Commutative Algebra 41

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’. ♦

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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.

swapping_lemma

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. ♦

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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<n and j<m.

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. ♦

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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.

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