Transcendence degree
From Academic Kids

In abstract algebra, the transcendence degree of a field extension L/K is a certain rather coarse measure of the "size" of the extension. Specifically, it is defined as the largest cardinality of an algebraically independent subset of L over K.
A subset S of L is a transcendence basis of L/K if it is algebraically independent over K and if furthermore L is an algebraic extension of the field K(S) obtained by adjoining the elements of S to K. One can show that every field extension has a transcendence basis, and that all transcendence bases have the same cardinality; this cardinality is equal to the transcendence degree of the extension.
If no field K is specified, the transcendence degree of a field L is its degree relative to the prime field of the same characteristic, i.e., Q if L is of characteristic 0 and F_{p} if L is of characteristic p.
The field extension L/K is purely transcendental if there is a subset S of L that's algebraically independent over K and such that L = K(S).
Contents 
Examples
 Every algebraic extension has transcendence degree 0; the empty set serves as a transcendence basis here.
 The field of rational functions in n variables K(x_{1},...,x_{n}) is a purely transcendental extension with transcendence degree n over K; we can for example take {x_{1},...,x_{n}} as a transcendence base
 More generally, the transcendence degree of the function field L of an ndimensional algebraic variety over a ground field K is n.
 Q(√2, π) has transcendence degree 1 over Q because √2 is algebraic while π is transcendental.
 The transcendence degree of C or R over Q is the cardinality of the continuum.
 The transcendence degree of Q(π, e) is either 1 or 2; the precise answer is unknown because we don't know whether π and e are algebraically independent.
Analogy with vector space dimensions
There is an analogy with the theory of vector space dimensions. The dictionary matches algebraically independent sets with linearly independent sets; sets S such that L is algebraic over K(S) with spanning sets; transcendence bases with bases; and transcendence degree with dimension. The fact that transcendence bases always exist (like the fact that bases always exist in linear algebra) requires the axiom of choice. The proof that any two bases have the same cardinality depends, in each setting, on an exchange lemma.
Facts
If M/L is a field extension and L/K is another field extension, then the transcendence degree of M/K is equal to the sum of the transcendence degrees of M/L and L/K. This is proven by showing that a transcendence basis of M/K can be obtained by taking the union of a transcendence basis of M/L and one of L/K.
Applications
Transcendence bases are a useful tool to prove various existence statements about field homomorphisms. Here is an example: Given an algebraically closed field L, a subfield K and a field automorphism f of K, there exists a field automorphism of L which extends f (i.e. whose restriction to K is f). For the proof, one starts with a transcendence basis S of L/K. The elements of K(S) are just quotients of polynomials in elements of S with coefficients in K; therefore the automorphism f can be extended to one of K(S) by sending every element of S to itself. The field L is the algebraic closure of K(S) and algebraic closures are unique up to isomorphism; this means that the automorphism can be further extended from K(S) to L.
As another application, we show that there are (many) proper subfields of the complex number field C which are (as fields) isomorphic to C. For the proof, take a transcendence basis S of C/Q. S is an infinite (even uncountable) set, so there exist (many) maps f : S → S which are injective but not surjective. Any such map can be extended to a field homomorphism Q(S) → Q(S) which is not surjective. Such a field homomorphism can in turn be extended to the algebraic closure C, and the resulting field homomorphisms C → C are not surjective.