The Stacks project

Proof. Let $A$ be an algebraically independent subset of $E$. Let $G$ be a subset of $E$ containing $A$ that generates $E/F$. We claim we can find a transcendence basis $B$ such that $A \subset B \subset G$. To prove this, consider the collection $\mathcal{B}$ of algebraically independent subsets whose members are subsets of $G$ that contain $A$. Define a partial ordering on $\mathcal{B}$ using inclusion. Then $\mathcal{B}$ contains at least one element $A$. The union of the elements of a totally ordered subset $T$ of $\mathcal{B}$ is an algebraically independent subset of $E$ over $F$ since any algebraic dependence relation would have occurred in one of the elements of $T$ (since polynomials only involve finitely many variables). The union also contains $A$ and is contained in $G$. By Zorn's lemma, there is a maximal element $B \in \mathcal{B}$. Now we claim $E$ is algebraic over $F(B)$. This is because if it wasn't then there would be an element $f \in G$ transcendental over $F(B)$ since $F(G) = E$. Then $B \cup \{ f\}$ would be algebraically independent contradicting the maximality of $B$. Thus $B$ is our transcendence basis.

Let $B$ and $B'$ be two transcendence bases. Without loss of generality, we can assume that $|B'| \leq |B|$. Now we divide the proof into two cases: the first case is that $B$ is an infinite set. Then for each $\alpha \in B'$, there is a finite set $B_{\alpha } \subset B$ such that $\alpha$ is algebraic over $F(B_{\alpha })$ since any algebraic dependence relation only uses finitely many indeterminates. Then we define $B^* = \bigcup _{\alpha \in B'} B_{\alpha }$. By construction, $B^* \subset B$, but we claim that in fact the two sets are equal. To see this, suppose that they are not equal, say there is an element $\beta \in B \setminus B^*$. We know $\beta$ is algebraic over $F(B')$ which is algebraic over $F(B^*)$. Therefore $\beta$ is algebraic over $F(B^*)$, a contradiction. So $|B| \leq |\bigcup _{\alpha \in B'} B_{\alpha }|$. Now if $B'$ is finite, then so is $B$ so we can assume $B'$ is infinite; this means

because each $B_\alpha$ is finite and $B'$ is infinite. Therefore in the infinite case, $|B| = |B'|$.

Now we need to look at the case where $B$ is finite. In this case, $B'$ is also finite, so suppose $B = \{ \alpha _1, \ldots , \alpha _ n\}$ and $B' = \{ \beta _1, \ldots , \beta _ m\}$ with $m \leq n$. We perform induction on $m$: if $m = 0$ then $E/F$ is algebraic so $B = \emptyset$ so $n = 0$. If $m > 0$, there is an irreducible polynomial $f \in F[x, y_1, \ldots , y_ n]$ such that $f(\beta _1, \alpha _1, \ldots , \alpha _ n) = 0$ and such that $x$ occurs in $f$. Since $\beta _1$ is not algebraic over $F$, $f$ must involve some $y_ i$ so without loss of generality, assume $f$ uses $y_1$. Let $B^* = \{ \beta _1, \alpha _2, \ldots , \alpha _ n\}$. We claim that $B^*$ is a basis for $E/F$. To prove this claim, we see that we have a tower of algebraic extensions

since $\alpha _1$ is algebraic over $F(B^*)$. Now we claim that $B^*$ (counting multiplicity of elements) is algebraically independent over $F$ because if it weren't, then there would be an irreducible $g\in F[x, y_2, \ldots , y_ n]$ such that $g(\beta _1, \alpha _2, \ldots , \alpha _ n) = 0$ which must involve $x$ making $\beta _1$ algebraic over $F(\alpha _2, \ldots , \alpha _ n)$ which would make $\alpha _1$ algebraic over $F(\alpha _2, \ldots , \alpha _ n)$ which is impossible. So this means that $\{ \alpha _2, \ldots , \alpha _ n\}$ and $\{ \beta _2, \ldots , \beta _ m\}$ are bases for $E$ over $F(\beta _1)$ which means by induction, $m = n$. $\square$

Proof. Choose a transcendence basis $A \subset K$ of $K$ over $k$. Choose a transcendence basis $B \subset L$ of $L$ over $K$. Then it is straightforward to see that $A \cup B$ is a transcendence basis of $L$ over $k$. $\square$

Proof. By multiplicativity of degrees of extensions (Lemma 9.7.7) it suffices to prove this when $k'$ is generated by a single element $\alpha \in k'$ over $k$. Let $f \in k[T]$ be the minimal polynomial of $\alpha$ over $k$. Then $k'(x_1, \ldots , x_ r)$ is generated by $\alpha , x_1, \ldots , x_ r$ over $k$ and hence $k'(x_1, \ldots , x_ r)$ is generated by $\alpha$ over $k(x_1, \ldots , x_ r)$. Thus it suffices to show that $f$ is still irreducible as an element of $k(x_1, \ldots , x_ r)[T]$. We only sketch the proof. It is clear that $f$ is irreducible as an element of $k[x_1, \ldots , x_ r, T]$ for example because $f$ is monic as a polynomial in $T$ and any putative factorization in $k[x_1, \ldots , x_ r, T]$ would lead to a factorization in $k[T]$ by setting $x_ i$ equal to $0$. By Gauss' lemma we conclude. $\square$

Proof. Let $x_1, \ldots , x_ r \in K$ be a transcendence basis for $K$ over $k$. Then $n = [K : k(x_1, \ldots , x_ r)] < \infty$. Suppose that $k \subset k' \subset K$ with $k'/k$ finite. In this case $[k'(x_1, \ldots , x_ r) : k(x_1, \ldots , x_ r)] = [k' : k] < \infty$, see Lemma 9.26.10. Hence

In other words, the degrees of finite subextensions are bounded and the lemma follows. $\square$