D. J. Bernstein
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# The Zariski-Goldman-Krull theorem (the generalized Nullstellensatz)

Here's one way to state the theorem: If K is a field, R is a subring of K, and gen_R K is finite, then there exists r in R-{0} such that R[1/r] is a field and len_{R[1/r]} K is finite. Here len means length as a module (e.g., dimension as a vector space), and gen means minimum number of generators as an algebra.

In particular, if K is a field, R is a subring of K, gen_R K is finite, and R is a quotient of a Jacobson ring, then R is a field and len_R K is finite. (Definition of a Jacobson ring: if R is a quotient domain of a Jacobson ring, and R[1/r] is a field for some r in R-{0}, then R is a field.)

Here's the canonical proof of the Zariski-Goldman-Krull theorem for, e.g., gen_R K=3. By hypothesis K = R[x1,x2,x3] for some x1,x2,x3 in K. Write R0 = R, R1 = R[x1], R2 = R[x1,x2], and R3 = R[x1,x2,x3]. Define F0 as the field of fractions of R0; F1 as the field of fractions of R1; F2 as the field of fractions of R2; and F3 as the field of fractions of R3. Please take out a felt-tip pen and draw an inclusion diagram on your screen at this point.

The idea in a nutshell is that each field of fractions can be obtained by inverting a single element. I will construct r3 in R3-{0} such that F3 = R3[1/r3]; and then r2 in R2-{0} such that F2 = R2[1/r2]; and then r1 in R1-{0} such that F1 = R1[1/r1]; and finally the desired r0 in R0-{0} such that F0 = R0[1/r0]. Along the way I will show that len_F3 K, len_F2 F3, len_F1 F2, and len_F0 F1 are finite; hence len_F0 K is finite.

To get from r1 to r0: First R0[x1][1/r1] = R1[1/r1] = F1 so F0[x1][1/r1] = F1. Also r1 is in F0[x1]; so x1 is F0-integral by the lemma below. Thus F0[x1] is a field, and len_F F0[x1] is finite. This field contains r1, so it contains 1/r1, so 1/r1 is F0-integral. Hence len_F0 F1 = len_F0 F0[x1][1/r1] is finite. Now clear denominators: both x1 and 1/r1 are R0[1/r0]-integral for some r0 in R0-{0}. Hence the field F1 = R0[1/r0][x1][1/r1] is R0[1/r0]-integral; so R0[1/r0] is a field; so F0 = R0[1/r0].

Exactly the same construction gets from r2 to r1, and from r3 to r2. The initial construction of r3 is trivial: R3 = K so simply take r3 = 1. Q.E.D.

Lemma: if F is a field, F[x] is the polynomial ring over F, and r is in F[x]-{0}, then F[x][1/r] is not a field. Proof: If F[x][1/r] is a field then (1-xr)^{-1} = g/r^n for some g in F[x], so r^n = (1-xr)g in F[x]; but r^n and 1-xr are coprime, so 1-xr is a unit, so r = 0, contradiction.

Here are several common ways to slow down the proof of the Zariski-Goldman-Krull theorem:

• Prove the lemma by proving that there are infinitely many maximal ideals in F[x].
• Work with rings that have R0, R1, etc. as quotients, instead of working inside K.
• Get from r3 to r2 as a special case (taking advantage of r3 = 1), rather than giving a unified proof as above.
• Merge various polynomial manipulations into the proof, instead of giving a name to the concept of integrality.
• Define Jacobson ring as ``every prime ideal is an intersection of maximal ideals,'' and apply Zorn's lemma to prove equivalence with the definition given above. For even more impressive slowdowns, thread intersections of maximal ideals through the entire proof.

Literature:

1. 1947 Zariski: A new proof of Hilbert's Nullstellensatz.
2. 1951 Goldman: Hilbert Rings and the Hilbert Nullstellensatz, Math. Z. 54 (1951), 136-140.
3. 1952 Krull: Jacobsonsches Radikal und Hilbertscher Nullstellensatz, ICM Proceedings.
4. 1974 Kaplansky
5. 1995 Eisenbud: Commutative algebra with a view toward algebraic geometry, pages 131-134.
6. 1998 Bernstein (DVI)
7. 1999 Munshi: Hilbert's Nullstellensatz.
8. 2000 Stallings (PS)
9. 2001 Grayson (PS)
10. 2003 May (PDF)
11. 2003 Putnam (PS)