Chapter 0 — Preliminaries

§1. Rings of fractions

1.0. Rings and algebras

(1.0.1) Every ring considered in this treatise will possess a unit element; every module over such a ring will be assumed unitary; every ring homomorphism will be assumed to send the unit element to the unit element; unless explicitly stated otherwise, a subring of a ring $A$ is taken to contain the unit element of $A$. We shall consider above all commutative rings, and when we speak of a ring without further qualification, it is to be understood that we mean a commutative ring. If $A$ is a ring, not necessarily commutative, by an $A$-module we shall always mean a left module, unless explicit mention to the contrary is made.

(1.0.2) Let $A$, $B$ be two rings, not necessarily commutative, and let $\varphi :A\to B$ be a homomorphism. Every left (resp. right) $B$-module $M$ can be equipped with the structure of a left (resp. right) $A$-module by setting $a\cdot m=\varphi (a)\cdot m$ (resp. $m\cdot a=m\cdot \varphi (a)$); when it is necessary to distinguish the $A$-module and $B$-module structures on $M$, we shall denote by $M_{\varphi }$ the left (resp. right) $A$-module so defined. If $L$ is an $A$-module, a homomorphism $u:L\to M_{\varphi }$ is therefore a homomorphism of abelian groups such that $u(a\cdot x)=\varphi (a)\cdot u(x)$ for $a\in A$, $x\in L$; one also calls such a $u$ a φ-homomorphism $L\to M$, and one calls the pair $(\varphi ,u)$ (or, by abuse of language, $u$ itself) a di-homomorphism from $(A,L)$ to $(B,M)$. The pairs $(A,L)$ consisting of a ring $A$ and an $A$-module $L$ therefore form a category, whose morphisms are the di-homomorphisms.

(1.0.3) Under the hypotheses of (1.0.2), if $\mathfrak{J}$ is a left (resp. right) ideal of $A$, we shall denote by $B\mathfrak{J}$ (resp. $\mathfrak{J}B$) the left (resp. right) ideal $B\varphi (\mathfrak{J})$ (resp. $\varphi (\mathfrak{J})B$) of $B$ generated by $\varphi (\mathfrak{J})$; this is also the image of the canonical homomorphism $B{\otimes }_A\mathfrak{J}\to B$ (resp. $\mathfrak{J}{\otimes }_{A}B\to B$) of left (resp. right) $B$-modules.

(1.0.4) If $A$ is a (commutative) ring and $B$ is a ring not necessarily commutative, giving a structure of $A$-algebra on $B$ amounts to giving a ring homomorphism $\varphi :A\to B$ such that $\varphi (A)$ is contained in the center of $B$. For every ideal $\mathfrak{J}$ of $A$, the ideal $\mathfrak{J}B=B\mathfrak{J}$ is then a two-sided ideal of $B$, and for every $B$-module $M$, $\mathfrak{J}M$ is a $B$-submodule equal to $(B\mathfrak{J})M$.

(1.0.5) We shall not return to the notions of module of finite type and (commutative) algebra of finite type; to say that an $A$-module $M$ is of finite type means that there is an exact sequence $A^p\to M\to 0$. An $A$-module $M$ is said to admit a finite presentation if it is isomorphic to the cokernel of a homomorphism $A^p\to A^q$; in other words, if there is an exact sequence $A^p\to A^q\to M\to 0$. Note that over a Noetherian ring $A$, every $A$-module of finite type admits a finite presentation.

Recall that an $A$-algebra $B$ is said to be integral over $A$ if every element of $B$ is a root in $B$ of a monic polynomial with coefficients in $A$; this is the same as saying that every element of $B$ is contained in a subalgebra of $B$ that is an $A$-module of finite type. When this is so and $B$ is commutative, the subalgebra of $B$ generated by a finite subset is an $A$-module of finite type; the commutative algebra $B$ is therefore integral and of finite type over $A$ if and only if $B$ is an $A$-module of finite type. One then also says that $B$ is a finite integral $A$-algebra (or simply finite, if no confusion results). We note that in these definitions the homomorphism $A\to B$ defining the $A$-algebra structure is not assumed to be injective.

(1.0.6) An integral ring [modern: integral domain] is a ring in which the product of a finite family of nonzero elements is nonzero; equivalently, in such a ring $0\ne 1$ and the product of two nonzero elements is nonzero. A prime ideal of a ring $A$ is an ideal $\mathfrak{p}$ such that $A/\mathfrak{p}$ is integral; this therefore implies $\mathfrak{p}\ne A$. For a ring $A$ to have at least one prime ideal, it is necessary and sufficient that $A\ne 0$.

(1.0.7) A local ring is a ring $A$ in which there is a single maximal ideal, which is then the complement of the set of invertible elements and contains every ideal $\ne A$. If $A$ and $B$ are local rings with maximal ideals $\mathfrak{m}$ and $\mathfrak{n}$, a homomorphism $\varphi :A\to B$ is called local if $\varphi (\mathfrak{m})\subset \mathfrak{n}$ (equivalently, if ${\varphi }^{-1}(\mathfrak{n})=\mathfrak{m}$). By passage to the quotients, such a homomorphism then defines a monomorphism of the residue field $A/\mathfrak{m}$ into the residue field $B/\mathfrak{n}$. The composite of two local homomorphisms is local.

1.1. Radical of an ideal. Nilradical and radical of a ring

(1.1.1) Let $\mathfrak{a}$ be an ideal of a ring $A$; the radical of $\mathfrak{a}$, written $r(\mathfrak{a})$, is the set of $x\in A$ such that $x^n\in \mathfrak{a}$ for some integer $n>0$; it is an ideal containing $\mathfrak{a}$. One has $r(r(\mathfrak{a}))=r(\mathfrak{a})$; the relation $\mathfrak{a}\subset \mathfrak{b}$ entails $r(\mathfrak{a})\subset r(\mathfrak{b})$; the radical of a finite intersection of ideals is the intersection of their radicals. If $\varphi$ is a homomorphism from a ring $A^{\prime }$ to $A$, one has $r({\varphi }^{-1}(\mathfrak{a}))={\varphi }^{-1}(r(\mathfrak{a}))$ for every ideal $\mathfrak{a}\subset A$. For an ideal to be the radical of an ideal, it is necessary and sufficient that it be an intersection of prime ideals. The radical of an ideal $\mathfrak{a}$ is the intersection of those prime ideals minimal among the prime ideals containing $\mathfrak{a}$; if $A$ is Noetherian, these minimal primes are finite in number.

The radical of the zero ideal (0) is also called the nilradical of $A$; it is the set $\mathfrak{N}$ of nilpotent elements of $A$. A ring $A$ is said to be reduced if $\mathfrak{N}=(0)$; for every ring $A$, the quotient $A/\mathfrak{N}$ is reduced.

(1.1.2) Recall that the radical $\mathfrak{R}(A)$ of a ring $A$ (not necessarily commutative) is the intersection of the maximal left ideals of $A$ (and also of the maximal right ideals). The radical of $A/\mathfrak{R}(A)$ is (0).

1.2. Modules and rings of fractions

(1.2.1) A subset $S$ of a ring $A$ is called multiplicative if $1\in S$ and the product of any two elements of $S$ lies in $S$. The examples that will matter most in what follows are:

1. the set $S_f$ of powers $f^n$ ($n\ge 0$) of an element $f\in A$;
2. the complement $A-\mathfrak{p}$ of a prime ideal $\mathfrak{p}$ of $A$.

(1.2.2) Let $S$ be a multiplicative subset of a ring $A$ and $M$ an $A$-module. On the set $M\times S$, the relation between pairs $(m_1,s_1)$ and $(m_2,s_2)$

"there exists $s\in S$ such that $s(s_1{m}_2-s_2{m}_1)=0$"

is an equivalence relation. We write $S^{-1}M$ for the quotient set, and $m/s$ for the canonical image in $S^{-1}M$ of the pair $(m,s)$; the canonical map of $M$ into $S^{-1}M$ is the map $i_M^s:m\mapsto m/1$ (also written iˢ). This map is in general neither injective nor surjective; its kernel is the set of $m\in M$ such that $sm=0$ for some $s\in S$.

On $S^{-1}M$ we define an additive group law by

(m₁/s₁) + (m₂/s₂) = (s₂m₁ + s₁m₂)/(s₁s₂)

(one checks that this is independent of the representatives chosen). On $S^{-1}A$ we define in addition a multiplicative law by $(a_1/s_1)(a_2/s_2)=(a_1{a}_2)/(s_1{s}_2)$, and an external law on $S^{-1}M$ with ring of operators $S^{-1}A$ by $(a/s)(m/s^{\prime })=(am)/(s{s}^{\prime })$. One checks that $S^{-1}A$ is thereby a ring (called the ring of fractions of $A$ with denominators in $S$) and $S^{-1}M$ an $S^{-1}A$-module (called the module of fractions of $M$ with denominators in $S$); for every $s\in S$, the element $s/1$ is invertible in $S^{-1}A$, with inverse $1/s$. The canonical map $i_A^s$ (resp. $i_M^s$) is a ring homomorphism (resp. an $A$-module homomorphism, where $S^{-1}M$ is viewed as an $A$-module via $i_A^s:A\to S^{-1}A$).

(1.2.3) If $S_f={f^n}_{n\ge 0}$ for some $f\in A$, we write $A_f$ and $M_f$ in place of $S_f^{-1}A$ and $S_f^{-1}M$; viewing $A_f$ as an algebra over $A$, we may write $A_f=A[1/f]$. The ring $A_f$ is isomorphic to the quotient algebra $A[T]/(fT-1)A[T]$. When $f=1$, $A_f$ and $M_f$ are canonically identified with $A$ and $M$; if $f$ is nilpotent, $A_f$ and $M_f$ reduce to 0.

When $S=A-\mathfrak{p}$ for a prime ideal $\mathfrak{p}$ of $A$, we write $A_{\mathfrak{p}}$ and $M_{\mathfrak{p}}$ in place of $S^{-1}A$ and $S^{-1}M$; here $A_{\mathfrak{p}}$ is a local ring, whose maximal ideal $\mathfrak{q}$ is generated by $i_A^s(\mathfrak{p})$, and one has $(i_A^s{)}^{-1}(\mathfrak{q})=\mathfrak{p}$. By passage to the quotients, $i_A^s$ gives a monomorphism of the integral ring $A/\mathfrak{p}$ into the field $A_{\mathfrak{p}}/\mathfrak{q}$, which is canonically identified with the field of fractions of $A/\mathfrak{p}$.

(1.2.4) The ring of fractions $S^{-1}A$ and the canonical homomorphism $i_A^s$ solve a universal mapping problem: every homomorphism $u$ from $A$ to a ring $B$ such that $u(S)$ consists of invertible elements of $B$ factors uniquely as

u : A ──iˢ_A──→ S⁻¹A ──u*──→ B

where $u\ast$ is a ring homomorphism. Under the same hypotheses, let $M$ be an $A$-module, $N$ a $B$-module, and $v:M\to N$ an $A$-module homomorphism (for the $B$-module structure on $N$ defined by $u:A\to B$); then $v$ factors uniquely as

v : M ──iˢ_M──→ S⁻¹M ──v*──→ N

where $v\ast$ is an $S^{-1}A$-module homomorphism (for the $S^{-1}A$-module structure on $N$ defined by $u\ast$).

(1.2.5) One defines a canonical isomorphism $S^{-1}A{\otimes }_{A}M\stackrel{\sim }{\to }{S}^{-1}M$ of $S^{-1}A$-modules by sending $(a/s)\otimes m$ to $(am)/s$; the inverse isomorphism sends $m/s$ to $(1/s)\otimes m$.

(1.2.6) For every ideal ${\mathfrak{a}}^{\prime }$ of $S^{-1}A$, $\mathfrak{a}=(i_A^s{)}^{-1}({\mathfrak{a}}^{\prime })$ is an ideal of $A$, and ${\mathfrak{a}}^{\prime }$ is the ideal of $S^{-1}A$ generated by $i_A^s(\mathfrak{a})$, which is canonically identified with $S^{-1}\mathfrak{a}$ (1.3.2). The map ${\mathfrak{p}}^{\prime }\mapsto (i_A^s{)}^{-1}({\mathfrak{p}}^{\prime })$ is an order-isomorphism from the set of prime ideals of $S^{-1}A$ onto the set of prime ideals $\mathfrak{p}$ of $A$ such that $\mathfrak{p}\cap S=\varnothing$. Moreover, the local rings $A_{\mathfrak{p}}$ and $(S^{-1}A{)}_{S^{-1}\mathfrak{p}}$ are then canonically (1.5.1) isomorphic.

(1.2.7) When $A$ is an integral ring with field of fractions $K$, the canonical map $i_A^s:A\to S^{-1}A$ is injective for every multiplicative subset $S$ not containing 0, and $S^{-1}A$ is canonically identified with a subring of $K$ containing $A$. In particular, for every prime ideal $\mathfrak{p}$ of $A$, $A_{\mathfrak{p}}$ is a local ring containing $A$, with maximal ideal $\mathfrak{p}{A}_{\mathfrak{p}}$, and one has $\mathfrak{p}{A}_{\mathfrak{p}}\cap A=\mathfrak{p}$.

(1.2.8) If $A$ is a reduced ring (1.1.1), so is $S^{-1}A$: indeed, if $(x/s{)}^n=0$ for $x\in A$, $s\in S$, this means that there is some $s^{\prime }\in S$ with $s^{\prime }{x}^n=0$, whence $(s^{\prime }x{)}^n=0$; the hypothesis then gives $s^{\prime }x=0$, so $x/s=0$.

1.3. Functorial properties

(1.3.1) Let $M$, $N$ be two $A$-modules and $u:M\to N$ an $A$-homomorphism. If $S$ is a multiplicative subset of $A$, one defines an $S^{-1}A$-homomorphism $S^{-1}M\to S^{-1}N$, written $S^{-1}u$, by $(S^{-1}u)(m/s)=u(m)/s$; if $S^{-1}M$ and $S^{-1}N$ are canonically identified with $S^{-1}A{\otimes }_{A}M$ and $S^{-1}A{\otimes }_{A}N$ (1.2.5), $S^{-1}u$ is identified with $1\otimes u$. If $P$ is a third $A$-module and $v:N\to P$ an $A$-homomorphism, then $S^{-1}(v\circ u)=(S^{-1}v)\circ (S^{-1}u)$; in other words, $S^{-1}M$ is a covariant functor in $M$ from the category of $A$-modules to the category of $S^{-1}A$-modules ($A$ and $S$ being fixed).

(1.3.2) The functor $S^{-1}M$ is exact; that is, if

M ──u──→ N ──v──→ P

is exact, so is

S⁻¹M ──S⁻¹u──→ S⁻¹N ──S⁻¹v──→ S⁻¹P.

In particular, if $u:M\to N$ is injective (resp. surjective), so is $S^{-1}u$; if $N$ and $P$ are submodules of $M$, then $S^{-1}N$ and $S^{-1}P$ are canonically identified with submodules of $S^{-1}M$, and one has

S⁻¹(N + P) = S⁻¹N + S⁻¹P    and    S⁻¹(N ∩ P) = (S⁻¹N) ∩ (S⁻¹P).

(1.3.3) Let $(M_{\alpha },{\varphi }_{\beta \alpha })$ be an inductive limit [modern: direct limit] system of $A$-modules; then $(S^{-1}{M}_{\alpha },S^{-1}{\varphi }_{\beta \alpha })$ is an inductive system of $S^{-1}A$-modules. Expressing $S^{-1}{M}_{\alpha }$ and $S^{-1}{\varphi }_{\beta \alpha }$ via tensor products (1.2.5 and 1.3.1), and using the fact that tensor product commutes with inductive limits, one obtains a canonical isomorphism

S⁻¹(lim⃗ M_α) ⥲ lim⃗ S⁻¹M_α,

which one also expresses by saying that the functor $S^{-1}M$ (in $M$) commutes with inductive limits.

(1.3.4) Let $M$, $N$ be two $A$-modules; there is a canonical functorial isomorphism (in $M$ and $N$)

(S⁻¹M) ⊗_{S⁻¹A} (S⁻¹N) ⥲ S⁻¹(M ⊗_A N)

sending $(m/s)\otimes (n/t)$ to $(m\otimes n)/(st)$.

(1.3.5) Similarly one has a functorial homomorphism (in $M$ and $N$)

S⁻¹Hom_A(M, N) → Hom_{S⁻¹A}(S⁻¹M, S⁻¹N)

sending $u/s$ to the homomorphism $m/t\mapsto u(m)/(st)$. When $M$ has a finite presentation, this homomorphism is an isomorphism: this is immediate when $M$ is of the form $A^r$, and one reduces from there to the general case by using the exact sequence $A^p\to A^q\to M\to 0$, the exactness of the functor $S^{-1}M$, and the left exactness of ${\mathrm{Hom}}_A(M,N)$ in $M$. Note that this situation always obtains when $A$ is Noetherian and $M$ is of finite type.

1.4. Change of multiplicative subset

(1.4.1) Let $S$, $T$ be multiplicative subsets of a ring $A$ with $S\subset T$; there is a canonical homomorphism ${\rho }_A^{T,S}$ (or simply ${\rho }^{T,S}$) from $S^{-1}A$ to $T^{-1}A$ sending the element $a/s$ of $S^{-1}A$ to the element $a/s$ of $T^{-1}A$; one has $i_A^T={\rho }_A^{T,S}\circ i_A^s$. For every $A$-module $M$, one similarly has a map ${\rho }_M^{T,S}:S^{-1}M\to T^{-1}M$ (the latter viewed as an $S^{-1}A$-module via ${\rho }_A^{T,S}$) sending $m/s$ to $m/s$; this map is written ${\rho }_M^{T,S}$, or simply ${\rho }^{T,S}$, and one has $i_M^T={\rho }_M^{T,S}\circ i_M^s$. Under the canonical identification (1.2.5), ${\rho }_M^{T,S}$ is identified with ${\rho }_A^{T,S}\otimes 1$. The map ${\rho }_M^{T,S}$ is a functorial morphism (or natural transformation) from the functor $S^{-1}M$ to the functor $T^{-1}M$; that is, the diagram

S⁻¹M ─S⁻¹u─→ S⁻¹N
  │            │
ρ^{T,S}_M     ρ^{T,S}_N
  │            │
  ↓            ↓
T⁻¹M ─T⁻¹u─→ T⁻¹N

commutes for every homomorphism $u:M\to N$. Note moreover that $T^{-1}u$ is entirely determined by $S^{-1}u$, since for $m\in M$ and $t\in T$ one has

$$(T^{-1}u)(m/t)=(t/1{)}^{-1}\cdot {\rho }^{T,S}((S^{-1}u)(m/1)).$$

(1.4.2) With the same notation, for two $A$-modules $M$, $N$, the diagrams (cf. (1.3.4) and (1.3.5))

(S⁻¹M) ⊗_{S⁻¹A} (S⁻¹N) ⥲ S⁻¹(M ⊗_A N)       S⁻¹Hom_A(M, N) → Hom_{S⁻¹A}(S⁻¹M, S⁻¹N)
        │                       │                       │                            │
        ↓                       ↓                       ↓                            ↓
(T⁻¹M) ⊗_{T⁻¹A} (T⁻¹N) ⥲ T⁻¹(M ⊗_A N)       T⁻¹Hom_A(M, N) → Hom_{T⁻¹A}(T⁻¹M, T⁻¹N)

commute.

(1.4.3) There is an important case in which the homomorphism ${\rho }^{T,S}$ is bijective: namely, when every element of $T$ divides some element of $S$; one then identifies $S^{-1}M$ and $T^{-1}M$ via ${\rho }^{T,S}$. A multiplicative set $S$ is called saturated if every divisor in $A$ of an element of $S$ is itself in $S$; replacing $S$ by the set $T$ of all divisors of elements of $S$ (a multiplicative and saturated set), one sees that one may always, if one wishes, restrict to modules of fractions $S^{-1}M$ with $S$ saturated.

(1.4.4) If $S$, $T$, $U$ are three multiplicative subsets of $A$ with $S\subset T\subset U$, one has

$${\rho }^{U,S}={\rho }^{U,T}\circ {\rho }^{T,S}.$$

(1.4.5) Consider an increasing filtered family $(S_{\alpha })$ of multiplicative subsets of $A$ (we write $\alpha \le \beta$ for $S_{\alpha }\subset S_{\beta }$), and let $S={\bigcup }_{\alpha }{S}_{\alpha }$, again multiplicative. Set ${\rho }_{\beta \alpha }={\rho }_A^{S_{\beta },S_{\alpha }}$ for $\alpha \le \beta$; by (1.4.4), the homomorphisms ${\rho }_{\beta \alpha }$ define a ring $A^{\prime }$ as the inductive limit of the inductive system $(S_{\alpha }^{-1}A,{\rho }_{\beta \alpha })$. Let ${\rho }_{\alpha }:S_{\alpha }^{-1}A\to A^{\prime }$ be the canonical map, and set ${\varphi }_{\alpha }={\rho }_A^{S,S_{\alpha }}$; since ${\varphi }_{\alpha }={\varphi }_{\beta }\circ {\rho }_{\beta \alpha }$ for $\alpha \le \beta$ by (1.4.4), there is a unique homomorphism $\varphi :A^{\prime }\to S^{-1}A$ such that the diagram

                  S_α⁻¹A
                ρ_α ↙ ↘ φ_α
              A′ ────φ──── S⁻¹A
                ↖  ρ_β    φ_β ↗
                  S_β⁻¹A
                    (α ≤ β)

commutes. In fact $\varphi$ is an isomorphism: it is plainly surjective by construction. On the other hand, if ${\rho }_{\alpha }(a/s_{\alpha })\in A^{\prime }$ satisfies $\varphi ({\rho }_{\alpha }(a/s_{\alpha }))=0$, then $a/s_{\alpha }=0$ in $S^{-1}A$, so there is $s\in S$ with $sa=0$; pick $\beta \ge \alpha$ with $s\in S_{\beta }$; then ${\rho }_{\alpha }(a/s_{\alpha })={\rho }_{\beta }(sa/(s{s}_{\alpha }))=0$, showing $\varphi$ injective. One handles an $A$-module $M$ in the same way, obtaining canonical isomorphisms

lim⃗_α S_α⁻¹A ⥲ (lim⃗ S_α)⁻¹A,    lim⃗_α S_α⁻¹M ⥲ (lim⃗ S_α)⁻¹M,

the second functorial in $M$.

(1.4.6) Let $S_1$, $S_2$ be multiplicative subsets of $A$; then $S_1{S}_2$ is also multiplicative. Let $S_2^{\prime }$ denote the canonical image of $S_2$ in $S_1^{-1}A$, a multiplicative subset of that ring. For every $A$-module $M$ there is a functorial isomorphism

$$S_2^{\prime -1}(S_1^{-1}M)\stackrel{\sim }{\to }(S_1{S}_2{)}^{-1}M$$

sending $(m/s_1)/(s_2/1)$ to $m/(s_1{s}_2)$.

1.5. Change of ring

(1.5.1) Let $A$, $A^{\prime }$ be two rings, $\varphi :A^{\prime }\to A$ a homomorphism, and let $S$ (resp. $S^{\prime }$) be a multiplicative subset of $A$ (resp. $A^{\prime }$) such that $\varphi (S^{\prime })\subset S$. The composite $A^{\prime }\to A\to S^{-1}A$ factors as $A^{\prime }\to S^{\prime -1}{A}^{\prime }\stackrel{{\varphi }^{S^{\prime }}}{\to }{S}^{-1}A$ by (1.2.4), with ${\varphi }^{S^{\prime }}(a^{\prime }/s^{\prime })=\varphi (a^{\prime })/\varphi (s^{\prime })$. If $A=\varphi (A^{\prime })$ and $S=\varphi (S^{\prime })$, then ${\varphi }^{S^{\prime }}$ is surjective. If $A^{\prime }=A$ and $\varphi$ is the identity, then ${\varphi }^{S^{\prime }}$ is none other than the homomorphism ${\rho }_A^{S,S^{\prime }}$ of (1.4.1).

(1.5.2) Under the hypotheses of (1.5.1), let $M$ be an $A$-module. There is a canonical functorial homomorphism

$$\sigma :S^{\prime -1}(M_{\varphi })\to (S^{-1}M{)}_{{\varphi }^{S^{\prime }}}$$

of $S^{\prime -1}{A}^{\prime }$-modules sending $m/s^{\prime }\in S^{\prime -1}(M_{\varphi })$ to $m/\varphi (s^{\prime })\in (S^{-1}M{)}_{{\varphi }^{S^{\prime }}}$; one checks at once that this is well-defined. When $S=\varphi (S^{\prime })$, this homomorphism $\sigma$ is bijective. When $A^{\prime }=A$ and $\varphi$ is the identity, $\sigma$ is the homomorphism ${\rho }_M^{S,S^{\prime }}$ of (1.4.1).

When in particular $M=A$, the homomorphism $\varphi$ makes $A$ an $A^{\prime }$-algebra; $S^{\prime -1}A$ is then a ring (identified with $(\varphi (S^{\prime }){)}^{-1}A$), and $\sigma :S^{\prime -1}(A_{\varphi })\to S^{-1}A$ is a homomorphism of $S^{\prime -1}{A}^{\prime }$-algebras.

(1.5.3) Let $M$, $N$ be two $A$-modules. Composing (1.3.4) with (1.5.2) yields a homomorphism

(S⁻¹M ⊗_{S⁻¹A} S⁻¹N)_[φ^{S′}] ← S′⁻¹((M ⊗_A N)_[φ])

which is an isomorphism when $\varphi (S^{\prime })=S$. Similarly, composing (1.3.5) with (1.5.2) gives a homomorphism

S′⁻¹((Hom_A(M, N))_[φ]) → (Hom_{S⁻¹A}(S⁻¹M, S⁻¹N))_[φ^{S′}]

which is an isomorphism when $\varphi (S^{\prime })=S$ and $M$ admits a finite presentation.

(1.5.4) Now consider an $A^{\prime }$-module $N^{\prime }$, and form the tensor product $N^{\prime }{\otimes }_{A^{\prime }}{A}_{\varphi }$, which becomes an $A$-module via $a\cdot (n^{\prime }\otimes b)=n^{\prime }\otimes (ab)$. There is a functorial isomorphism of $S^{-1}A$-modules

τ : (S′⁻¹N′) ⊗_{S′⁻¹A′} (S⁻¹A)_[φ^{S′}] ⥲ S⁻¹(N′ ⊗_{A′} A_[φ])

sending $(n^{\prime }/s^{\prime })\otimes (a/s)$ to $(n^{\prime }\otimes a)/(\varphi (s^{\prime })s)$; one checks separately that replacing $n^{\prime }/s^{\prime }$ (resp. $a/s$) by another expression of the same element leaves $(n^{\prime }\otimes a)/(\varphi (s^{\prime })s)$ unchanged. An inverse for $\tau$ is given by sending $(n^{\prime }\otimes a)/s$ to $(n^{\prime }/1)\otimes (a/s)$; one uses the canonical isomorphism $S^{-1}(N^{\prime }{\otimes }_{A^{\prime }}{A}_{\varphi })\simeq (N^{\prime }{\otimes }_{A^{\prime }}{A}_{\varphi }){\otimes }_A{S}^{-1}A\simeq N^{\prime }{\otimes }_{A^{\prime }}(S^{-1}A{)}_{\psi }$ (1.2.5), where $\psi$ is the composite $a^{\prime }\mapsto \varphi (a^{\prime })/1$ of $A^{\prime }\to S^{-1}A$.

(1.5.5) If $M^{\prime }$ and $N^{\prime }$ are two $A^{\prime }$-modules, composing (1.3.4) with (1.5.4) yields an isomorphism

S′⁻¹M′ ⊗_{S′⁻¹A′} S′⁻¹N′ ⊗_{S′⁻¹A′} S⁻¹A ⥲ S⁻¹(M′ ⊗_{A′} N′ ⊗_{A′} A).

Similarly, if $M^{\prime }$ admits a finite presentation, then by (1.3.5) and (1.5.4) one has an isomorphism

Hom_{S′⁻¹A′}(S′⁻¹M′, S′⁻¹N′) ⊗_{S′⁻¹A′} S⁻¹A ⥲ S⁻¹(Hom_{A′}(M′, N′) ⊗_{A′} A).

(1.5.6) Under the hypotheses of (1.5.1), let $T$ (resp. $T^{\prime }$) be a second multiplicative subset of $A$ (resp. $A^{\prime }$) such that $S\subset T$ (resp. $S^{\prime }\subset T^{\prime }$) and $\varphi (T^{\prime })\subset T$. Then the diagram

S′⁻¹A′ ──φ^{S′}──→ S⁻¹A
  │                  │
ρ^{T′,S′}          ρ^{T,S}
  │                  │
  ↓                  ↓
T′⁻¹A′ ──φ^{T′}──→ T⁻¹A

commutes. For an $A$-module $M$, the diagram

S′⁻¹(M_[φ]) ──σ──→ (S⁻¹M)_[φ^{S′}]
    │                    │
 ρ^{T′,S′}             ρ^{T,S}
    │                    │
    ↓                    ↓
T′⁻¹(M_[φ]) ──σ──→ (T⁻¹M)_[φ^{T′}]

commutes. Finally, for an $A^{\prime }$-module $N^{\prime }$, the diagram

(S′⁻¹N′) ⊗_{S′⁻¹A′} (S⁻¹A)_[φ^{S′}] ──τ──→ S⁻¹(N′ ⊗_{A′} A_[φ])
    │                                              │
    ↓                                            ρ^{T,S}
    ↓                                              ↓
(T′⁻¹N′) ⊗_{T′⁻¹A′} (T⁻¹A)_[φ^{T′}] ──τ──→ T⁻¹(N′ ⊗_{A′} A_[φ])

commutes, where the left-hand vertical arrow is obtained by applying ${\rho }_{N^{\prime }}^{T^{\prime },S^{\prime }}$ to $S^{\prime -1}{N}^{\prime }$ and ${\rho }_A^{T,S}$ to $S^{-1}A$.

(1.5.7) Let $A^{\prime \prime }$ be a third ring, ${\varphi }^{\prime }:A^{\prime \prime }\to A^{\prime }$ a ring homomorphism, and $S^{\prime \prime }$ a multiplicative subset of $A^{\prime \prime }$ with ${\varphi }^{\prime }(S^{\prime \prime })\subset S^{\prime }$. Set ${\varphi }^{\prime \prime }=\varphi \circ {\varphi }^{\prime }$; then

$${\varphi }^{\prime \prime S^{\prime \prime }}={\varphi }^{S^{\prime }}\circ {\varphi }^{\prime S^{\prime \prime }}.$$

For an $A$-module $M$, one has $M_{{\varphi }^{\prime \prime }}=(M_{\varphi }{)}_{{\varphi }^{\prime }}$; if ${\sigma }^{\prime }$ and ${\sigma }^{\prime \prime }$ are defined from ${\varphi }^{\prime }$ and ${\varphi }^{\prime \prime }$ as $\sigma$ is defined from $\varphi$ in (1.5.2), one has the transitivity formula

$${\sigma }^{\prime \prime }=\sigma \circ {\sigma }^{\prime }.$$

Finally, for an $A^{\prime \prime }$-module $N^{\prime \prime }$, the $A$-module $N^{\prime \prime }{\otimes }_{A^{\prime \prime }}{A}_{{\varphi }^{\prime \prime }}$ is canonically identified with $(N^{\prime \prime }{\otimes }_{A^{\prime \prime }}{A}_{{\varphi }^{\prime }}^{\prime }){\otimes }_{A^{\prime }}{A}_{\varphi }$, and likewise the $S^{-1}A$-module $(S^{\prime \prime -1}{N}^{\prime \prime }){\otimes }_{S^{\prime \prime -1}{A}^{\prime \prime }}(S^{-1}A{)}_{{\varphi }^{\prime \prime S^{\prime \prime }}}$ is canonically identified with $((S^{\prime \prime -1}{N}^{\prime \prime }){\otimes }_{S^{\prime \prime -1}{A}^{\prime \prime }}(S^{\prime -1}{A}^{\prime }{)}_{{\varphi }^{\prime S^{\prime \prime }}}){\otimes }_{S^{\prime -1}{A}^{\prime }}(S^{-1}A{)}_{{\varphi }^{S^{\prime }}}$. Under these identifications, if ${\tau }^{\prime }$ and ${\tau }^{\prime \prime }$ are defined from ${\varphi }^{\prime }$ and ${\varphi }^{\prime \prime }$ as $\tau$ is defined from $\varphi$ in (1.5.4), one has the transitivity formula

$${\tau }^{\prime \prime }=\tau \circ ({\tau }^{\prime }\otimes 1).$$

(1.5.8) Let $A$ be a subring of a ring $B$; for every minimal prime ideal $\mathfrak{p}$ of $A$, there exists a minimal prime ideal $\mathfrak{q}$ of $B$ with $\mathfrak{p}=A\cap \mathfrak{q}$. Indeed, $A_{\mathfrak{p}}$ is a subring of $B_{\mathfrak{p}}$ (1.3.2) and has exactly one prime ideal ${\mathfrak{p}}^{\prime }$ (1.2.6); since $B_{\mathfrak{p}}\ne 0$, it has at least one prime ideal ${\mathfrak{q}}^{\prime }$, and necessarily ${\mathfrak{q}}^{\prime }\cap A_{\mathfrak{p}}={\mathfrak{p}}^{\prime }$. The prime ideal ${\mathfrak{q}}_1$ of $B$ lying over ${\mathfrak{q}}^{\prime }$ then satisfies ${\mathfrak{q}}_1\cap A=\mathfrak{p}$, and a fortiori $\mathfrak{q}\cap A=\mathfrak{p}$ for every minimal prime $\mathfrak{q}$ of $B$ contained in ${\mathfrak{q}}_1$.

1.6. Identification of $M_f$ with an inductive limit

(1.6.1) Let $M$ be an $A$-module and $f$ an element of $A$. Consider a sequence $(M_n)$ of $A$-modules, all equal to $M$, and for $m\le n$ let ${\varphi }_{nm}:M_m\to M_n$ be the homomorphism $z\mapsto f^{n-m}z$. The pair $((M_n),({\varphi }_{nm}))$ is an inductive system of $A$-modules; let $N=\underrightarrow{\lim }{M}_n$. We shall define a canonical functorial $A$-isomorphism of $N$ onto $M_f$. To do so, note that for each $n$, ${\theta }_n:z\mapsto z/f^n$ is an $A$-homomorphism $M_n=M\to M_f$, and that ${\theta }_n\circ {\varphi }_{nm}={\theta }_m$ for $m\le n$. There is therefore an $A$-homomorphism $\theta :N\to M_f$ with ${\theta }_n=\theta \circ {\varphi }_n$ for every $n$, where ${\varphi }_n:M_n\to N$ is the canonical map. Since by hypothesis every element of $M_f$ has the form $z/f^n$ for some $n$, $\theta$ is surjective. On the other hand, if $\theta ({\varphi }_n(z))=0$, that is $z/f^n=0$, then $f^{k}z=0$ for some integer $k>0$, so ${\varphi }_{n+k,n}(z)=0$, whence ${\varphi }_n(z)=0$. Thus $M_f$ and $\underrightarrow{\lim }{M}_n$ are identified via $\theta$.

(1.6.2) Write $M_{f,n}$, ${\varphi }_{nm}^f$, and ${\varphi }_n^f$ in place of $M_n$, ${\varphi }_{nm}$, and ${\varphi }_n$. Let $g$ be a second element of $A$. Since $f^n$ divides $f^n{g}^n$, there is a functorial homomorphism

ρ_{fg, f} : M_f → M_{fg}    (1.4.1 and 1.4.3);

if $M_f$ and $M_{fg}$ are identified with $\underrightarrow{\lim }{M}_{f,n}$ and $\underrightarrow{\lim }{M}_{fg,n}$ respectively, then ${\rho }_{fg,f}$ is the inductive limit of the maps ${\rho }_{fg,f}^n:M_{f,n}\to M_{fg,n}$ defined by ${\rho }_{fg,f}^n(z)=g^{n}z$. Indeed, this follows at once from commutativity of the diagram

M_{f,n} ──ρⁿ_{fg,f}──→ M_{fg,n}
   │                       │
 φ^f_n                  φ^{fg}_n
   │                       │
   ↓                       ↓
  M_f ───ρ_{fg,f}───→  M_{fg}

1.7. Support of a module

(1.7.1) Given an $A$-module $M$, the support of $M$, written $Supp(M)$, is the set of prime ideals $\mathfrak{p}$ of $A$ such that $M_{\mathfrak{p}}\ne 0$. One has $M=0$ if and only if $Supp(M)=\varnothing$: indeed, if $M_{\mathfrak{p}}=0$ for every $\mathfrak{p}$, the annihilator of any element $x\in M$ is not contained in any prime ideal, hence equals $A$.

(1.7.2) If $0\to N\to M\to P\to 0$ is an exact sequence of $A$-modules, then

Supp(M) = Supp(N) ∪ Supp(P),

since for every prime $\mathfrak{p}$ of $A$ the sequence $0\to N_{\mathfrak{p}}\to M_{\mathfrak{p}}\to P_{\mathfrak{p}}\to 0$ is exact (1.3.2), and $M_{\mathfrak{p}}=0$ iff $N_{\mathfrak{p}}=P_{\mathfrak{p}}=0$.

(1.7.3) If $M$ is the sum of a family $(M_{\lambda })$ of submodules, then $M_{\mathfrak{p}}$ is the sum of the $(M_{\lambda }{)}_{\mathfrak{p}}$ for every prime $\mathfrak{p}$ of $A$ (1.3.3 and 1.3.2), so $Supp(M)={\bigcup }_{\lambda }Supp(M_{\lambda })$.

(1.7.4) If $M$ is an $A$-module of finite type, then $Supp(M)$ is the set of prime ideals containing the annihilator of $M$. Indeed, if $M$ is monogenic, generated by $x$, then $M_{\mathfrak{p}}=0$ iff there is $s\notin \mathfrak{p}$ with $sx=0$, that is, iff $\mathfrak{p}$ does not contain the annihilator of $x$. If now $M$ has a finite generating system $(x_i{)}_{1\le i\le n}$ and ${\mathfrak{a}}_i$ is the annihilator of $x_i$, then by (1.7.3) $Supp(M)$ is the set of $\mathfrak{p}$ containing some ${\mathfrak{a}}_i$, equivalently the set of $\mathfrak{p}$ containing $\mathfrak{a}={\bigcap }_i{\mathfrak{a}}_i$, the annihilator of $M$.

(1.7.5) If $M$ and $N$ are two $A$-modules of finite type, then

Supp(M ⊗_A N) = Supp(M) ∩ Supp(N).

It suffices to see, using (1.3.4), that for every prime $\mathfrak{p}$ of $A$, the condition $M_{\mathfrak{p}}{\otimes }_{A_{\mathfrak{p}}}{N}_{\mathfrak{p}}\ne 0$ is equivalent to "$M_{\mathfrak{p}}\ne 0$ and $N_{\mathfrak{p}}\ne 0$". In other words, we must show that if $P$, $Q$ are nonzero modules of finite type over a local ring $B$, then $P{\otimes }_{B}Q\ne 0$. Let $\mathfrak{m}$ be the maximal ideal of $B$. By Nakayama, the vector spaces $P/\mathfrak{m}P$ and $Q/\mathfrak{m}Q$ are nonzero, so their tensor product $(P/\mathfrak{m}P){\otimes }_{B/\mathfrak{m}}(Q/\mathfrak{m}Q)=(P{\otimes }_{B}Q){\otimes }_B(B/\mathfrak{m})$ is nonzero, whence the conclusion.

In particular, if $M$ is an $A$-module of finite type and $\mathfrak{a}$ is an ideal of $A$, then $Supp(M/\mathfrak{a}M)$ is the set of prime ideals containing both $\mathfrak{a}$ and the annihilator $\mathfrak{n}$ of $M$ (1.7.4) — equivalently, the set of prime ideals containing $\mathfrak{a}+\mathfrak{n}$.