
Best Answer
True.
False: $\operatorname{ann}_k(A)=0$. The considered Krull dimension is that of $A$ as a commutative ring.
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?).