1.Localization at prime ideals. 2.Morphisms and localization. 3.Reduction to fields.
4.Endomorphisms of finitely generated modules. 5.R as a direct summand. 6.Central idempotents and nilpotent elements. 7.Localization over regular rings. 8.Tensor prod-uct over regular rings. 9.Associative regular algebras. 10.Associative strongly regular algebras. 11.Exercises.
As special types of multiplicative subsets of R we can take S = R \p for any prime ideal p ⊂ R. Then the ring of fractions RS−1 is denoted by Rp and is called the localization of R at p.
We denote by P the set of all prime ideals of R (the prime spectrum), and by M the set of all maximal ideals ofR (the maximal spectrum).
17.1 Localization at prime ideals. We use the above notation.
(1) For every p∈ P, Rp is a local ring with maximal ideal pRp, and Rp/pRp 'Q(R/p), the quotient field of R/p.
The prime ideals in Rp correspond to the prime ideals of R contained inp.
(2) For every m∈ M, Rm is a local ring with maximal ideal mRm and Rm/mRm 'R/m.
(3) FP =Lp∈PRp and FM=Lm∈MRm are faithfully flat R-modules.
(4) For any R-module M, we have a canonical monomorphism M →Y
MMm, k 7→([k,1]m).
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Proof. (1) and (2) follow from 16.2.
(3) The Rp are flat by 16.2 and direct sums of flat modules are again flat ([40, 36.1]). By (2), for every maximal ideal m⊂R,mRm 6=Rm. Hence mFM 6=FM and FM is faithfully flat by 16.1.
The same argument applies to FP.
(4) For the kernel K of this map, we have K ⊗RRm = 0 for all m ∈ M, hence
K = 0 by (3). 2
From 17.1(3) we immediately deduce the behaviour of morphisms:
17.2 Morphisms and localization.
Let f :M →N be a morphism of R-modules.
(1) f is monic if and only if, for every m ∈ M, fm :M⊗Rm →N⊗Rm is monic.
(2) f is epic if and only if, for every m∈ M, fm is epic.
(3) f is an isomorphism if and only if, for every m∈ M, fm is an isomorphism.
In some cases the test of a module to be zero can also be achieved by tensoring with certain factor rings:
17.3 Reduction to fields. Let M and N be R-modules.
(1) Assume R is a local ring with maximal ideal m and M is finitely generated, or m is a t-nilpotent ideal.
Then M ⊗RR/m= 0 implies M = 0.
(2) Assume M is finitely generated, or mRm is t-nilpotent for every m∈ M.
(i) Then M ⊗RR/m= 0 for all m∈ M implies M = 0.
(ii) For f :N →M the following are equivalent:
(a) f is surjective;
(b) for every m∈ M, f ⊗id:N⊗RR/m→M ⊗RR/mis surjective.
Proof. (1) AssumeM⊗R/m= 0, i.e.,M =mM. SinceRis local,mis the Jacobson radical of R. If M is finitely generated, the Nakayama Lemma ([40, 21.13]) implies M = 0. If m is a t-nilpotent ideal, then mM =M is only possible for M = 0 (e.g., [40, 43.5]).
(2)(i) Assume M ⊗R/m = 0 for all m ∈ M. By R/m ' Rm/mRm (see 17.1), M ⊗RRm/mRm = 0. Since Rm is a local ring with maximal ideal mRm, we know from (1) that M⊗RRm = 0 for all m∈ M. By 17.1(3), this implies M = 0.
(ii) (a)⇒(b) is obvious.
(b)⇒(a) Put L=M/N f. The functor − ⊗RR/myields the exact sequence N ⊗RR/m−→f⊗idM ⊗RR/m−→L⊗RR/m−→0.
Since allf ⊗id are surjective, we obtain L⊗R/m= 0 for all m ∈ Mand L= 0 by
(2). Hence f is surjective. 2
The next proposition collects characterizations of automorphisms of finitely gen-erated modules:
17.4 Endomorphisms of finitely generated modules.
Let M be a finitely generated R-module. Then:
(1) If I is an ideal in R with IM=M, then (1−r)M = 0 for somer ∈I.
(2) Every surjective endomorphism of M is an isomorphism.
(3) For f ∈EndR(M), the following are equivalent:
(a) f is an isomorphism;
(b) f is an epimorphism;
(c) f is a pure monomorphism;
(d) for every m∈ M, f ⊗id:M⊗R/mR→M ⊗R/mR is surjective;
(e) for every m∈ M, f ⊗id:M⊗R/mR →M ⊗R/mR is injective;
(f ) for every m∈ M, f ⊗id:M ⊗Rm →M ⊗Rm is surjective;
(g) for every m∈ M, f ⊗id:M⊗Rm →M⊗Rm is a pure monomorphism.
Proof. (1) This is a well-known property of modules over commutative rings (e.g., [40, 18.9]).
(2) Consider some f ∈ EndR(M) with M f = M, and denote by R[f] the R-subalgebra of EndR(M) generated by f and idM. M is a faithful module over the commutative ring R[f], and for the ideal < f >⊂ R[f] generated by f, we have
< f > M =M. By (1), (idM −gf)M = 0, for some g ∈R[f], and hence gf =idM, i.e., f is an isomorphism.
(3) Obviously, (a) implies all the other assertions.
(b)⇒(a) is shown in (2). (d)⇒(b) follows from 17.3.
(c) ⇒ (e) ⇒ (g) Pure monomorphisms remain (pure) monomorphisms under tensor functors (e.g. [40, 34.5]).
(e)⇒(d) This is well-known for the finite dimensional R/mR-vector space M ⊗R/mR.
(f)⇒(b) follows from 17.2.
(g)⇒(e) Apply the functor− ⊗RmRm/mRm and Rm/mRm 'R/m. 2
17.5 R as a direct summand.
Let M be a finitely generated projective R-module. Assume M contains R as a submodule. Then R is a direct summand of M.
Proof. The inclusion i:R →M splits if and only if the map Hom(i, R) :HomR(M, R)→HomR(R, R)'R is surjective. Tensoring withR/m, m∈ M, we obtain (using 15.8)
HomR(M, R)⊗RR/m'HomR/m(M⊗RR/m, R/m)→R/m.
SinceM is faithful, M⊗RR/m6= 0 (e.g., [40, 18.9]) and hence this map is surjective for all m∈ M. By 17.3, this implies that Hom(i, R) is surjective and isplits. 2 If A is an R-algebra, Ap ' A⊗RRp is an Rp-algebra, for every p∈ P (see 16.4).
We list some properties of these scalar extensions:
17.6 Central idempotents and nilpotent elements.
Let A be an R-algebra.
(1) Assume for every m ∈ M, all idempotents in Am are central. Then all idempo-tents in A are central.
(2) If A is power-associative, then the following are equivalent:
(a) A is reduced;
(b) for every m∈ M, Am is reduced.
Proof. (1) The canonical map A → QMAm, a 7→ ([a,1]m), is an injective algebra morphism (see 17.1). For every idempotent f ∈ A, fm is an idempotent in Am and hence is central by our assumption. So the image of f under the above map is in the centre of QMAm. This implies thatf is in the centre ofA.
(2) (a)⇒(b) This follows from 16.4(5).
(b)⇒(a) Assume in the proof of (1) that f ∈Ais nilpotent. Then fm = 0 for all
m∈ M (because of (b)) and hencef = 0. 2
In general, the mapR →Rm need not be surjective and hence it may be difficult to transfer properties from R to Rm. However, over regular rings the situation is different:
17.7 Localization over regular rings.
Let M be an R-module, m⊂R a maximal ideal and J =J ac(R).
(1) Assume R is regular. Then:
(i) The canonical map R →Rm is surjective with kernel m and Rm 'R/m'lim
−→ {R/eR|e2 =e∈m}.
(ii) The kernel of the canonical map M →Mm ismM and Mm 'M/mM 'M⊗RRm 'lim
−→{M/eM |e2 =e∈m}.
The map M → Mm, a 7→ a + mM, is a direct limit of splitting R-module epimorphisms and hence is pure in σ[M].
(2) Assume R/J is regular. Then (M/J M)m 'M/mM.
Proof. (1)(i) Consider [a, s] ∈ Rm, a ∈ R, s ∈ R \ m. Then for some t ∈ R, s(1−ts) = 0, and s(a−tsa) = 0 implies [a, s] = [at,1], i.e., the map is surjective.
Therefore Rm is regular and so its Jacobson radical mRm is zero. By 17.1, this meansRm 'R/m. m is maximal and contained in the kernel of R→Rm. Hence the two ideals coincide.
SinceRis regular,m'lim
−→{Re|e2 =e∈m}. This implies the other isomorphisms with direct limits.
(ii) By (i), we have Mm ' M ⊗Rm ' M ⊗ R/m ' M/mM. Hence mM is contained in the kernel of M →Mm.
Assume a ∈ M is in this kernel. Then sa = 0 for some s ∈ R\m. Choose t∈ R with s(1−ts) = 0. Then 1−ts∈m and a= (1−ts)a ∈mM.
(2) M/J M is an R/J-module and (since J ⊂ m) we may identify R \ m and (R/J)\(m/J). Putting m =m/J we have by (1),
(M/J M)m '(M/J M)/m(M/J M)'M/mM.
2 17.8 Tensor product over regular rings.
Let R be a regular ring and A an associative unital R-algebra. AssumeM andM0 are left A-modules with M0 ∈σ[M] and M finitely presented in σ[M]. Then the map from 15.7,
νM :HomA(M, M0)⊗RRm →HomA(M, M0 ⊗RRm), is an isomorphism for every m ∈ M. In particular,
EndA(M)⊗RRm 'EndAm(M⊗RRm).
Proof. Taking into account that the canonical map M →Mm is a pure epimorphism in σ[M] (see 17.7), the assertion is shown with the same proof as the corresponding
result for Pierce stalks in 18.5. 2
As a first local-global result we characterize (von Neumann) regularR-algebras by their localizations at maximal ideals of R:
17.9 Associative regular algebras.
Let A be an associative R-algebra with unit.
(1) The following statements are equivalent:
(a) A is regular;
(b) for every m∈ M, Am is a regular algebra.
(2) If A is a central algebra, then the following are equivalent:
(a) A is regular;
(b) R is regular and for every m∈ M, A/mA is regular;
(c) for every m∈ M, Am is regular.
(3) R is regular if and only if Rm is a field, for every m∈ M.
Proof. (1) (a)⇒(b) This follows from 16.6.
(b)⇒(a) Assume all Am 'A⊗Rm to be regular. Then all A⊗Rm-modules are flat and hence, by 17.1, allA-modules are flat, i.e., A is regular (see 7.4).
(2) (a)⇒(b) R is the centre of the regular ring A and hence is regular (e.g., [40, 3.16]). Obviously, all factor rings A/mA are regular.
(b)⇒(c) SinceR is regular, Am 'A/mA by 17.7.
(c)⇒(a) This is shown in (1).
(3) Observing 17.7 this is a special case of (2). 2
Strongly regular algebras are regular algebras without nilpotent elements or with all idempotents central. They can also be described by their algebras of fractions.
Further characterizations will be obtained in 18.4.
17.10 Associative strongly regular algebras.
Let A be an associative R-algebra with unit.
(1) Assume for every m ∈ M, Am is a division algebra. Then A is a strongly regular algebra.
(2) If A is a central R-algebra, then the following are equivalent:
(a) A is strongly regular;
(b) R is regular and A/mA is a division algebra, for every m∈ M;
(c) for every m∈ M, Am is a division algebra.
Proof. (1) If all Am are division algebras, then A is regular by 17.9, and is reduced by 17.6. HenceA is strongly regular.
(2) (a)⇒(b) As centre of the regular algebraA, R is regular. Assume K ⊂A is a finitely generated left ideal such that K +mA/mA is a non-zero ideal in A/mA.
Then K = eA for some idempotent e ∈ R. Since K 6⊂ mA we have e 6∈ m. This impliesK +mA⊃(Re+m)A=A and A/mA has no non-trivial left ideal, i.e., it is a division algebra.
(b)⇒(c) This follows from 17.7.
(c)⇒(a) is shown in (1). 2
17.11 Exercises.
(1) Let A be an associative central R-algebra with unit. Prove:
(i)A is fully idempotent if and only if Am is fully idempotent for every m∈ M.
(ii) A is a left V-ring (every simple left A-module is injective, [40, 23.5]) if and only if Am is a left V-ring for every m∈ M ([55, Theorem 6]).
(2) Suppose that inR every prime ideal is maximal, and letm ⊂R be a maximal ideal. Prove thatRm 'R/I, where the ideal I ⊂R is generated by the idempotents inm ([253]).
References: Armendariz-Fisher-Steinberg [55], Bourbaki [7], Matsumura [25], Burkholder [99], v.Oystaeyen-v.Geel [218], Szeto [252, 253], Villamayor-Zelinsky [260], Wisbauer [276].