## Algebraic Geometry, Finitely Generated Algebra as a Field

### UFD Field Lemma

Let R be a ufd with fraction field F.  Assume R has infinitely many primes.  This is the case when R = Z, or when R = K[x] for some field K.  If K is infinite, each monomial x-c is prime, and if K is finite, there are plenty of irreducible polynomials over K.

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.

### Finitely Generated Algebra as a Field

Let A be a finitely generated K algebra.  If A is a field then A is a finite extension of K.

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.

### Finite Fields

If K is a finite field and the finitely generated K algebra A is also a field then A is a finite field.  This follows directly from the previous theorem.

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.

### Mapping A into C

Let A be a finitely generated K algebra and let M be a maximal ideal in A.  Mod out by M and find a finitely generated K algebra that happens to be a field.  As shown above, such a field is a finite extension of K, hence algebraic over K.  If C is the closure of K, embed K in C, map M to 0 in C, and carry the quotient field A/M into C, thus giving a ring homomorphism from A into C.  The image is L, where L/K is a finite field extension of K inside C.

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.

### Weak Nullstellensatz

One can use these theorems to prove weak nullstellensatz.  (I'll deal with weak and strong nullstellensatz in the next section; this is merely a foretaste.)  Let H be a proper ideal in the polynomial ring K adjoin finitely many indeterminants, and let C be the algebraic closure of K. The algebraic set H′ in Cn is nonempty.

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.