The ideas presented in this page can be applied to fields or commutative rings, though they are usually introduced as part of field theory. I'll try to present the subject in both contexts simultaneously; I hope it's not too confusing.
In an earlier section we saw that an element v is algebraic over the base field K if it satisfies a polynomial (in one variable) with coefficients drawn from K. Failing this, v is transcendental. These definitions remain valid when K is a ring. Furthermore, v might be integral, if the aforementioned polynomial is monic, i.e. its lead coefficient is 1. Thus integral implies algebraic. Algebraic and integral are synonymous when K is a field, since one can always divide p through by its lead coefficient to make it monic. An entire branch of mathematics is devoded to integral extensions, but that is beyond the scope of this page.
A set of elements S in a ring/field extension F/K is algebraically independent if there is no multivariable polynomial p(x1,x2,…xn), with coefficients in K, that, when evaluated at s1, s2, … sn, gives 0. Polynomials constructed from the elements of S are never 0, nor are they equal, else their difference would be a polynomial that evaluates to 0. In other words, the algebraically independent elements in S act just like so many indeterminants, and generate a polynomial ring K[S].
If our rings are fields, or integral domains that we might want to embed into fields, polynomials extend to rational functions, i.e. quotients of polynomials. If a rational function in s1 s2 s3 … sn is 0, its numerator (a polynomial) has to be 0, and that is a contradiction. Nor can distinct quotients of polynomials be equal, else their difference is 0. The field generated by independent elements is isomorphic to fractions of polynomials, denoted K(S).
If v is algebraic over K[S], it is algebraic over K(S). How about the converse? Each coefficient on p(v) is a quotient of polynomials from K[S]. Multiply through by a common denominator, and find a polynomial in v with coefficients in K[S] that equals 0. This makes v algebraic over K[S]. It doesn't matter whether S generates a ring or a field; the algebraic elements over K[S] or K(S) are the same.
Are there any new independent elements when we extend K[S] to K(S)? Since everything in K(S) is algebraic over K(S), it is also algebraic over K[S]. (See the previous paragraph.) There are no new transcendentals in the fraction field. We've gone from rings to fields, from K[S] to K(S), and the elements of S remain algebraically independent, and no new independent elements are introduced. (Remember, we're only jumping from rings to fields if K is an integral domain.)
Every permutation of the elements of S extends to a ring/field automorphism of K[S] or K(S).
Verify that u is transcendental over K[S] iff S union u is an algebraically independent set. A polynomial p(u) over K[S] implies a polynomial p(S,u) over K, and conversely. Since algebraic over K[S] and algebraic over K(S) are synonymous, we can make the same claim for fields: u is transcendental over K(S) iff S union u is an algebraically independent set. In either rings or fields, algebraically independent sets grow by adding transcendental elements.
If u and v are transcendental over K, v is transcendental over K(u) iff u is transcendental over K(v) iff u and v are algebraically independent.
A basis is a maximal algebraically independent set under set theoretic inclusion, which must exist by zorn's lemma. The basis S generates a transcendent space within the field extension F/K, and F becomes algebraic over K(S). Adjoin the indeterminants to K to build the transcendent space, the cake, then bring in the remaining algebraic elements, the icing on top.
This applies to rings as well as fields. The transcendent space is a subring of F, and all of F is algebraic over K[S].
We call K[S] or K(S) a transcendent space because, when K is a field, the transcended space is a vector space built from transcendental elements. Even when K is a ring, K[S] is a free K module, spanned by products of powers of indeterminants drawn from S.
The field extension F/K does not have one transcendent space; it may have many. Let me illustrate with K(x), polynomials over x. We could adjoin x2 , giving a smaller transcendent space. This is a smaller cake, and it has more icing. The element x is algebraic over x2 , namely its square root. The field F can be entirely transcendental, or it can be described as a transcendental extension followed by an algebraic extension.
It may not be possible to get rid of the icing on the cake, making F entirely transcendental. Start with K(x), where x is an indeterminant, and adjoin v1, the square root of x. Then adjoin v2, the square root of v1. Then adjoin v3, the square root of v2. Continue this forever, building a field F that is algebraic over K(x). The intermediate field generated by vj looks like the polynomials, and quotients thereof, in vj, with coefficients in K. This brings in x, and K(x). We'll see below that a ring or field based on one indeterminant, such as x, cannot contain two algebraically independent elements. We can select one and only one indeterminant, x or something else. Let w be some other transcendental element in F. Now w lies in one of these intermediate subfields, generated by vj, for some j. It is trapped in the jth extension, and cannot generate all of F. The rest of F is an infinite algebraic extension of K(w). F always consists of a transcendent space in one indeterminant, and an infinite algebraic extension on top.
If the dimension of a transcendent space is the size of its basis, i.e. the number of indeterminants, does F/K have the invariant dimension property? Remember, there are many ways to build a transcendent space within F.
Let's consider the finite case first. Let the set A, consisting of a1 a2 a3 … an, be a maximal independent set, a basis for a transcendent space within F/K, and suppose B is any set of algebraically independent elements b1 b2 b3 etc, with more than n elements.
Write b1 as the root of a polynomial over K[A], which is possible since everything in F is algebraic over K(A). Since b1 is transcendental over K, its polynomial contains something from A, say a1. If we include b1, instead of a1, in our basis, and leave everything else the same, a1 becomes algebraic over K[b1,a2,a3,…an]. With a1 in hand, all of A is at our disposal, and the rest of F is algebraic. This does not guarantee b1 is independent from the rest of A. There may be some polynomial satisfied by b1 a7 and a9. That's ok though; because b1 is transcendental, and the remaining generators from A are algebraically independent, and together they span a subfield that leaves only icing - only algebraic elements in F.
Next write b2 algebraic over K[b1,a2,a3,…an]. Its polynomial must contain some a2, since b1 and b2 are algebraically independent over K. Replace a2 with b2. The resulting set has a2 algebraic, whence the rest of F becomes algebraic. this reproduces the conditions described after we replaced a1 with b1. Continue this process through b3, b4, and so on up to bn. Now bn+1 is both algebraic and transcendental over its predecessors. This is a contradiction.
If a basis is finite it fixes the transcendent degree for all of F/K, and every maximal transcendent space within F has degree n.
What happens when A and B are infinite? Let c be the cardinality of A. Look at K[A] as a K vector space, or as a free K module (if K is not a field). Each monomial, each finite product of powers of indeterminants from A, becomes a member in the basis of K[A]. There are c monomials of degree 1, c choose 2 monomials of degree 2, c choose 3 monomials of degree 3, and so on. Add these terms up and get c. Therefore the dimension of K[a], as a K vector space, is c. But how large is K[A] itself? If d is the cardinality of K, a vector is a finite set of basis elements, with coefficients drawn from K. If the vector has length j, we're talking about dj possibilities. This remains finite if d is finite, or it equals d if d is infinite. Add this up over all finite sets drawn from c, and the size of K[A] is c or d, whichever is larger. Let c = d if d is larger. Now, the fraction field is no more than the square of the underlying integral domain, hence K(A) has cardinality c. Finally, put the icing on the cake. There are cj monic polynomials of degree j, over K(A), and each can introduce at most j algebraic elements. The algebraic closure of K(A) has size c, and F lies somewhere in between this closure and K(A), hence |F| = c. What does all this tell us? If c = |A|, and |K| < c, then |F| = c, and any other transcendent basis would give the wrong size for F. The transcendent degree has to be c. The theorem is also valid when K is countable, such as a finite field or the rationals; because we already dealt with finite transcendent degree, and if F has two different infinite transcendent degrees, one of them is uncountable, and larger than K.
I believe transcendent degree is fixed, even when |K| exceeds |A|, but I don't have a proof for that. Thoughts welcome.
Let F be an extension of E, which is an extension of K. Let x1 x2 x3 etc form a transcendent basis for E over K, and let y1 y2 y3 form a transcendent basis for F over E. We will show that x∪y gives a transcendent basis for F over K. Suppose p is a polynomial in x∪y that is equal to 0. This becomes a polynomial in y, with coefficients in E, that is equal to 0, and that is a contradiction. We have algebraic independence.
If X is the set of indeterminants in E, and A is the set of follow-on algebraic elements, the icing on the cake, then E = K(x,A). Everything in F is algebraic over E(Y), is algebraic over K(X,Y,A), is algebraic over K(X,Y). This makes X∪Y a maximal set of independent elements, i.e. a basis. The transcendent degree of F/K is the transcendent degree of E/K plus the transcendent degree of F/E.
A larger vector space cannot fit inside a smaller one, and the same holds for transcendent spaces. Let F/K have transcendent degree n, and for E inside F, let E/K have a higher transcendent degree, possibly infinite. Extend the basis of the transcendent space of E to become a basis for the transcendent space of F. Now F has degree beyond n, which is a contradiction.