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# B. Comparison with classical'' definition

In order to compare the given definition of schemes with the classical'' one, we will prove the following theorem:

Theorem 18   Let f : RS be a homomorphism between finitely generated rings so that for every finite ring A, the induced map Hom(S, A)Hom(R, A) is a bijection. Then f is an isomorphism.

In the paragraphs below R and S will always denote rings satisfying the conditions of the theorem. We first prove a special case:

Lemma 19   Let f : RS be a homomorphism of finite rings so that for every finite ring A the induced map Hom(S, A)Hom(R, A) is a bijection. Then f is an isomorphism.

Proof. Taking A = R we see that there is a homomorphism g : SR such that the composite gof : RSR is identity. For any finite ring A, consider the chain of maps

Hom(R, A)Hom(S, A)Hom(R, A)

The second map is a bijection by assumption. The composite is the identity and in particular, a bijection. It follows that Hom(R, A)Hom(S, A) is a bijection as well. Now, taking A = S we see that we also have a homomorphism h : RS so that the composite homomorphism SgRhS is the identity. We then have

f = idSof = hogof = hoidR = h

Thus fog = idS and gof = idR, hence f and g are isomorphisms.

Next, we show that the above condition is inherited'' by quotients.

Lemma 20   Let f : RS be as above. Let I be an ideal in R, then we obtain a homomorphism R/IS/f (I)S. For any finite ring A, the induced map Hom(S/f (I)S, A)Hom(R/I, A) is a bijection.

Proof. Consider the diagram

 Hom(S, A) Hom(R, A) Hom(S/f (I)S, A) Hom(R/I, A)

The top row is a bijection. Let g : S/f (I)SA be any element in the bottom left corner then the corresponding element h : SA in the top left corner satisfies h(f (I)S) = 0. Thus hof : RA satisfies hof (I) = 0. Thus it factors through a homomorphism e : R/IA. Thus we see that the elements in the bottom left corner are mapped to elements in the bottom right corner. Conversely, let g : R/IA be an element in the bottom right corner and h : RA be its image in the top right corner; then h(I) = 0. By assumption there is a homomorphism e : SA such that h = eof. It follows e(f (I)) = 0 and thus e(f (I)S) = 0. Thus e factors through an element d : S/f (I)SA in the bottom left corner. In other words we have a bijection Hom(S/f (I)S, A)Hom(R/I, A).

Combining the above two lemmas we see that if I is any ideal in R such that R/I and S/f (I)S are finite, then the map R/IS/f (I)S is an isomorphism. We will now show that if R/I is finite then S/f (I)S is automatically'' finite as well.

Lemma 21   Let f : RS be as in the theorem. For any maximal ideal m in R, the ideal f (m)S in S also a maximal ideal.

Proof. Since R is finitely generated R/m is a finite field by Hilbert's Nullstellensatz. Thus Hom(S, R/m)Hom(R, R/m) is a bijection and so the homomorphism RR/m must factor through S; moreover, this factorisation is unique. Let n be the kernel of this factorisation. Then n is a maximal ideal containing f (m)S such that R/mS/n is an isomorphism. Now, let n' be any maximal ideal in S containing f (m)S. Then, the composite RSS/n' factors through R/m. Thus, S/n' is a finite field extension of R/m. If this extension has degree > 1 then if q is the cardinality of R/m, the map x xq is a non-trivial automorphism of S/n' which is identity on R/m. Thus we obtain two maps SS/n' which restrict to the same map RS/n' contradicting the hypothesis. Thus R/mS/n' is an isomorphism. But then this isomorphism gives a map SR/m which restricts to the natural map RR/m; there is a unique such map by hypothesis. Since that map has kernel n, we see that n' = n.

In other words, we see that f (m)S is contained in a unique maximal ideal n in S. Thus S/f (m)S is an Artinian ring. By the earlier discussion we see that R/mS/f (m)S is an isomorphism. In other words f (m)S = n is a maximal ideal for every maximal ideal m in R. Conversely, if n is any maximal ideal in S, then f-1(n) = m is the kernel of the composite RSS/n which is a map to a finite field; hence m is a maximal ideal. It follows that every maximal ideal in S is of the form f (m)S for a maximal ideal m in R.

Now, if I is any ideal such that R/I is finite then there are finitely many maximal ideals m1, ..., mk and positive integers r1, ..., rk such that I m1r1 . m2r2 ... mkrk. As seen above ni = f (mi)S is a maximal ideal. The relations

f (I)S f (m1r1 ... mkrk)S = n1r1 ... nkrk

shows that the ring S/f (I)S is finite as well. It follows that for any ideal I such that R/I is finite, the map R/IS/f (I)S is an isomorphism.

On the other hand suppose J is any ideal in S such that S/J is finite and let I = f-1(J); then R/I is a subring of S/J and thus also finite. We have seen above that this implies that R/IS/f (I)S is an isomorphism. But the inverse image of J/f (I)S under this is the zero ideal in R/I. Thus we have J = f (I)S. To summarise,

Lemma 22   Let f : RS be as in the conditions of the theorem. The map I f (I)S is a one-one correspondence between ideals of finite index in R and ideals of finite index in S. The map J f-1(J) is the inverse correspondence from ideals J in S to ideals in R. Moreover, the natural homomorphism R/IS/f (I)S is an isomorphism for such ideals.

Thus the original condition has been re-stated intrinsically in terms of ideals. Next we wish to prove that the given homomorphism is closed''. That is to say given a prime ideal Q in S, let m be a maximal ideal in R that contains the prime ideal P = f-1(Q). We wish to prove that there is a maximal ideal n in S which contains Q and satisfies f-1(n) = m. To do this we can restrict our attention to R/PS/f (Q)S. Since f-1(f (P)S) f-1(Q) = P, the latter homomorphism is also injective.

Lemma 23   Let f : RS be an injective homomorphism of finitely generated rings with R a domain. We have a factoring of f as follows

RR[X1,..., Xa] = R1R1[t1,..., tb] = R2S

where
1. R1 is a polynomial ring over R.
2. There is a non-zero element r of R1 such that for each i, the element rti R2 satisfies a monic polynomial over R1. Other than this relation there are no further relations among the ti in R2.
3. R2S is the quotient by an ideal that intersects R1 in the zero ideal.

Proof. Since S is finitely generated we can choose a maximal collection of elements X1, ..., Xa of S that are algebraically independent over (the quotient field of) R. Then R1 = R[X1,..., Xa] is the polynomial ring over R and is a subring of S. The remaining generators of S are algebraically dependent on the Xi's. Thus each of them satisfies an equation of the form r0Td + r1Td - 1 + ... + rd for some elements rj in R1. Moreover, we can assume that r0 is non-zero in such an equation. Let r be the product in R1 of polynomials r0 corresponding to different generators of S. Since R is a domain, so is R1 and the polynomial r is non-zero. For each generator S choose a polynomial of the above form with leading coefficient r (one such such clearly exists) and let R2 be the ring obtained from R1 by adjoining the roots of these equations. We have a natural map R2S; let be the kernel. Since R1S factors through R2 and is injective, it follows that intersects R1 in the zero ideal.

Let Q1, ..., Qr be the minimal primes in S or equivalently a minimal primes in R2 that contains the kernel of R2S. Since R1 meets this kernel in the zero ideal, the intersection of the prime ideals Qi R1 in R1 is a nilpotent ideal. Since R1 is a domain there is an index i such that Qi R1 = (0). Let Q denote the prime ideal Qi for any such index i.

Let m be a maximal ideal in R such that r is not contained in the prime ideal m[X1,..., Xa] of R1. Since R1 is a domain we see that Qr (R1)r is the zero ideal. Now, (R2)r is a finite free module over (R1)r and so (by the going up theorem) there is a prime ideal Q' in R2 which contains Q and restricts to m[X1,..., Xa] in R1. Similarly, for any maximal ideal n' in R1 that contains m[X1,..., Xa] and does not contain r, there is a maximal ideal n in R2 that contains Q' (and hence Q) that lies over n'.

Now, if a > 0 (i. e. R R1) then there are at least two (in fact infinitely many) such maximal ideals n'. But then we see that we have at least two maximal ideals in S that lie over a given maximal ideal m in R contradicting lemma 22. Thus we must have R = R1.

Again, if is another minimal prime in R2 that contains the kernel of R2S and such that R1 = (0), then as above we can find a prime ideal which contains and lies over m and is distinct from Q'. Now there are distinct maximal ideals n' and in R2, that contain Q' and respectively. This again contradicts lemma 22. It follows that there is a unique minimal prime Q containing the kernel of R2S such that Q R = (0).

Now suppose that Q0 is another miminal prime in S, or equivalently a minimal prime in R2 that contains the kernel of R2S. We must have Q0 R (0). However, we have the lemma

Lemma 24   Let f : RS be a homomorphism of finitely generated rings with R a domain. Let Q be a minimal prime in S such that f-1(Q) is non-zero. Then there is a maximal ideal n in S and an integer k such that if m = f-1(n), then R/mkS/nk is not an isomorphism.

Proof. Let x be an element of all the minimal primes of S other than Q. Replacing S by its localisation Sx at x, we can assume that Q is the unique minimal prime in S. Then Q consists of nilpotent elements. Since f-1(Q) is non-zero and R is a domain it follows that RS has a non-zero kernel. Now let n be any maximal ideal in S and m = f-1(m). The homomorphism of local rings RmSn has a non-zero kernel. The result follows by the Artin-Rees lemma.

On the other hand, for our given homomorphism RS we know that R/mkS/nk must be an isomorphism for all k. It follows that there is no such prime ideal Q0 in S.

We have thus proved that there is a unique prime ideal Q in S that lies over a given prime ideal P in R and f-1(Q) = P. The closed''-ness condition is an immediate corollary.

Let us note that if R[X] is the polynomial ring over a ring R, then Hom(R[X], A) is naturally identified with Hom(R, AA. Thus, if g : R[X]S[X] denotes the natural extension of the above homomorphism to the corresponding polynomial rings then, for any finite ring the induced map Hom(S[X], A)Hom(R[X], A) is a bijection whenever Hom(S, A)Hom(R, A) is a bijection. In particular, we can apply the above lemmas to the homomorphism g as well.

Lemma 25   Let f : RS be as in the theorem and g : R[X]S[X] be the induced homomorphism on polynomial rings in one variable. Let be any element of S and be the ideal (X - )S[X] in S[X]. Let be the ideal g-1((X - a)S[X]). Then contains a monic polynomial.

Proof. Let A be any ring and be an ideal in the polynomial ring A[X]. Let denote the ideal . A[X, X-1] in the ring A[X, X-1]. We have

= {P(X) . X-n| P(X)  and n 0 an integer }

Let be the restriction A[X-1] of this ideal to A[X-1]. We have

= {P(X) . X-d| P(X)  and d = deg(P(X))}

The content c() of the ideal is defined as the image of in A[X-1]/(X-1) = A. Clearly,

c() = {a A|P(X)  such that P(X) = aXd +  lower degree terms }

Returning to the rings R and S let us use the subscripts 1 and 2 to denote the above constructions applied to ideals in R[X] and S[X]; specifically to the ideals and .

We want to show that the content c() of the ideal in R[X] is the unit ideal. Suppose that c() m for some maximal ideal m in R. The ideal = m[X-1] + X-1R[X-1] is then a maximal ideal in R[X-1] which contains . Moreover, by the above description of it is clear that = g2-1(), where g2 : R[X-1]S[X-1] is the natural homomorphism. Applying the going-up'' which has been proved above, it follows that there should exist a prime ideal containing such that g2-1() = . But is the ideal generated by 1 - X-1 and X-1 lies in . Thus would have to be the unit ideal which contradicts its primality. It follows that c() is the unit ideal.

From this lemma we see that S is integral over R. Now the result that R/mS/f (m)S is an isomorphism for all maximal ideals m implies theorem 18 by Nakayama's lemma.

Next: Bibliography Up: Some Lectures on Number Previous: A..1 Counterexample
Kapil Hari Paranjape 2002-10-20