## 9.26 Transcendence

We recall the standard definitions.

Definition 9.26.1. Let $k \subset K$ be a field extension.

A collection of elements $\{ x_ i\} _{i \in I}$ of $K$ is called *algebraically independent* over $k$ if the map

\[ k[X_ i; i\in I] \longrightarrow K \]

which maps $X_ i$ to $x_ i$ is injective.

The field of fractions of a polynomial ring $k[x_ i; i \in I]$ is denoted $k(x_ i; i\in I)$.

A *purely transcendental extension* of $k$ is any field extension $k \subset K$ isomorphic to the field of fractions of a polynomial ring over $k$.

A *transcendence basis* of $K/k$ is a collection of elements $\{ x_ i\} _{i \in I}$ which are algebraically independent over $k$ and such that the extension $k(x_ i; i\in I) \subset K$ is algebraic.

Example 9.26.2. The field $\mathbf{Q}(\pi )$ is purely transcendental because $\pi $ isn't the root of a nonzero polynomial with rational coefficients. In particular, $\mathbf{Q}(\pi ) \cong \mathbf{Q}(x)$.

Lemma 9.26.3. Let $E/F$ be a field extension. A transcendence basis of $E$ over $F$ exists. Any two transcendence bases have the same cardinality.

**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 of algebraically independent subsets $\mathcal{B}$ 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\} $ wold 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 $B_{\alpha }$ 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

\[ |B| \leq |\bigcup \nolimits _{\alpha \in B'} B_{\alpha }| = |B'| \]

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

\[ E/ F(B^*, \alpha _1) / F(B^*) \]

since $\alpha _1$ is algebraic over $F(B^*)$. Now we claim that $B^*$ (counting multiplicity of elements) is algebraically independent over $E$ 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, $m = n$.
$\square$

Definition 9.26.4. Let $k \subset K$ be a field extension. The *transcendence degree* of $K$ over $k$ is the cardinality of a transcendence basis of $K$ over $k$. It is denoted $\text{trdeg}_ k(K)$.

Lemma 9.26.5. Let $k \subset K \subset L$ be field extensions. Then

\[ \text{trdeg}_ k(L) = \text{trdeg}_ K(L) + \text{trdeg}_ k(K). \]

**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 $k$.
$\square$

Example 9.26.6. Consider the field extension $\mathbf{Q}(e, \pi )$ formed by adjoining the numbers $e$ and $\pi $. This field extension has transcendence degree at least $1$ since both $e$ and $\pi $ are transcendental over the rationals. However, this field extension might have transcendence degree $2$ if $e$ and $\pi $ are algebraically independent. Whether or not this is true is unknown and whence the problem of determining $trdeg(\mathbf{Q}(e, \pi ))$ is open.

Example 9.26.7. Let $F$ be a field and $E = F(t)$. Then $\{ t\} $ is a transcendence basis since $E = F(t)$. However, $\{ t^2\} $ is also a transcendence basis since $F(t)/F(t^2)$ is algebraic. This illustrates that while we can always decompose an extension $E/F$ into an algebraic extension $E/F'$ and a purely transcendental extension $F'/F$, this decomposition is not unique and depends on choice of transcendence basis.

Example 9.26.8. Let $X$ be a compact Riemann surface. Then the function field $\mathbf{C}(X)$ (see Example 9.3.6) has transcendence degree one over $\mathbf{C}$. In fact, *any* finitely generated extension of $\mathbf{C}$ of transcendence degree one arises from a Riemann surface. There is even an equivalence of categories between the category of compact Riemann surfaces and (non-constant) holomorphic maps and the opposite of the category of finitely generated extensions of $\mathbf{C}$ of transcendence degree $1$ and morphisms of $\mathbf{C}$-algebras. See [Forster].

There is an algebraic version of the above statement as well. Given an (irreducible) algebraic curve in projective space over an algebraically closed field $k$ (e.g. the complex numbers), one can consider its “field of rational functions”: basically, functions that look like quotients of polynomials, where the denominator does not identically vanish on the curve. There is a similar anti-equivalence of categories (insert future reference here) between smooth projective curves and non-constant morphisms of curves and finitely generated extensions of $k$ of transcendence degree one. See [H].

Definition 9.26.9. Let $k \subset K$ be a field extension.

The *algebraic closure of $k$ in $K$* is the subfield $k'$ of $K$ consisting of elements of $K$ which are algebraic over $k$.

We say $k$ is *algebraically closed in $K$* if every element of $K$ which is algebraic over $k$ is contained in $k$.

Lemma 9.26.10. Let $k'/k$ be a finite extension of fields. Let $k'(x_1, \ldots , x_ r)/k(x_1, \ldots , x_ r)$ be the induced extension of purely transcendental extensions. Then $[k'(x_1, \ldots , x_ r) : k(x_1, \ldots , x_ r)] = [k' : k] < \infty $.

**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.
$\square$

Lemma 9.26.11. Let $k \subset K$ be a finitely generated field extension. The algebraic closure of $k$ in $K$ is finite over $k$.

**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

\[ [k' : k] = [k'(x_1, \ldots , x_ r) : k(x_1, \ldots , x_ r)] < [K : k(x_1, \ldots , x_ r)] = n. \]

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

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