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 **Z**p 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 **Z**p 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.