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:


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  10. 2003 May (PDF)
  11. 2003 Putnam (PS)