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 : R$\to$S be a homomorphism between finitely generated rings so that for every finite ring A, the induced map Hom(S, A)$\to$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 : R$\to$S be a homomorphism of finite rings so that for every finite ring A the induced map Hom(S, A)$\to$Hom(R, A) is a bijection. Then f is an isomorphism.

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

Hom(R, A)$$\to$$Hom(S, A)$$\to$$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)$\to$Hom(S, A) is a bijection as well. Now, taking A = S we see that we also have a homomorphism h : R$\to$S so that the composite homomorphism S$\xrightarrow$gR$\xrightarrow$hS is the identity. We then have

f = id_Sof = hogof = hoid_R = h

Thus fog = id_S and gof = id_R, hence f and g are isomorphisms. $\qedsymbol$

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

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

Proof. Consider the diagram

Hom(S, A)	$$\to$$	Hom(R, A)
$$\uparrow$$		$$\uparrow$$
Hom(S/f (I)S, A)		Hom(R/I, A)

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

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/I$\to$S/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 : R$\to$S 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)$\to$Hom(R, R/m) is a bijection and so the homomorphism R$\to$R/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/m$\to$S/n is an isomorphism. Now, let n' be any maximal ideal in S containing f (m)S. Then, the composite R$\to$S$\to$S/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 $\mapsto$ x^q is a non-trivial automorphism of S/n' which is identity on R/m. Thus we obtain two maps S$\to$S/n' which restrict to the same map R$\to$S/n' contradicting the hypothesis. Thus R/m$\to$S/n' is an isomorphism. But then this isomorphism gives a map S$\to$R/m which restricts to the natural map R$\to$R/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/m$\to$S/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 R$\to$S$\to$S/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. $\qedsymbol$

Now, if I is any ideal such that R/I is finite then there are finitely many maximal ideals m₁, ..., m_k and positive integers r₁, ..., r_k such that I $\supset$ m₁^{r₁ .} m₂^{r₂ ...} m_k^r_k. As seen above n_i = f (m_i)S is a maximal ideal. The relations

f (I)S $$\supset$$ f (m₁^{r₁ ...} m_k^r_k)S = n₁^{r₁ ...} n_k^r_k

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/I$\to$S/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/I$\to$S/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 : R$\to$S be as in the conditions of the theorem. The map I $\mapsto$ 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 $\mapsto$ f^-1(J) is the inverse correspondence from ideals J in S to ideals in R. Moreover, the natural homomorphism R/I$\to$S/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/P$\to$S/f (Q)S. Since f^-1(f (P)S) $\subset$ f^-1(Q) = P, the latter homomorphism is also injective.

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

R$$\to$$R[X₁,..., X_a] = R₁$$\to$$R₁[t₁,..., t_b] = R₂$$\to$$S

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 rt_i $\in$ R₂ satisfies a monic polynomial over R₁. Other than this relation there are no further relations among the t_i in R₂.
3. R₂$\to$S 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, ..., X_a of S that are algebraically independent over (the quotient field of) R. Then R₁ = R[X₁,..., X_a] is the polynomial ring over R and is a subring of S. The remaining generators of S are algebraically dependent on the X_i's. Thus each of them satisfies an equation of the form r₀T^d + r₁T^{d - 1} + ... + r_d for some elements r_j 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 R₂$\to$S; let $\mathfrak{a}$ be the kernel. Since R₁$\to$S factors through R₂ and is injective, it follows that $\mathfrak{a}$ intersects R₁ in the zero ideal. $\qedsymbol$

Let Q₁, ..., Q_r be the minimal primes in S or equivalently a minimal primes in R₂ that contains the kernel of R₂$\to$S. Since R₁ meets this kernel in the zero ideal, the intersection of the prime ideals Q_i $\cap$ R₁ in R₁ is a nilpotent ideal. Since R₁ is a domain there is an index i such that Q_i $\cap$ R₁ = (0). Let Q denote the prime ideal Q_i 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,..., X_a] of R₁. Since R₁ is a domain we see that Q_r $\cap$ (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,..., X_a] in R₁. Similarly, for any maximal ideal n' in R₁ that contains m[X₁,..., X_a] 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 $\neq$ 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 $\tilde{Q}$ is another minimal prime in R₂ that contains the kernel of R₂$\to$S and such that $\tilde{Q}$ $\cap$ R₁ = (0), then as above we can find a prime ideal $\tilde{Q'}$ which contains $\tilde{Q}$ and lies over m and is distinct from Q'. Now there are distinct maximal ideals n' and $\tilde{n'}$ in R₂, that contain Q' and $\tilde{Q'}$ respectively. This again contradicts lemma 22. It follows that there is a unique minimal prime Q containing the kernel of R₂$\to$S such that Q $\cap$ R = (0).

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

Lemma 24   Let f : R$\to$S 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/m^k$\to$S/n^k 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 S_x 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 R$\to$S has a non-zero kernel. Now let n be any maximal ideal in S and m = f^-1(m). The homomorphism of local rings R_m$\to$S_n has a non-zero kernel. The result follows by the Artin-Rees lemma. $\qedsymbol$

On the other hand, for our given homomorphism R$\to$S we know that R/m^k$\to$S/n^k 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, A)×A. Thus, if g : R[X]$\to$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)$\to$Hom(R[X], A) is a bijection whenever Hom(S, A)$\to$Hom(R, A) is a bijection. In particular, we can apply the above lemmas to the homomorphism g as well.

Lemma 25   Let f : R$\to$S be as in the theorem and g : R[X]$\to$S[X] be the induced homomorphism on polynomial rings in one variable. Let $\alpha$ be any element of S and $\mathfrak{b}$ be the ideal (X - $\alpha$)S[X] in S[X]. Let $\mathfrak{a}$ be the ideal g^-1((X - a)S[X]). Then $\mathfrak{a}$ contains a monic polynomial.

Proof. Let A be any ring and $\mathfrak{a}$ be an ideal in the polynomial ring A[X]. Let $\mathfrak{a}_{1}^{}$ denote the ideal $\mathfrak{a}$ ^. A[X, X^-1] in the ring A[X, X^-1]. We have

$$\mathfrak{a}_{1}^{}$$ = {P(X) ^. X^-n| P(X) $$\in$$ $$\mathfrak{a}$$ and n $$\geq$$ 0 an integer }

Let $\mathfrak{a}_{2}^{}$ be the restriction $\mathfrak{a}_{1}^{}$ $\cap$ A[X^-1] of this ideal to A[X^-1]. We have

$$\mathfrak{a}_{2}^{}$$ = {P(X) ^. X^-d| P(X) $$\in$$ $$\mathfrak{a}$$ and d = deg(P(X))}

The content c($\mathfrak{a}$) of the ideal $\mathfrak{a}$ is defined as the image of $\mathfrak{a}_{2}^{}$ in A[X^-1]/(X^-1) = A. Clearly,

c($$\mathfrak{a}$$) = {a $$\in$$ A|$$\exists$$P(X) $$\in$$ $$\mathfrak{a}$$ such that P(X) = aX^d +  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 $\mathfrak{a}$ and $\mathfrak{b}$.

We want to show that the content c($\mathfrak{a}$) of the ideal $\mathfrak{a}$ in R[X] is the unit ideal. Suppose that c($\mathfrak{a}$) $\subset$ m for some maximal ideal m in R. The ideal $\tilde{m}$ = m[X^-1] + X^-1R[X^-1] is then a maximal ideal in R[X^-1] which contains $\mathfrak{a}_{2}^{}$. Moreover, by the above description of $\mathfrak{a}_{2}^{}$ it is clear that $\mathfrak{a}_{2}^{}$ = g₂^-1($\mathfrak{b}_{2}^{}$), where g₂ : R[X^-1]$\to$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 $\tilde{p}$ containing $\mathfrak{b}_{2}^{}$ such that g₂^-1($\tilde{p}$) = $\tilde{m}$. But $\mathfrak{b}_{2}^{}$ is the ideal generated by 1 - $\alpha$X^-1 and X^-1 lies in $\tilde{m}$. Thus $\tilde{p}$ would have to be the unit ideal which contradicts its primality. It follows that c($\mathfrak{a}$) is the unit ideal. $\qedsymbol$

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