Flat Modules We Recall Here Some Properties of Flat Modules As

Flat modules

We recall here some properties of flat modules as exposed in Bourbaki, AlgébreCommutative, ch. 1. The aim of these pages is to expose the proofs of some of the characterizations of flat and faithfully flat modules given in Matsumura’s book Commutative Algebra. This part is mainly devoted to the exposition of a proof of the following Theorem. Let A be a commutative ring with identity and E an A-module. The following conditions are equivalent α β (a) for any exact sequence of A-modules 0 → M 0 / M / M 00 → 0 , the sequence

0 1E ⊗α 1E ⊗β 00 0 → E ⊗A M / E ⊗A M / E ⊗A M → 0

is exact; ∼ (b) for any ﬁnitely generated ideal I of A, the sequence 0 → E ⊗A I → E ⊗A A = E is exact;  b1   x1  t Pr . r . r (c) If bx = i=1 bixi = 0 for some b =  .  ∈ A and x =  .  ∈ E , there exist a matrix br xr s t A ∈ Mr×s(A) and y ∈ E such that x = Ay and bA = 0.

α β We recall that, for any A-module, N, and for any exact sequence 0 → M 0 / M / M 00 → 0 , one gets the exact sequence 0 1N ⊗α 1N ⊗β 00 N ⊗A M / N ⊗A M / N ⊗A M → 0 . Moreover, if αM 0 is a direct summand in M (which is equivalent to the existence of a morphism πM → M 0 0 1N ⊗α such that πα = 1M 0 ), then the morphism N ⊗A M / N ⊗A M is injective and the image of 1N ⊗ α is a direct summand in N ⊗A M. Definition. Let A be a commutative ring with identity and E and M two A-modules. We say that E is M-flat if, for any A-module, M 0, and for any injective homomorphism α : M 0 → M, the homomorphism 0 1E ⊗ α : E ⊗A M → E ⊗A M is injective. Keeping the notations above, we remark that E is M-flat if the above condition holds for any finitely 0 0 Pr generated A-module M . In fact, for a general A-module, M , and an element z = i=1 ei ⊗yi in ker(1E ⊗α), 0 0 0 one can take the submodule, N , generated by y1, . . . , yr and the natural homomorphism E⊗AN → E⊗AM . 0 0 If z = 0 in E ⊗A N , the same holds for its image in E ⊗A M . Lemma 1. Let M and E be A-modules and suppose E to be M-flat. (a) For any submodule, N, of M, E is N-flat. (b) For any quotient module, Q, of M, E is Q-flat.

Proof. (a) Let j : N → M be the natural inclusion map. Starting with the diagram, whose rows and columns are exact,

0 0

0 α 0 1E ⊗α 0 / N / N E ⊗A N / E ⊗A N and tensoring by E, one gets j 1E ⊗j 0 0 / N / M 0 j◦α 0 / E ⊗A N / E ⊗A M 1E ⊗(j◦α)

and rows and columns are again exact, because E is M-ﬂat. This implies the injectivity of 1E ⊗ α.

1 (b) Let p : M → Q be the projection map and let i : K → M the inclusion of its kernel. Given an injection β : Q0 → Q, let M 0 = p−1(imβ) and consider the following diagram, whose rows and columns are exact.

0 0 0

K K E ⊗A K E ⊗A K

0 0 i i 1E ⊗i 1E ⊗i j Tensoring by E, one gets 0 0 1E ⊗j 0 / M / M 0 / E ⊗A M / E ⊗A M

0 p p 0 1E ⊗p 1E ⊗p 0 0 / Q / Q 0 β E ⊗A Q / E ⊗A Q 1E ⊗β

0 0 0 0

whose rows and columns are again exact, because E is M-ﬂat. Using the snake lemma, or by a direct check on elements, one deduces that ker(1E ⊗ β) = 0.

Lemma 2. Let (Mi)i∈I a family of A modules and E be an A-module Mi-ﬂat for each i ∈ I. Let M = L i∈I Mi, then E is M-ﬂat.

i p Proof. Suppose I = {1, 2} and consider the sequence 0 → M1 / M / M2 → 0 where M = M1 ⊕M2. 0 0 0 0 0 let M be a submodule of M and set M1 = M ∩ M1 and M2 = p(M ). In this way one gets the commutative diagram, with exact rows and columns,

0 0 0

0 v1 0 1E ⊗v1 0 / M1 / M1 0 / E ⊗A M1 / E ⊗A M1

0 0 i i 1E ⊗i 1E ⊗i Tensoring by E, one gets 0 v 0 1E ⊗v 0 / M / M E ⊗A M / E ⊗A M

0 p 0 p 1E ⊗p 1E ⊗p 0 0 0 / M / M2 2 v2 0 / E ⊗A M2 / E ⊗A M2 1E ⊗v2

0 0 0 0

whose rows and columns are exact. This allows to conclude that 1E ⊗ v is an injective homomorphism. By induction one can conclude that the claim is true for any finite sum of modules. In order to conclude in the general case, we observe that even if M is a direct sum of a general family of modules, it suffices to verify the condition for a finitely generated submodule M 0 of M and actually M 0 is contained in a finite sum of modules of the family.

Now we can prove a characterization of flat modules (one can take as the definition of flatness, that an A module E is flat if, and only if, it satisfies one of the equivalent conditions in the following proposition).

2 Proposition. Let E be an A module. The following facts are equivalent. (a) E is A-ﬂat (i.e. for any ideal I of A, the map I ⊗A E → A ⊗A E is injective); (b) For any A-module, M, E is M-ﬂat; α β (c) For any exact sequence of A-modules M 0 / M / M 00 , the sequence

0 α⊗1E β⊗1E 00 M ⊗A E / M ⊗A E / M ⊗A E

is exact.

Proof. (a) ⇒ (b); Any A-module, M, is isomorphic to a quotient of a free A-module. By applying Lemma 1 and Lemma 2, one deduces from (a) that E is M-ﬂat. (c) ⇒ (a); is straightforward. (b) ⇒ (c); is a consequence from the following lemma.

α β Lemma 3. Let M 0 / M / M 00 be an exact sequence of A-modules and E be an A module, M 00-ﬂat. Then the sequence 0 α⊗1E β⊗1E 00 M ⊗A E / M ⊗A E / M ⊗A E is exact.

α β Proof. Let N 00 = imβ ⊆ M 00 from the exact sequence M 0 / M / N 00 → 0 one deduces the exactness of 0 α⊗1E β⊗1E 00 M ⊗A E / M ⊗A E / N ⊗A E → 0 for general properties of tensor product. The fact that E is M 00-ﬂat, implies the injectivity of the map 00 00 N ⊗A E → M ⊗A E induced by the inclusion. This means precisely that the sequence in the above proclaim is exact.

t Pr Corollary. An A-module, E, is ﬂat if, and only if, the following condition holds. If bx = i=1 bixi = 0 for  b1   x1  . r . r s some b =  .  ∈ A and x =  .  ∈ E , then there exist a matrix A ∈ Mr×s(A) and y ∈ E such that br xr x = Ay and tbA = 0.

Pr Proof. Let I be the ideal of A generated by b1, . . . , br. An element, η = i=1 bi ⊗ xi, in the kernel of  x1  . r t Pr I ⊗A E → A ⊗A E = E determines an r-uple x =  .  ∈ E such that bx = i=1 bixi = 0. The existence xr s t of a matrix A and y ∈ E such that x = Ay and bA = 0, means exactly that η = 0 in I ⊗A E; hence E satisﬁes the condition (a) of the above Proposition.

 b1  t Pr . r Now suppose that E is a ﬂat module and bx = i=1 bixi = 0 for some b =  .  ∈ A and x = br  x1  . r r  .  ∈ E . The map A → A which sends u1, . . . , ur to b1u1 + ··· + brur, has a kernel, K, and we can xr write the exact sequence K → Ar → A, which remains exact after tensoring by E. This means that the Ps r r-uple x = (x1, . . . , xr) is in the image of K ⊗A E hence x = j=1 αj ⊗ yj for some αj ∈ K ⊂ A and Ps Pr yj ∈ E. Writing αj = (a1,j, . . . , ar,j) one has xi = j=1 ai,jyj for i = 1, . . . , r, and i=1 biai,j = 0 for j = 1, . . . , s.

3 We have given a complete proof of the Theorem stated at the beginning of the section. We can observe here that if φ : M → N is a morphism of A-modules, then for any A-module E, one has coker(1E ⊗ φ) = E ⊗A cokerφ, because the sequence M → N → cokerφ → 0 remains exact after tensoring by E. If E is a flat module, even the sequence 0 → kerφ → M → N remains exact after tensoring by E; then ker(1E ⊗ φ) = E ⊗A kerφ. Tensoring by E (flat) the exact sequence 0 → imφ → N → cokerφ → 0, one deduces even that im(1E ⊗ φ) = E ⊗A imφ. In that sense tensor product with flat modules is very similar to tensor product of vector spaces (over a field, every module is free hence flat). We gather here some facts about flat modules. Proposition. Let A be a commutative ring (with identity). (a) Let E be a flat A-module and a an element of A not dividing zero. Then ax = 0 for some x ∈ E if and only if x = 0. (b) Let A be an integral domain in which any finitely generated ideal is principal (i.e. a Bezout ring, in particular a PID when A is noetherian). An A-module, E, is flat if, and only if, E is torsion-free. (c) Let B a ring and φ : A → B a (unitary) ring-homomorphism, which makes B an A-module. For any homogeneous linear system My = 0 with coefficents in A, the solutions in B of the system are linear combinations with coefficents in B of the solutions in A if, and only, if B is a flat A-module.

µa Proof. (a) The sequence 0 → A / A is exact; Thus, tensoring by E, one gets the exact sequence

1E ⊗µa 0 → E ⊗A A / E ⊗A A . It suffices to remark that E ⊗A A = E and that in this identification 1E ⊗ µa induces the multiplication by a in E. (b) That a flat A-module has no torsion, follows from (a). On the other hand, let I be a finitely generated ideal in A and denote by a an element that generates I. This means that the inclusion of I in A can be µa identified with the map 0 → A / A of multiplication by a that is injective, because A has no zero- µa divisors. Tensoring by E, one gets the map 0 → E / E of multiplication by a in F , which is injective, because E has no torsion. (c) Let B be a flat A-module and M be a s × r-matrix with entries in A. There is an exact sequence M 0 → K / Ar / As , in which K is the set of solutions in Ar of the linear system My = 0. Tensoring r M s r r by B, one gets the exact sequence 0 → B ⊗A K / B / B , in which we identify B ⊗A A with B s s and B ⊗A A with B . The exactness of this sequence means precisely the assertion about the solution of the linear system in Br. j On the other side let I be a finitely generated ideal in A and let 0 / I / A be the exact 1B ⊗j ∼ sequence given by its inclusion in A. Tensoring by B one gets the map B ⊗A I / B ⊗A A = B and let Pr i=1 bi ⊗ xi an element of its kernel, where x1, . . . , xr is a set of generators of I in A. This means that Pr r b = (b1, . . . , br) is a solution of the linear equation i=1 yixi = 0; then, there exist α1, . . . , αs ∈ A and r Pr c1, . . . , cs in B such that b = c1α1 + ... + csαs and one has αj = (a1,j, . . . , ar,j) ∈ A , with i=i ai,jxi = 0. The matrix A = (ai,j) ∈ Mr×s(A) is exactly what appear in corollary above in order to verify the flatness of B.

The following proposition shows the relations between flatness and T or functors. Proposition. Let E be an A-module. The following facts are equivalent: (a) E is flat. A (b) For any A-module M and any integer n ≥ 1, one has T orn (E,M) = 0. A (c) For any A-module M, one has T or1 (E,M) = 0. A (d) For any ideal I, finitely generated as A-module, one has T or1 (E, A/I) = 0.

Proof. Let us start with (a)⇒(b). Given a resolution of M by free A-modules,

· · · → Ln / Ln−1 / ··· / L0 / M → 0

4 Tensoring by the ﬂat module E, one gets the exact sequence

· · · → E ⊗A Ln / E ⊗A Ln−1 / ··· / E ⊗A L0 / E ⊗A M → 0 .

A The homology groups of this sequence are the modules T orn (E,M), hence they vanish for any n ≥ 1. It’s straightforward that (b)⇒(c)⇒(d). It remains to prove that (d)⇒(a). From the exact sequence 0 → I / A / A/I → 0 , one deduces the exact sequence

A T or1 (E, A/I) / E ⊗A I / E ⊗A A

and (d) implies that the homomorphism E ⊗A I → E ⊗A A is injective, which allows us to conclude that E is ﬂat.

From this characterization and from the exact sequence of T or, one deduces that, given an exact sequence 0 → E0 / E / E00 → 0 of A-modules, with E00 flat, the A-module E0 is flat if, and only if, E is flat.

Faithfully flat modules The tensor product by a flat A-module an exact sequence of A-modules gives an exact sequence. Now we want to characterize the A-modules, E, for which a sequence of A-modules is exact if, and only if, it is exact after tensoring by E. For example, this property holds for the base ring A and then for any free A-module, but it is not true for any flat module; e.g. Q is a flat Z-module, but for any integer n > 1 the n sequence 0 → Z / Z → 0 is exact only after tensoring by Q (over Z there is a nontrivial cokernel). Proposition. Let A be a commutative ring with identity and E be an A-module. The following are equivalent: 0 00 0 00 (a) the sequence of A-modules, N → N → N , is exact if, and only if, E ⊗A N → E ⊗A N → E ⊗A N is exact. (b) E is flat and for any A-module, N, E ⊗A N = 0 implies N = 0. 0 ν (c) E is flat and for any A-module homomorphism, N / N , 1E ⊗ ν = 0 implies ν = 0. (d) E is flat and for any maximal ideal m of A, E 6= Em.

Proof. (a⇒b) It is obvious that condition (a) implies that E is a flat A-module. Let N be a module such that E ⊗A N = 0; then the sequence 0 → E ⊗A N → 0 is exact and, after (a), it follows that even the sequence 0 → N → 0 is exact, and then N = 0. 0 (b⇒c) E is flat and then, for any morphism ν : N → N, one has im(1E ⊗ ν) = E ⊗A imν. Then, 1E ⊗ ν = 0 implies E ⊗A imν = 0 and, after (b), this gives imν = 0, which means ν = 0. v w (c⇒a) As E is flat, the exactness of the sequence (1) : N 0 / N / N 00 implies the exactness of the 0 1E ⊗v 1E ⊗w 00 sequence (2) : E ⊗A N / E ⊗A N / E ⊗A N . Now suppose (2) exact, which implies that 0 = (1E ⊗ w) ◦ (1E ⊗ v) = 1E ⊗ (w ◦ v) and, after (c), this means w ◦ v = 0. Now, let K = kerw, I = imv and i : I → K be the inclusion map. We get the exact i p sequence 0 → I / K / K/I → 0 . Tensoring by the flat modules E, we have an exact sequence

1E ⊗i 1E ⊗p 0 → E ⊗A I / E ⊗A K / E ⊗A (K/I) → 0

and E ⊗A (K/I) = (E ⊗A K)/(E ⊗A I) = ker(1E ⊗ w)/im(1E ⊗ v) = 0, because the sequence (2) is exact. This gives 1E ⊗ p = 0 and then, by (c) we get p = 0 which gives the exactness of the sequence (1). ∼ (b⇒d) E/Em = E ⊗A (A/m) 6= 0, because A/m 6= 0 and (b) holds. (d⇒b) Let I be an ideal of A and let m be a maximal ideal containing I. E 6= Em implies that E 6= EI and then E ⊗A (A/I) 6= 0. This means that for any A-modules, N 6= 0, generated by 1 element (monogenous)

5 0 we have E ⊗A N 6= 0. Given a genral A-module, N 6= 0, let N be the submodule generated by a non zero element x, and take the exact sequence 0 → N 0 → N. Tensoring by the ﬂat module E we obtain the exact 0 sequence 0 → E ⊗A N → E ⊗A N, which gives E ⊗A N 6= 0, which allows to conclude.

A module is faithfully flat if it satisfies one (hence all) of the equivalent conditions of the above Propo- sition. In particular, a faithfully flat A-module, E, is faithful, and this means that is injective the structural morphism A → EndE , sending any element a of A to the map µa : E → E, which maps x ∈ E to ax. Denote by ma : A → A the analogous map of multiplication by a in A and observe that µa = 1E ⊗ ma. If E is faithfully flat and µa(x) = 0 for all x ∈ E, this implies ma = 0 and then ab = 0 for all b ∈ A; in particular this holds for b = 1 and then a = 0. Over a PID, a module, E, is faithfully flat if it has no torsion and E 6= Ep for any irreducible element p of A (an element p of a principal ring, A, is irreducible [extremal] if for any x in A the ideal (p, x) is either (p) or A). In particular a module which is faithful and flat is not necessarily faithfully flat, and the Z-module Q gives an example. Proposition. Let A and B be two rings, E a faithfully flat A-module and F an (A, B)-bimodule. Then F is flat (resp. faithfully flat) over B if, and only if, E ⊗A F is flat (resp. faithfully flat) over B.

Proof. Suppose F to be flat over B. Given an exact sequence of B-modules, N 0 → N → N 00, the sequence 0 00 F ⊗B N → F ⊗B N → F ⊗B N is an exact sequence of B-modules and remains exact if we see it as a 0 sequence of A-modules. The tensor product by E gives an exact sequence E ⊗A F ⊗B N → E ⊗A F ⊗B N → 00 E ⊗A F ⊗B N and then E ⊗A F is flat over B. 0 Now suppose E ⊗A F flat over S and take an injective homomorphism ν : N → N of B-modules. Then 0 0 0 → E ⊗A F ⊗B N → E ⊗A F ⊗B N is exact and then F ⊗B N → F ⊗B N is exact, because E is faithfully flat. This prove that F is flat over B.

Let F be faithfully flat over B and we have already proved that E⊗A F is flat over S. Given a B-module, N, we have that E ⊗A F ⊗B N = 0 implies F ⊗B N = 0, because E is faithfully flat and that F ⊗B N = 0 implies N = 0, because F is faithfully flat. Conversely, if E ⊗A F is faithfully flat and N is a B-module, F ⊗B N = 0 implies that E ⊗A F ⊗B N = 0 and then N = 0.

We state some consequences of the above Proposition. Corollary. Let ρ : A → B be a ring-homomorphism. (a) If E and F are faithfully flat A-modules, then E ⊗A F is faithfully flat over A. ∗ (b) If E is a faithfully flat A-module, then ρ E = E ⊗A B is faithfully flat over B. (c) If E is a faithfully flat A-module and I is an ideal of A, then E/EI is faithfully flat over A/I. (d) If B is faithfully flat over A, then an A-module E is flat (resp. faithfully flat) over A if, and only if, E ⊗A B is flat (resp. faithfully flat) over B. (e) If F is a faithfully flat B-module, then ρ∗F is flat (resp. faithfully flat) over A if, and only if, B is flat (resp faithfully flat) over A.

Proof. (a) It suffices to apply the above Proposition with A = B. (b) It is again a consequence of the above Proposition, because B is a faithfully flat B-module. (c) It follows from (b) if we take B = A/I and the canonical projection as ρ : A → A/I. (d) Suppose E to be flat over A and let ν : N 0 → N be an injective homomorphism of B-modules. We have a commutative diagram 0 E ⊗A B ⊗B N / E ⊗A B ⊗B N 1E ⊗1B ⊗ν

0 E ⊗A N / E ⊗A N 1E ⊗ν where the vertical arrows identify e ⊗ b ⊗ n with e ⊗ bn. The bottom homomorphism is injective, because E is ﬂat, so the same holds for the above homomorphism and then E ⊗A B is ﬂat over B. Moreover, if E

6 is faithfully flat and E ⊗A B ⊗B N = 0, then by the above identification, we have E ⊗A N = 0 and then N = 0. This proves the “only if” part. 0 Conversely, if E ⊗A B is flat over B and ν : N → N is an injective homomorphism of A-modules, we 0 have that 1B ⊗ν : B⊗A N → B⊗A N is injective, because B is faithfully flat. Tensoring by E⊗A B it remains ∼ injective. We observe that, for any A-module, M, one has E ⊗A B ⊗B B ⊗A M = E ⊗A M ⊗A B, and then we 0 have the injective homomorphism 1E ⊗ ν ⊗ 1B : E ⊗A N ⊗A B → E ⊗A N ⊗A B. The faithful flatness of B 0 over A allows to conclude that 1E ⊗ ν : E ⊗A N → E ⊗A N is injective and then E is flat. Finally, if E ⊗A B ∼ is faithfully flat and N is an A-module such that E ⊗A N = 0, then E ⊗A B ⊗B B ⊗A N = E ⊗A N ⊗A B = 0, implies that B ⊗A N = 0 and again the faithful flatness of B over A allows to conclude that N = 0. (e) We conclude by applying the above Proposition, with B, A, F and B in place of A, B, E, and F in the ∼ statement of the Proposition (F ⊗B B = F ).

We can observe that the claim (e) above, implies that B is faithfully flat over A if there exists a B- module faithfully flat over both rings. In particular, if ρ : A → B and σ : B → C are ring-homomorphisms, the fact that B is faithfully flat over A and C is faithfully flat over B implies that C is faithfully flat over A; but the fact that B and C are both faithfully flat over A, doesn’t imply that C is faithfully flat over B. As an example, if K is a field, then K[X] and K(X) are both faithfully flat over K (every non-zero module over a field is faithfully flat), but K(X) isn’t faithfully flat over K[X] exactly as Q is not faithfully flat over Z (K(X) has no torsion, but K(X)P (X) = K(X) for any irreducible polynomial, P (X)). In what follows, we want to give some characterizations of morphism of rings ρ : A → B for which B is a faithfully flat A-module. ∗ For any homomorphism ρ : A → B one can consider the two (adjoint) functors ρ and ρ∗, defined by ∗ sending an A-module, N, to the B-module ρ N = B ⊗A N and sending a B-module, M, to the A-module ∗ ρ∗M = M. In particular, given a B-module, M, you can consider the homomorphism i : M → ρ ρ∗M = B ⊗A M, defined by m 7→ 1 ⊗ m. It is an injective homomorphism of A-modules and has a left-inverse µ : B ⊗A M → M, defined by b ⊗ m 7→ bm. This means that the image i(M) is a direct factor of the A-module B ⊗A M. We will use these facts in the proof of the following Proposition. Let ρ : A → B be a ring homomorphism and suppose that it exists a B-module, E, such that ρ∗E s a faithfully flat A-module. Then one has ∗ (a) For any A-module, F , the canonical map j : F → ρ F = B ⊗A F , which sends x ∈ F to 1 ⊗ x, is injective. (b) For any ideal I of A, one has ρ−1(IB) = I. (c) ρ : A → B is injective. (d) For any maximal ideal m of A, there exists a maximal ideal n of B such that ρ−1(n) = m.

Proof. (a) We recalled above that the map i : E → E ⊗A B, defined by i(x) = x ⊗ 1, is injective and the image of E is a direct factor in the tensor product. Then, for any A-module, F , the map i ⊗ 1F : E ⊗A F → E ⊗A B ⊗A F is again injective. Now observe that i ⊗ 1F (y ⊗ x) = y ⊗ 1 ⊗ x = 1E ⊗ j(y ⊗ x) and recall that E is faithfully flat as A-module and conclude that j is injective. (b) We can apply (a) to the A-module A/I and conclude that the map j : A/I → B ⊗A (A/I) = B/IB is injective. (c) It suffices to apply (b) to the ideal (0) of A. (d) If m is a maximal ideal in A, we have ρ−1(mB) = m and then mB 6= B. Hence there exists a maximal ideal, n in B, containing mB and we have m ⊆ ρ−1(n) ⊂ A and the last inclusion is proper, because ρ is −1 injective and 1 ∈/ n. Then m = ρ (n) because the first is a maximal ideal.

When B itself is a faithfully flat A-module we can strenghten the above results. Proposition. Let ρ : A → B be a ring homomorphism. The following facts are equivalent: (a) B is a faithfully flat A-module. (b) ρ is injective and the A-module B/ρ(A) is flat. ∗ (c) B is flat over A and for any A-module, F , the map j : F → ρ F = B ⊗A F is injective.

7 (d) B is ﬂat over A and for any ideal I of A one has ρ−1(IB) = I. (e) B is ﬂat over A and for any maximal ideal m of A, there exists a maximal ideal n of B such that ρ−1(n) = m.

Proof. (c) ⇒ (d) It suﬃces to take F = A/I. (d) ⇒ (b) It suﬃces to take I = (0) to see that ρ is injective. In the exact sequence

0 → A → B → B/ρ(A) → 0,

B is a flat A-module and for any finitely generated ideal I of A, tensoring by A/I one gets the exact sequence −1 0 → A/I → B/IB → B/ρ(A) ⊗A A/I → 0, because I = ρ (IB). From the exact sequence of T or and A the flatness of B, one deduce that T or1 (B/ρ(A), A/I) = 0 and this allows to conclude that B/ρ(A) is a flat A-module. (b) ⇒ (a) In the exact sequence 0 → A → B → B/ρ(A) → 0, B/ρ(A) and A are flat modules so from the exact sequence of T or functors we deduce the flatness of B. Moreover, for any A-module, N, we have A T or1 (B/ρ(A),N) = 0 and then the exact sequence

0 → A ⊗A N → B ⊗A N → B/ρ(A) ⊗A N → 0.

Thus, if B ⊗A N = 0, this implies 0 = A ⊗A N = N and then B is faithfully ﬂat over A. It follows from the previous Proposition that (a) implies (c) and (e). Finally, if (e) holds, for any maximal ideal m in A, one has mB 6= B and B is faithfully ﬂat over A; so (e) ⇒ (a) and the proof is complete.

In order to prove a characterization of fully faithful ring extension in terms of prime ideals, we recall some easy results about multiplicative subsets of rings. A subset S of a ring A is a multiplicative set if 1 ∈ S and x, y ∈ S ⇒ xy ∈ S. As an example, if p is a prime ideal in A, then its complement, A r p, is a multiplicative set. If ρ : A → B is a ring-homomorphism and S is a multiplicative set in AS, then ρ(S) is a multiplicative set in B. Lemma 1. Let A be a ring and S a multiplicative set in A with 0 ∈/ S. Then for any ideal I of A such that I ∩ S = ∅, a maximal element in the set of ideals J such that I ⊆ J and J ∩ S = ∅ is a prime ideal.

Proof. The set of ideals, J of A such that I ⊆ J and J ∩ S = ∅ contains I and satisﬁes to the hypothesis of the Zorn’ Lemma. Then it has a maximal element that we denote by p. Given two elements x and y such that xy ∈ p, if no one of them belongs to p, the ideals p+(x) and p+(y) strictly contains p and then intersect S. This means that there exist elements a, a0 in p and b, b0 in A such that a + bx ∈ S and a0 + b0y ∈ S. This implies that S contains the product (a + bx)(a0 + b0y) = aa0 + ab0y + a0bx + bb0xy which belongs to p, beacause xy ∈ p. This proves that one at least between x and y must be in p and then p is a prime ideal.

Lemma 2. Let ρ : A → B be a homomorphism of rings and p a prime ideal in A. There exists a prime ideal p0 of B such that ρ−1(p0) = p if, and only if, ρ−1(pB) = p.

Proof. Suppose that there exists such a p0. Then pB ⊆ p0 and then ρ−1(pB) ⊆ p. The opposite inclusion is trivial. Conversely, let’s consider the multiplicative set Arp in A and set S = ρ(Arp). S is a multiplicative set in B and it doesn’t intersect the ideal pB, because ρ−1(pB) = p. Then, by the Lemma 1, there exists a maximal element, p0, in the set of ideals J of B such that pB ⊆ J and J ∩ S = ∅ and this maximal element 0 0 −1 0 is a prime ideal. Then pB ⊆ p and p ∩ ρ(A r p) = ∅, implies that ρ (p ) = p.

Proposition. Let ρ : A → B be a homomorphism of rings. (a) If there exists a B-module E such that ρ∗E is faithfully ﬂat over A then for any prime ideal p in A, there exists a prime ideal p0 in B such that p = ρ−1(p0).

8 (b) If B is ﬂat over A and for any prime ideal p in A, there exists a prime ideal p0 in B such that p = ρ−1(p0), then B is faithfully ﬂat over A.

Proof. (a) In a Proposition above we have proved that the existence of a B-module E such that ρ∗E is faithfully flat over A implies that for any ideal, I, in A one has I = ρ−1(IB). Then the above Lemma 2 allows to conclude. (b) By one of the above Propositions, it suffices to prove that for any maximal ideal, m in A, there exists a maximal ideal, n in B, such that m = ρ−1(n). By hypothesis there exists a prime ideal, p0, such that p = ρ−1(p0), then it suffices to take a maximal ideal, n, containing p0 in order to conclude, because ρ−1(n) is a proper ideal containing the maximal ideal m.

Now, we are able to give a characterization of faithfully flat ring extensions in terms of resolution of linear systems. We have already seen that, if an homomorphism ρ : A → B makes B a faithfully flat modules over A, then ρ is injective. In what follows we suppose to have a couple of rings (A, B), with A a subring of B. Let’s start with the following definition. Definition. Let A be a subring of B. We say that the couple (A, B) has the linear extension property if n any solution y = (y1, . . . , yn) ∈ B of a linear system Cy = d with coefficents C = (cij) 1≤i≤m ∈ Mm×n(A), 1≤j≤n m n d ∈ A , can be written in the form y = x + b1z1 + ··· + bpzp, where x ∈ A is a solution of the above system, n z1, . . . , zp ∈ A are solutions of the associated homogeneous system, Cz = 0, and the coefficents b1, . . . , bp are in B. If K is a subfield of a field L, then the couple (K,L) has the property of linear extension, as follows from elementary linear algebra. We want to prove the following Proposition. Let A be a subring of B. The couple (A, B) has the linear extension property if, and only if, B is faithfully flat over A.

Proof. If B is faithfully flat over A, in particular it is flat and we have already seen that the solutions in Bn of a homogeneous linear system Cz = 0 with coefficents in C = (cij) 1≤i≤m ∈ Mm×n(A) are linear combination, 1≤j≤n with coefficents in B, of the solutions in An. So it suffices to prove that the existence of a solution y ∈ Bn m n m to Cy = d, d ∈ A , implies the existence of a solution x ∈ A to the same system. Denote by cj ∈ A the j-th column of the matrix C and observe that the existence of a solution y ∈ Bn to Cy = d, can be written in the form n X m m yk ⊗ ck = 1 ⊗ d in B ⊗A A = B . k=1 m If we denote by M the submodule of A spanned by the columns ck, k = 1, . . . , n, we have to prove that m n the element d = 1⊗d, which belongs to A ∩ B ⊗ M, because of the existence of a solution y ∈ B , actually belongs to (the image of) M. This means that there exists a solution x ∈ An. We can prove it in this way. j p We have the inclusion i : M → Am and the exact sequence of flat modules 0 → A / B / B/A → 0 . By tensor products we obtain the following commutative diagram with exact rows

0 0 0

j⊗1M p⊗1M 0 / M / B ⊗A M / B/A ⊗A M / 0

i 1B ⊗i 1B/A⊗i

m j⊗1 m p⊗1 m 0 / A / B ⊗A A / B/A ⊗A A / 0

Pn If we set z = k=1 yk ⊗ ck, the above equality can be written as j ⊗1 (d) = 1B ⊗ i(y). Then, one has (p⊗1) ◦ (1B⊗i)(y) = (p⊗1) ◦ (j⊗1)(d) = 0 and, by commutativity, one gets (1B/A⊗i) ◦ (p ⊗ 1M )(y) = 0. The

9 map 1B/A⊗i is injective and then we have p ⊗ 1M (y) = 0. The exactness of the ﬁrst row implies that there exists an element x ∈ M such that y = j ⊗ 1M (x) and then

j ⊗1 (d) = 1B ⊗ i(y) = (1B ⊗ i) ◦ (j ⊗ 1M )(x) = (j⊗1) ◦ i(x).

The map j ⊗ 1 is injective and then d = i(x) which is exactly what we need to conclude. Conversely, suppose now that the couple (A, B) satisfies the linear extension property. This gives the flatness of B by loking only to homogeneous linear systems. Now, let I be an ideal of A and take an element d ∈ IB ∩ A. These means that there exist a1, . . . , an ∈ I and b1, . . . , bn ∈ B such that d = b1a1 + ··· + bnan. By the linear extension properties, there exist elements x1, . . . , xn in A such that d = x1a1 + ··· + xnan ∈ I. One concludes that I = IB ∩ A and that this equality occurs for any ideal I in A. Then B is faithfully flat over A.

10 Exercises from Chapter 1 of Bourbaki, Alg`ebreCommutative § 1 diagram chasing Exercise 1. Let u v A / B / C

a b c A0 / B0 / C0 u0 v0 be a commutative diagram such that v ◦ u = 0 and imu0 = kerv0 (exactness of the second row). Prove that imb ∩ imu0 = b(ker(c ◦ v)).

Solution. Let x ∈ imb∩imu0. Then x = b(y) for some y ∈ B and v0(b(y)) = v0(x) = 0, because imu0 = kerv0. But v0(b(y)) = c(v(y)) and this implies that x = b(y) ∈ b(ker(c ◦ v)). Now, let x = b(y) ∈ b(ker(c ◦ v)). The commutativity of the diagram implies v0(b(y)) = c(v(y)) = 0 and then x ∈ imb ∩ kerv0. The equality imu0 = kerv0 suﬃces for the conclusion. (It seems that the hypothesis 0 0 v ◦ u = 0 suﬃces to conclude, without any assumptions on the couple (v, u).)

Exercise 2. Let u v A / B / C

a b c A0 / B0 / C0 u0 v0 a0 b0 A00 / B00 u00 be a commutative diagram of abelian groups such that: imb = kerb0, imu0 = kerv0, v0 ◦ u0 = 0, a0 ◦ a = 0, c and u0 injective and a0 surjective. Prove that u00 is injective.

Solution. Let x00 ∈ keru00. The surjectivity of a0 means that x00 = a0(x0) for some x0 ∈ A0 and b0u0(x0) = u00a0(x0) = u00(x00) = 0. Hence u0(x0) ∈ kerb0 = imb and there exists y ∈ B such that b(y) = u0(x0). This gives cv(y) = v0b(y) = v0u0(x0) = 0 and the injectivity of c allows to conclude that y ∈ kerv = imu. Let x ∈ A with u(x) = y and observe that u0a(x) = bu(x) = b(y) = u0(x0). The injectivity of u0 allows to 0 00 0 0 0 conclude that a(x) = x and then x = a (x ) = a a(x) = 0.

Exercise 3. Let v w B / C / D

b c d A0 / B0 / C0 / D0 u0 v0 w0 a0 b0 c0 A00 / B00 / C00 u00 v00 a00 b00 A000 / B000 u000 be a commutative diagram of abelian groups having exact rows and colums and d, u00 injective and a00 surjective. Prove that u000 is injective and generalize.

Solution. As in the previous exercise we have to prove that the only element of keru000 is 0. Let’s start with x000 ∈ keru000. The surjectivity of a00 means that x000 = a00(x00) for some x00 ∈ A00 and b00u00(x00) = u000a00(x00) =

11 u000(x000) = 0. Hence u00(x00) ∈ kerb00 = imb0 and there exists y0 ∈ B0 such that b0(y0) = u00(x00). This gives c0v0(y0) = v00b0(y0) = v00u00(x00) = 0; then v0(y0) ∈ kerc0 = imc and there exists z ∈ C such that c(z) = v0(y0). One has dw(z) = w0c(z) = w0v0(y0) = 0; then the injectivity of d allows to conclude that z ∈ kerw = imv. Let y ∈ B such that v(y) = z and observe that v0b(y) = cv(y) = c(z) = v0(y0). Then y0 = b(y) + u0(x0) for some x0 ∈ A0. This implies u00(x00) = b0(y0) = b0b(y) + b0u0(x0) = u00a0(x0). The injectivity of u00 allows to conclude that a0(x0) = x00 and then x000 = a00(x00) = a00a0(x0) = 0. The generalization is clear (... but I don’t want to draw the big diagram that its statement needs).

Exercise 4. Let v w A u / B / C / D

a b c d A0 / B0 / C0 / D0 u0 v0 w0 be a commutative diagram of abelian groups having exact rows. (a) Prove that if a is surjective and b, d are injectives then c is injective. (b) Prove that if d is injective and a, c are surjectives then b is surjective.

Solution. (a) Let z ∈ kerc. Then dw(z) = w0c(z) = 0 and the injectivity of d implies that z ∈ kerw = imv. Then z = v(y) for some y ∈ B and v0b(y) = cv(y) = c(z) = 0. This means that b(y) ∈ kerv0 = imu0, and b(y) = u0(x0) for some x0 ∈ A0. Let x ∈ A such that a(x) = x0 (it exists because a is surjective), and observe that bu(x) = u0a(x) = u0(x0) = b(y). The injectivity of b implies that y = u(x) and then z = v(y) = vu(x) = 0 as we have to prove. (b) The same game with cokernels. Let y0 ∈ B0 and observe that the surjectivity of c implies that v0(y0) = c(z) for some z ∈ C. Then dw(z) = w0c(z) = w0v0(y0) = 0 and the injectivity of d implies that z ∈ kerw = imv. Let y ∈ B such that v(y) = z and observe that v0b(y) = cv(y) = c(z) = v0(y0). From the exactness of the second row, we deduce that y0 = b(y) + u0(x0) for some x0 ∈ A0 and the surjectivity of a allows to conclude that x0 = a(x) for some x in A. Then y0 = b(y) + u0a(x) = b(y) + bu(x) = b(y + u(x)) ∈ imb which means the surjectivity of b.

0 v 00 Exercise 5. Let A u / A / A / 0 be an exact sequence of modules over a ring R and let a0 a00 B0 / A0 , B00 / A00 two surjective homomorphisms of R-modules. If B00 is a projective module, prove that there exists a surjective homomorphism a : B0 ⊕ B00 → A such that the following diagram commutes i p B0 / B0 ⊕ B00 / B00 / 0

a0 a a00

0 00 A u / A v / A / 0 where i and p are canonical maps deﬁning B0 ⊕ B00 =∼ B0 × B00.

Solution. Since B00 is a projective module and the morphism v is surjective, there exists a (unique) map ν : B00 → A such that v ◦ ν = a00. Denoting by j : B00 → B0 ⊕ B00 the canonical map to the direct sum, there exists a unique map a : B0 ⊕ B00 → A such that aj = ν and ai = ua0. This gives the commutativity of the ﬁrst square and the commutativity of the second follows from the remark that va = vajp = vνp = a00p, 0 because idB0⊕B00 = jp + iq and vaiq = vua q = 0. Finally, the surjectivity of a follows by a direct check or using part (b) of the previous exercise.

0 v 00 Exercise 6. Let 0 / A u / A / A be an exact sequence of modules over a ring R and let a0 a00 A0 / C0 , A00 / C00 two injective homomorphisms of R-modules. If C0 is an injective module, prove

12 that there exists an injective homomorphism a : A → C0 ⊕ C00 such that the following diagram commutes

0 00 0 / A u / A v / A

a0 a a00 i p 0 / C0 / B0 ⊕ B00 / B00

where i and p are canonical maps deﬁning C0 ⊕ C00 =∼ C0 × C00.

Solution. The previous solution with reversed arrows. Since C0 is a injective module and the morphism u is surjective, then there exists a (unique) map ν : A → C0 such that ν ◦ u = a0. Denoting by q : C0 ⊕ C00 → C0 the canonical map from the direct product, there exists a unique map a : A → C0 ⊕ C00 such that qa = ν and pa = a00v. This gives the commutativity of the second square and the commutativity of the ﬁrst follows 0 00 from the remark that au = iqau = iνu = ia , because idC0⊕C00 = jp + iq and jpau = ja vu = 0. Finally, the injectivity of a follows by a direct check or using part (a) of exercise 4.

Exercise 7. Let U, V , W be abelian groups and f : U → V , g : V → W , homomorphisms. (a) Show that the diagram α β 0 / U / U × V / V / 0

f h −g 0 / V / V × W / W / 0 γ δ is commutative, with exact rows, where α(u) = (u, f(u)), β(u, v) = v − f(u), γ(v) = (v, g(v)), δ(v, w) = w − g(v), h(u, v) = (v, gf(u)). (b) Use the snake lemma to deduce the exact sequence

0 → kerf → ker(g ◦ f) → kerg → cokerf → coker(g ∩ f) → cokerg → 0.

Solution. (a) α is injective, because pU ◦ α = idU and, conversely, β is surjective because β ◦ iV = idV , where pU (u, v) = u and iV (v) = (0, v). It is easy to check that imα ⊆ kerβ and conversely, given (u, v) ∈ kerβ, one has v − f(u) = 0, which means v = f(u) and then (u, v) = (u, f(u)) ∈ imα. This proves the exactness of the ﬁrst row and the veriﬁcation for the second is analogous. The commutativity of the diagram follows by direct checks: γf(u) = (f(u), gf(u)) = hα(u) and −gβ(u, v) = −g(v − f(u)) = gf(u) − g(v) = δ(v, gf(u)) = δh(u, v). (b) The snake lemma applied to the above diagram gives the exact sequence

0 → kerf → kerh → kerg → cokerf → cokerh → cokerg → 0.

Now it suﬃces to observe that

kerh = { (u, 0) | gf(u) = 0 } =∼ ker(g ◦ f) and imh = { (v, gf(u)) | (u, v) ∈ U × V } =∼ V × im(g ◦ f) to conclude that cokerh =∼ V × W/V × im(g ◦ f) =∼ coker(g ◦ f). Taking into account the deﬁnition of the maps and the above identiﬁcations, one gets that in the exact sequence of the proclaim, the map kerf → ker(g ◦ f) is the inclusion, which is injective. The map ker(g ◦ f) → kerg is the restriction of f to this subset of U. The map kerg → cokerf maps an element z ∈ kerg to its image, z+imf, in cokerf. The map cokerf → coker(g◦f) sends an element v+imf ∈ cokerf to g(v) + im(g ◦ f). Finally, the map coker(g ◦ f) → cokerg sends an element t + im(g ◦ f) to the class t + img and it is obviously surjective, because im(g ◦ f) ⊆ img. This gives an explicit description of the above sequence.

13 § 2 flat modules Exercise 1. Show an exact sequence of A-modules, 0 → N 0 → N → N 00 → 0, and an A-module E such that E is N 0-flat and N 00-flat, but not N-flat.

Solution. It suﬃces to take the Z-modules N 0 = N 00 = E = Z/2Z, N = Z/4Z and the exact sequence

α β 0 / Z/2Z / Z/4Z / Z/2Z / 0 , where α(x + 2Z) = 2x + 4Z and β(y + 4Z) = y + 2Z. Tensoring by E = Z/2Z one gets E ⊗ N 0 =∼ E ⊗ N 00 =∼ ∼ Z Z E ⊗Z N = Z/2Z but the map 1E ⊗ α is the 0-morphism.

Exercise 2. Let M and N be submodules of an A-module E, such that M + N is flat. Prove that M and N are flat if, and only if, M ∩ N is flat.

Solution. Let’s take the exact sequence

α β 0 → M ∩ N / M × N / M + N → 0 where β(m, n) = m − n and α is the diagonal inclusion. Given an exact sequence 0 → E0 / E / E00 → 0 of A-modules, with E00 flat, the A-module E0 is flat if, and only if, E is flat (as one can see using the exact sequence of T or functors). Then, the conclusion ∼ follows by observing that M × N = M ⊕ N is flat if, and only if, M and N are both flat.

Exercise 3. Let A = K[X,Y ] be the ring of polynomials in two indeterminates over a field K. (a) Let b = (X), c = (Y ) the principal ideals in A, that are free A-modules and take their intersection b ∩ c = (XY ) which is again a free A-module. Show that a = b + c is an A-module without torsion but not flat. (b) Let R = { (a, −a) | a ∈ a } ⊂ A × A = A2. Let M and N be the images in the A-module A2/R of the two factors of A2. show that M and N are isomorphic to A, but the A-module M ∩ N is not flat.

Solution. (a) a has no torsion because A is an integral domain. The inclusion a → A is injective, but the ∼ induced map a ⊗A a → a ⊗A A = a sends both X ⊗ Y and Y ⊗ X to XY ⊗ 1 = XY and fails to be injective. (b)(x, 0) − (0, y) ∈ R if, and only if x = y ∈ a. This means that M ∩ N is isomorphic to the image of a in both components and a is not a ﬂat A-module.

Exercise 4. (a) Show an exact sequence, not split, 0 → E0 → E → E00 → 0, whose terms are all ﬂat A-modules. (b) Deduce from the previous sequence an exact sequence, not split, 0 → E0 → E → E00 → 0, whose terms are not ﬂat A-modules, but having the property that, for any A-modules M, the sequence 0 → 0 00 M ⊗A E → M ⊗A E → M ⊗A E → 0 is exact.

m Solution. (a) Let p be a prime in Z and denote by Z(p) the set Z(p) = n ∈ Q | (n, p) = 1 . Given two diﬀerent primes, p and q, we obtain the exact sequence

α 0 → Z(p) ∩ Z(q) / Z(p) × Z(q) / Q → 0

where α(x, y) = x − y. The sequence is not split and its terms are module without torsion over Z, hence they are flat. (b) It suffices to add to the terms of the previous sequence the terms of a split sequence of torsion (i.e. non-flat) Z-modules. As an example, the sequence α 0 → Z(p) ∩ Z(q) × Z/2Z / Z(p) × Z(q) × Z/2Z × Z/2Z / Q × Z/2Z → 0

have the required properties.