Explanation of Zariski's lemma's proof from Atiyah-MacDonald

Question

The proof of Zariski's lemma (Proposition 7.9) given in Atiyah-MacDonald is quite terse, and misses many details, for instance, where exactly is $r\ge 1$ being used? Can someone explain?
Here is the statement:

Let $k\subseteq E$ be a field extension with $E$ being f.g. (finitely generated) as $k$-algebra. Then $E$ is an algebraic extension of $k$, of finite degree.

Answer 1

Let's break down the proof.

1. We know that $E = k[\alpha_1, \ldots, \alpha_n]$ for some $\alpha_i\in E$, $n\ge 0$ (since f.g. as $k$-algebra). Now, we can re-number $\alpha_i$'s such that $\alpha_1, \ldots, \alpha_r$ are algebraically independent over $k$ and $\alpha_{r + 1}, \ldots, \alpha_n$ are algebraic over $F := k(\alpha_{1}, \ldots, \alpha_r)$, for $0\le r\le n$. (See Lemma 1 below.)

2. Since $E$ is a field, $E = k(\alpha_1, \ldots, \alpha_n)$. To show that $[E : k] < \infty$ (i.e., the extension $k\subseteq E$ is of finite degree), it suffices to show that each $\alpha_i$ is algebraic over $k$. (See Lemma 2 below.) This will follow if $r = 0$ in the above. So, for the sake of contradiction, let's assume $r\ge 1$.

3. $E = F(\alpha_{r + 1}, \ldots, \alpha_n)$ so that (by point (1.) and Lemma 2), we have that $[E : F] < \infty$, or in other words, $E$ is f.g. as a vector space over $F$. Now, applying Artin-Tate lemma (Proposition 7.8 in the book) on $k\subseteq F\subseteq E$, we can conclude that $F$ is f.g. as $k$-algebra, and hence we write $F = k[\beta_1, \ldots, \beta_k]$ for $\beta_i\in F$, i.e., $\beta_i = f_i/g_i$ with $f_i, g_i\in k[\alpha_1, \ldots, \alpha_r] =: D$, $g_i\ne 0$. (Note that $F$ is the field of fractions of $D$.)

4. Up till now, we haven't used $r\ge 1$. Now we will, in constructing an irreducible polynomial $p\in D$ coprime to each of $g_i$'s. (Note that I used "polynomial"; since $\alpha_1, \ldots, \alpha_r$ are algebraically independent, we have $k[x_1, \ldots, x_r]\cong D$ (simply via the evaluation $x_i\mapsto \alpha_i$) so that we'll identify their elements together.) I skip this, as Eric Wolsey has brilliantly done this here. (Spot where $r\ge 1$ is being used!)

5. Now, $p^{-1}\in F$ so that we must be able to express it as a polynomial evaluated at $\beta_i$'s (with coefficients from $k$). Clearing out the denominators will yield $$ (\text{product of $g_i$'s)} = p\cdot \text{(something from $D$)} $$ and this multiplication happens in $D$ as well, so that by primality of $p$ (in a GCD domain, irreducible are prime), $p$ divides some $g_i$, a contradiction.

Lemma 1. If $k\subseteq E$ is a field extension and $S\subseteq E$, then there exists a maximal subset $\tilde S\subseteq S$ such that $\tilde S$ is algebraically independent over $k$ and the rest $S\setminus\tilde S$ is algebraic over $k(\tilde S)$.

Lemma 2. If $k\subseteq E$ is a field extension and $\alpha_1, \ldots, \alpha_n\in E$ are algebraic over $k$, then the degree of the field extension $[k(\alpha_1, \ldots, \alpha_n) : k] < \infty$.