Maximal ideal

Let X be a smooth projective variety over the complex numbers C. Then X is already defined over a finitely generated extension ring R of Z, i.e. there is a variety Xr defined over R such that Xr xr C = X. For every prime ideal p of R let Xp = XR xr R/p. There is a nonempty open subset U C pec(R) such that Xp is smooth for all p U, and the /-adic Betti-numbers of Xp coincide with those of X for all primes / different from the characteristic of A/p (cf. [Kirwan (1) 15.], [Bialynicki-Birula, Sommese (1) 2.]. If m C R is a maximal ideal lying in U for which R/m is a finite field Fq of characteristic p /, we call Xm a good reduction of X modulo q. 

This ring is free over R so it is faithfully fiat over R. If mR denotes the maximal ideal of R then the unique maximal ideal of Rf is... 

Let v be a closed point of V and let u be the image of v in U. Let us take affine open subsets Spec(A) resp. Spec(B) in U resp. V containing u resp. v such that Spec(B) maps into Spec(A). Let p C A (resp. q C B) be the maximal ideal corresponding to u (resp. v). 

Situation 7.4. Here R is a Z/pn+1Z-flat local algebra which has the following property There exists a Z/pn+1Z-flat ideal m C R such that m -f pR is the maximal ideal of R. If we have this we put S = Spec( ) and So = Spec(R/pR). 

It is easily seen that R C R is finite flat and that R has only one maximal ideal. The field extension k C k is such that k = (k )p. Thus k k and k has a finite p-basis also. The ring R is Noetherian and complete because the ring extension R C R is finite flat. 

The fact that K ObCs0, i.e. that Q2 has no etale part, implies that A2 has exactly one maximal ideal m 2. Hence we must have mA,2 = y/m A2 where m denotes the maximal ideal of R and we must have Ia2 C m 2. Therefore, for all i = 1there exists a natural number Ni such that... 

IX) If A is a local noetherian ring with maximal ideal the m-adic completion A is a flat A-algebra. 

If the base ring k is not a field, then even maximal ideals turn out to be not quite all we want. The kernel of a homomorphism Z[X] - Z, for instance, is not maximal. The next natural generalization is to prime ideals, and these do indeed give a satisfactory theory. 

The spectrum Spec A of a ring A is the collection of its prime ideals. To see what topology it should have, consider fc. A closed set there is the set where a certain ideal I of functions vanishes the corresponding maximal ideals... 

If p in S corresponds to the maximal ideal P, then evaluating a function at p is the same as taking its image in A/P k. For a general A, then, one can intuitively think of elements of A as functions on Spec A, where the value of / at P is the image of / modulo P. It is possible for such a function to vanish at all P, but at least this condition (fef)P) forces / to be nilpotent (A.3). Using that remark we can now carry over the proofs in (5.1) almost verbatim to get the following results. 

Now suppose Z(I) is closed with closed complement Z(J). Then Z(I + J) is empty, so I + J equals A, since no maximal ideal contains it. Write... 

Corollary. Let A be a finitely generated algebra over a field. Let T be the set of maximal ideals. Then Spec A is connected iff its subset T is connected. 

Let A be finitely generated over a field. Show that X -+ X n T is a bijection from closed sets in Spec A to closed sets in the subspace of maximal ideals. 

Lemma. Let A be a finite-dimensional (commutative) k-algebra. Then A is a finite product of algebras A, each of which has a unique maximal ideal consisting of nilpotent elements. 

---