Suppose A is a finitely generated R algebra that happens to be a field. Let R embed in A, whence F also embeds in A. Finally, suppose A is algebraic over F.
Let A = R[y1,y2,…yn], where each yi is a generator of A, and the image of the corresponding indeterminant xi under the implied ring homomorphism. Each generator yi satisfies a monic polynomial over F. Let d be a common denominator for all the coefficients of all n polynomials. Since R is a ufd, d is well defined.
Let S be the fraction ring of R by d. Since R is an integral domain it embeds in S. Since A contains F it contains S. Extend our ring homomorphism to S[x1,x2,…xn], mapping onto S[y1,y2,…yn]. The latter ring is simply A.
Note that each generator yi is integral over S. Assume e is in F-S, so that e is brought in via the introduction of y1 through yn. Since there are prime denominators outside of d, there is no trouble finding such an e. Let e = 1/q where q is one of the primes not in d.
Now e lies in the integral extension of S, hence e is integral over S. Let e satisfy a monic polynomial p with coefficients in S. Clear denominators, so that p becomes a polynomnial in R, whose lead coefficient is a power of d, say dj. Let m be the degree of p. Multiply through by djŚ(m-1). Fold powers of d into e, and dj/q becomes integral over R. Yet R, a ufd, is integrally closed. This places dj/q in R, which is impossible, since q does not divide d. Therefore A cannot contain the field F. There is no finitely generated R algebra that is also an algebraic field extension of F.
Use induction on the number of generators. No generators means A = K, and we're done.
One generator means A = K[x] quotiented out by some maximal ideal. Since K[x] is a pid, the generator of the maximal ideal is an irreducible polynomial. The degree of this polynomial equals the dimension of the extension A/K, which is of course finite.
For multiple generators, replace the homomorphism from K[x1…xn] onto A with two successive homomorphisms. The first maps x1 onto its image y1 in A. The second maps the remaining indeterminants onto their images in A.
Let R = K[y1] and let F = K(y1), the fraction field of R. Thus R is a subring of A, and the second homomorphism takes R[x2…xn] onto R[y2…yn], which is, by assumption, a field. This field contains the fraction field of R, namely F.
Extend the ring R[x2…xn] to the ring F[x2…xn], and extend the homomorphism accordingly. For starters, F maps to F. The indeterminants map into A as they did before, and that establishes the extended homomorphism.
Now A is finitely generated over F, and by induction, A is a finite extension of F. If F is a finite extension of K we are done.
Suppose y1 is transcendental. Thus R is a ufd and F is its fraction field. Furthermore, a finitely generated R algebra becomes a field that algebraically extends F. This contradicts the ufd field lemma. Therefore y1 is algebraic over K.
There is some polynomial p such that p(y1) = 0. Since R embeds in a field it has no zero divisors. A factor of p would become a zero divisor, hence p is irreducible. The ring extension K[y1] is a finite field extension over K, and that completes the proof.
Assume F is a finitely generated field, with no base ring K. In other words, F is the quotient of Z[x1…xn].
If F has characteristic 0 it contains Q, the rational numbers. F is a finitely generated Q algebra that is also a field, F is a finite field extension of Q, and F is a finitely generated Z algebra. This contradicts the ufd field lemma. Thus F has characteristic p, and F becomes a finitely generated Zp algebra. By the above, F is a finite field. A finitely generated ring that is a field is a finite field.
Conversely, let f be a ring homomorphism from A into C. Since 1 maps to 1, the base field Q or Zp is fixed. As for the rest of K, combine f with an automorphism on C if necessary, so that f fixes K. Let L be the image of A in C. Now L is integral over K, and L is an integral domain. Since K is a field, L is also a field. The kernel has to be a maximal ideal.
Drive H up to a maximal ideal M. Mod out by M and the quotient ring is a field, and a finitely generated K algebra, hence it is a finite extension F/K. For each generator xi, let zi be the image of xi in F. Of course each zi is algebraic over K, thus building a point z in Cn. If p is a polynomial in our ring, then p, evaluated at z1 through zn, is the same as p(x1…xn) mapped into F. When p ∈ M, the result is 0. All of M vanishes on z, H vanishes on z, and H′ is nonempty.