# Basic Definitions

The objective of this article is to establish the theory of transcendence bases for field extensions. Readers who are already familiar with this may skip the article. We will focus on field extensions $K\subseteq L$ here.

Definition.

Let $K\subseteq L$ be a field extension. Elements $\alpha_1, \ldots, \alpha_n \in L$ are said to be algebraically dependent over K if

$\exists p \in K[X_1, \ldots, X_n] - \{0\}, \ p(\alpha_1, \ldots, \alpha_n) = 0.$

Otherwise, $\alpha_1, \ldots, \alpha_n$ are algebraically independent over K.

We say $(\alpha_1, \ldots, \alpha_n)$ forms a transcendence basis of L over K if it is algebraically independent over K and L is algebraic over $K(\alpha_1, \ldots, \alpha_n)$.

If such a basis exists for $K\subseteq L$, we say L has finite transcendence degree over K, and write

$n = \mathrm{trdeg} L/K$,

the transcendence degree of L over K.

Exercise A

Prove that if $A = \mathbb C[X, Y]/(Y^2 - X^3 + X)$ then $L := \mathrm{Frac} A \cong \mathbb C(X)[Y]/(Y^2 - X^3 + X)$ so $\alpha_1 = X$ forms a transcendentce basis of L over $\mathbb C$.

Note that a priori, it is not clear that the transcendence basis is well-defined: could we perhaps choose two transcendence bases of different sizes? The answer is no as we will see.

Lemma 1.

For $\alpha_1, \ldots, \alpha_n \in L$ write $L' = K(\alpha_1, \ldots, \alpha_n)$ for the field extension of K generated by $\alpha_1, \ldots, \alpha_n$.

If $\alpha_1, \ldots, \alpha_n$ are algebraically independent over K, then every element of L’ can be uniquely written as $f(\alpha_1, \ldots, \alpha_n)$ as $f$ runs through all rational functions $K(X_1, \ldots, X_n)$. Thus we have an isomorphism:

$K(X_1, \ldots, X_n) \cong L' = K(\alpha_1, \ldots, \alpha_n), \quad X_i \mapsto \alpha_i$.

Proof

Consider the K-algebra homomorphism $K[X_1, \ldots, X_n] \to L$ which takes $X_i \mapsto \alpha_i$ for each $1\le i \le n$. By definition of algebraic independence this map is injective so we get an embedding

$K(X_1, \ldots, X_n) = \mathrm{Frac}(K[X_1, \ldots, X_n]) \longrightarrow L$

of fields, whose image is clearly L’. ♦

# Swapping Lemma

The following will be repeatedly used throughout the article.

Key Observation.

Let $K\subseteq L$ be a field extension and $\alpha_1, \ldots, \alpha_n\in L$ be algebraically independent over K. Let $\beta \in L$.

Then $\alpha_1, \ldots, \alpha_n, \beta$ are algebraically dependent over K if and only if $\beta$ is algebraic over $K(\alpha_1, \ldots, \alpha_n)$.

Note

Compare this with the linear algebra variant: if V is a vector space and $v_1, \ldots, v_n \in V$ are linearly independent, then for $w\in V$, $v_1, \ldots, v_n, w$ are linearly dependent if and only if w is a linear combination of $v_1, \ldots, v_n$.

Proof

(⇒) If $\alpha_1, \ldots, \alpha_n, \beta$ are algebraically dependent over K we have a non-zero polynomial $p \in K[X_1, \ldots, X_n, Y]$ such that $p(\alpha_1, \ldots, \alpha_n, \beta) = 0$. This polynomial must involve Y or we would obtain a polynomial relation in $\alpha_1, \ldots, \alpha_n$, a contradiction. Taking

$q(Y) := p(\alpha_1, \ldots, \alpha_n, Y) \in K(\alpha_1, \ldots, \alpha_n)[Y] \cong K(X_1, \ldots, X_n)[Y]$,

we see that $q \ne 0$ and $q(\beta) = 0$. ♦

(⇐) If $\beta$ is algebraic over $K' := K(\alpha_1, \ldots, \alpha_n)$ pick $c_0, \ldots, c_d \in K'$, not all zero, such that

$c_d \beta^d + c_{d-1} \beta^{d-1} + \ldots + c_1 \beta + c_0 = 0.$

Clearing denominators, we assume each $c_i \in K[\alpha_1, \ldots, \alpha_n]$ so $p(\alpha_1, \ldots, \alpha_n, \beta) = 0$ for some non-zero polynomial $p(X_1, \ldots, X_n, Y)$ with coefficients in K. ♦

Here is the main result.

Swapping Lemma.

Let $\alpha_1, \ldots, \alpha_n$ be a transcendence basis of L over K and $\beta \in L$ be transcendental (i.e. not algebraic) over K. Then there exists an $\alpha_i$ such that

$\alpha_1, \ldots, \alpha_{i-1}, \beta, \alpha_{i+1}, \ldots \alpha_n$

form a transcendence basis of L over K.

Furthermore if $\beta, \alpha_1, \ldots, \alpha_j$ are algebraically independent over K, then we can pick $\alpha_i$ from $\alpha_{j+1}, \ldots, \alpha_n$.

Proof

By assumption $\beta$ is algebraic over $K(\alpha_1, \ldots, \alpha_n)$. Thus by the key observation, $\alpha_1, \ldots, \alpha_n, \beta$ are algebraically dependent. Assume

• $\beta, \alpha_1, \ldots, \alpha_j$ are algebraically independent but
• $\beta, \alpha_1, \ldots, \alpha_{j+1}$ are algebraically dependent.

We claim that swapping $\alpha_{j+1}$ with $\beta$ also gives a transcendence basis.

Indeed let $K' = K(\alpha_1, \ldots, \alpha_j, \beta, \alpha_{j+2}, \ldots, \alpha_n)$. By the key observation, $\alpha_{j+1}$ is algebraic over $K(\beta, \alpha_1, \ldots, \alpha_j)$ and hence over K’. So we have algebraic extensions $K' \subseteq K'(\alpha_{j+1}) \subseteq L$.

It remains to show that $\alpha_1, \ldots, \alpha_j, \alpha_{j+2}, \ldots, \alpha_n, \beta$ are algebraically independent. If not, by the key observation $\beta$ is algebraic over $K'' := K(\alpha_1, \ldots, \alpha_j, \alpha_{j+2}, \ldots, \alpha_n)$ and we have algebraic extensions $K'' \subseteq K''(\beta) = K' \subseteq K'(\alpha_{j+1})$, so $\alpha_{j+1}$ is algebraic over $K''$, a contradiction. ♦

# Transcendence Degree is Well-Defined

Proposition 1.

If $\alpha_1, \ldots, \alpha_n$ is a transcendence basis for $L/K$ and $\beta_1, \ldots, \beta_m \in L$ are algebraically independent over K, then

• $m\le n$ and
• we can replace m terms of $\alpha_1, \ldots, \alpha_n$ by $\beta_1, \ldots, \beta_m$ to obtain another transcendence basis for L over K.

Proof

We repeatedly apply swapping lemma. If m = 0, there is nothing to prove; otherwise by swapping lemma, we can swap out some element of $\alpha_1, \ldots, \alpha_n$ with $\beta_1$ to obtain another transcendence basis.

At each stage, suppose we have a transcendence basis $\beta_1, \ldots, \beta_j$ together with nj elements of $\alpha_i$. If jm we are done. If jn then $\beta_1, \ldots, \beta_j$ is already a transcendence basis so we also have jm. Hence suppose $j and $j.

Since $\beta_1, \ldots, \beta_j, \beta_{j+1}$ are algebraically independent, swapping lemma asserts we can swap out another $\alpha_i$ with $\beta_{j+1}$ so we get a transcendence basis $\beta_1, \ldots, \beta_{j+1}$ together with nj-1 elements of $\alpha_i$. Eventually, we obtain a transcendence basis comprising of $\beta_1,\ldots, \beta_m$ with n – m elements of $\alpha_i$. ♦

Corollary 1.

Any two transcendence bases of L/K have cardinality n so $\mathrm{trdeg} L/K$ is well-defined.

Proof

For any transcendence bases $\alpha_1, \ldots, \alpha_n$ and $\beta_1, \ldots, \beta_m$, proposition 1 says $m\le n$ and $n\le m$. ♦

# Transitivity

Lemma 2.

Let $K\subseteq L$ be a field extension and $\alpha_1, \ldots, \alpha_m \in L$ such that L is algebraic over $K(\alpha_1, \ldots, \alpha_m)$.

Then $(\alpha_1, \ldots, \alpha_m)$ contains a transcendence basis of L over K.

Proof

Pick a maximal algebraically independent subset of $(\alpha_1, \ldots, \alpha_m)$. Upon reordering suppose this is $(\alpha_1, \ldots, \alpha_n)$; set $K' := K(\alpha_1, \ldots, \alpha_n)$. By maximality of $(\alpha_1, \ldots, \alpha_n)$, each of $\alpha_{n+1}, \ldots, \alpha_m$ is algebraic over $K'$. Hence we get algebraic extensions

$K' \subseteq K'(\alpha_{n+1}, \ldots, \alpha_m) = K(\alpha_1, \ldots, \alpha_m) \subseteq L$

and we are done.

Proposition 2.

Let $K\subseteq L \subseteq M$ be field extensions. Then M has finite transcendence degree over K if and only if M has finite transcendence degree over L and L has finite transcendence degree over K, in which case

$\mathrm{trdeg} M/K = \mathrm{trdeg} M/L + \mathrm{trdeg} L/K.$

Proof

(⇒) Let $\alpha_1, \ldots, \alpha_n$ be a transcendence basis of M over K. In particular M is algebraic over $L(\alpha_1, \ldots, \alpha_n)$ so by lemma 2, there exists a subset of $\alpha_1, \ldots, \alpha_n$ which forms a transcendence basis of M over L.

Next suppose $\beta_1, \ldots, \beta_m \in L$ are algebraically independent over K. By proposition 1, $m\le n$ so there exists a maximal algebraically independent sequence $\beta_1, \ldots, \beta_m$ for L/K. By key observation it follows that any $\beta \in L$ is algebraic over $K(\beta_1, \ldots, \beta_m)$.

(⇐) Pick transcendence bases $\alpha_1, \ldots, \alpha_n$ for L over K, and $\beta_1, \ldots, \beta_m$ for M over L. We claim that $\alpha_1, \ldots, \alpha_n, \beta_1, \ldots, \beta_m$ form a transcendence basis of M over K, which would complete the proof.

Suppose $p(\alpha_1, \ldots, \alpha_n, \beta_1, \ldots, \beta_m) = 0$ for a non-zero $p\in K[X_1, \ldots, X_n, Y_1, \ldots, Y_m]$. Then $q(\alpha_1, \ldots, \alpha_n, Y_1, \ldots, Y_m) \in L[Y_1, \ldots, Y_m]$ is a polynomial relation for $\beta_i$; since these are algebraically independent we have $q=0$. But the coefficients of q are polynomials in $\alpha_j$; since $\alpha_j$ are algebraically independent, we also get $p=0$.

Finally, since $K(\alpha_1, \ldots, \alpha_n) \subseteq L$ is algebraic, so is

$K(\alpha_1, \ldots, \alpha_n, \beta_1, \ldots, \beta_m) \subseteq L(\beta_1, \ldots, \beta_m)$

since the RHS is generated as a field by L and $\beta_1, \ldots, \beta_m$, all algebraic over the LHS. Furthermore, $L(\beta_1, \ldots, \beta_m) \subseteq M$ is algebraic by assumption. Thus

$K(\alpha_1, \ldots, \alpha_n, \beta_1, \ldots, \beta_m) \subseteq M$

is an algebraic extension as desired. ♦

Exercise B

Let $A = K[X_1, \ldots, X_n]/(f)$ where $f\in K[X_1, \ldots, X_n]$ is irreducible. Prove that $\mathrm{Frac} A$ has transcendence degree n – 1 over K.

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### 2 Responses to Commutative Algebra 41

1. Vanya says:

Can you please check the proof of Proposition $1$. Especially the statement. “… Hence suppose j>n and j>m….” Also in the preceding statement how does it follow that $j = m$?

• limsup says:

You’re right it should be j < n and j < m.

Next on your question why j = m. If j < m, then $\beta_1, \ldots, \beta_j$ is a transcendence basis so L is algebraic over $K(\beta_1, \ldots, \beta_j)$ so $\beta_{j+1}$ is algebraic over $K(\beta_1, \ldots, \beta_j)$ which is a contradiction.