Commutative Algebra
Lecture 8: Flat Modules and Algebras
• Recall that an R-module M is flat if for each short exact sequence of R-modules
0 → A → B → C → 0,
the sequence obtained by tensoring with M is still exact:
0 → M ⊗R A → M ⊗R B → M ⊗R C → 0.
For example, free modules are always flat (it’s clear that R is flat as an R-module,
and easy to check that direct sums of flat modules are flat), and we’ve checked that if
S ⊆ R is a multiplicative subset, then S −1 R is a flat R-module. More generally, if A
is an R-algebra, we say A is flat (over R) is A is flat as an R-module. So localizations
are flat algebras; we will also see in Lecture 11 that completions of Noetherian rings
are flat.
• Flatness is one of those conditions that is equivalent to ostensibly stronger and weaker
properties:
Lemma 1. Let R be a ring and M an R-module. The following are equivalent:
i. If A → B is an injective R-module homomorphism with A and B finitely generated, then M ⊗R A → M ⊗R B is also injective.
ii. If A → B is any injective R-module homomorphism, then M ⊗R A → M ⊗R B
is also injective.
iii. M is flat, i.e. if 0 → A → B → C → 0 is a short exact sequence of R-modules,
then 0 → M ⊗R A → M ⊗R B → M ⊗R C → 0 is also exact.
iv. If A → B → C is any pair of composible R-module homomorphisms exact at B,
then M ⊗R A → M ⊗R B → M ⊗R C is exact at M ⊗R B.
Proof. We prove that (ii)⇒(iii)⇒(iv)⇒(ii) and that (i) ⇐⇒ (ii).
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(ii⇒iii) If 0 → A → B → C → 0 is exact, then we know that the sequence M ⊗R A →
M ⊗R B → M ⊗R C → 0 is exact without any flatness assumptions on M . Then
all that is left to check is exactness of 0 → M ⊗R A → M ⊗R B, i.e. injectivity
of M ⊗R A → M ⊗R B. But this holds if we assume that M ⊗R · preserves the
injectivity of A → B.
f
g
(iii⇒iv) If A
B
C is exact, construct the image factorization (surjection
followed by injection) for each of A → B and B → C:
im(f )
f
A
g
B
C
im(g)
By including the kernel of f and cokernel of g, we can extend this diagram to
include three short exact sequences:
0
0
0
im(f )
coker(g)
f
A
B
ker(f )
g
C
im(g)
0
0
0
(Exactness of im(f ) → B → im(g) at B is equivalent to exactness of A → B → C
at B.) Then tensoring with M , the short exact sequences stay short exact, so
we obtain the following injections and surjections
M ⊗ im(f )
M ⊗A
f
M ⊗B
g
M ⊗ im(g)
2
M ⊗C
with M ⊗ im(f ) → M ⊗ B → M ⊗ im(g) exact at M ⊗ B. But since image
factorization is unique, we find that M ⊗ im(f ) ∼
= im(M ⊗ f ) and M ⊗ im(g) ∼
=
im(M ⊗ g). Therefore im(M ⊗ f ) → M ⊗ B → im(M ⊗ g) is exact at M ⊗ B,
which implies that M ⊗ A → M ⊗ B → M ⊗ C is exact at M ⊗ B, as desired.
(iv⇒ii) If A → B is injective, then 0 → A → B is exact. Tensoring with M , then, we
obtain an exact sequence 0 → M ⊗ A → M ⊗ B, so M ⊗ A → M ⊗ B is injective.
(i ⇐⇒ ii) The implication (ii⇒i) is trivial. For the converse, suppose that f : A ֒→ B is an
injection such that M ⊗ A → M ⊗ B is not injective; we will construct a finitely
generated counterexample A′ ֒→ B ′ .
P
Since M ⊗ A → M P
⊗ B is not injective, there is a nonzero element i mi ⊗ ai ∈
M ⊗ A such that i mi ⊗ P
f (ai ) = 0. Define A′ ⊆ A as the submodule of A
generated by the ai . Then i mi ⊗ ai 6= P
0 as an element of M ⊗ A′ , since any
′
chain of elementary relations connecting
i mi ⊗ ai to 0 in M ⊗ A would also
P
constitute a proof in M ⊗ A that i mi ⊗ ai = 0.
P
Defining B ′ is a little trickier. Since i mi ⊗ f (ai ) = 0 in M ⊗ B, there does
exist a chain of equalities
X
X
mi ⊗ f (ai ) =
m′j ⊗ b′j = · · · = 0,
i
j
with each equality coming from one of the defining relations of the tensor product
M ⊗ B, such as m1 ⊗ b + m2 ⊗ b = (m1 + m2 ) ⊗ b. Then if we let B ′ ⊆ B be the
submodule generated by the f (ai ), the b′j , and all the other elements
of B used in
P
the chain of equalities above, then B ′ is finitely generated and i mi ⊗ f (ai ) = 0
′
′
in M ⊗ B ′ as well. Therefore f restricts to an injective homomorphism
PA → B
of finitely generated R-modules, but M ⊗f sends the nonzero element i mi ⊗ai
to 0.
• Geometrically, an R-algebra A corresponds to a morphism of schemes Spec(A) →
Spec(R). The “fiber” of Spec(A) at a point p ∈ Spec(R) (i.e. a prime ideal p of R)
is (Spec of) the tensor product A ⊗R κ(p), where κ(p) is the residue field of the local
ring Rp , or equivalently, the fraction field of the domain R/p. Roughly speaking, if A
is a flat R-algebra then the fiber “varies continuously” as we vary p.
Example 2. For example, let R = k[t] with k an algebraically closed field, and let
A = k[t, x]/(x2 − t). Then A is free of rank 2 as an R-module, so A is a flat R-algebra.
A prime ideal of k[t] of the form p = (t − a) has κ(p) = k, where the map k[t] → k
sends t 7→ a. Computing the fiber of A at this prime, we get
A ⊗R κ(p) ∼
= k[t, x]/(x2 − t) ⊗k[t] k ∼
= k[x]/(x2 − a).
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Now k[x]/(x2 − a) ∼
= k × k if a 6= 0, geometrically two disjoint points, and if a = 0
we get k[x]/(x2 ), geometrically a single point with multiplicity two. As a varies, the
two points merge and separate continuously.
Example 3. Another example has R = k[t] and A = k[t, x]/(tx − 1). Then A ∼
= Rt
is a localization, hence flat. The fiber over a prime (t − a) of R is
A ⊗R κ(p) ∼
= k[t, x]/(tx − 1) ⊗k[t] k ∼
= k[x]/(ax − 1),
which is isomorphic to k (geometrically, a single point) if a 6= 0, and isomorphic to the
zero ring (geometrically, the empty set) if a = 0. Looking at the graph of tx − 1 = 0 in
the tx-plane, we see that the fiber over (t − a) shoots off “to infinity” as a → 0. If we
projectivized this picture, there’d be a point in the fiber of A over (t) at “infinity.” But
generally, it’s not a problem if the fiber is smaller than you expect at some isolated
points; it’s a problem if the fiber is larger, as in the next example:
Example 4. Let R = k[t] and A = k[t, x]/(tx). This is not flat: because the Rmodule homomorphism R → R given by multiplication by t is injective, but after
tensoring with A, we get the homomorphism A → A multiplying by t, which sends
x 7→ 0. The fiber over a prime (t − a) is isomorphic to k (a point) if a 6= 0, but is
isomorphic to k[x] (a line) if a = 0. This dimension-jumping behavior is not possible
for flat families, as we will see in Lecture 10.
• More generally, flat modules are always torsion-free: If r ∈ R is a non-zerodivisor and
M is flat, then R r· R is injective, so we find that M r· M must be injective
too. This is what it means for M to be torsion-free.
— Bonus topic not covered in class —
• Even more generally, for any ideal I ⊆ R, we get an injection of R-modules I → R,
so if M is flat, then I ⊗R M → M is injective. (In other words, I ⊗R M → IM is an
isomorphism.) For example, M = Z/2 is not a flat R = Z-module: let I = (2). Then
I ⊗R M ∼
= R ⊗R M ∼
= M since I ∼
= R as an R-module, but IM = 0. Quotient rings
of R are almost never flat R-algebras; see the homework.
• In fact, this ideal-theoretic condition is equivalent to flatness:
Lemma 5. Let R be a ring, and M an R-module. Then the following are equivalent:
1. M is flat.
2. I ⊗R M → M is injective for all ideals I ⊆ R.
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Proof. We’ve shown that 1⇒2; we’ll prove now that 2⇒1. Assume that I ⊗R M → M
is injective for all ideals I of R. We will show that for all injections A → B of finitely
generated R-modules, the tensor product homomorphism M ⊗ A → M ⊗ B is also
injective.
Let A ֒→ B be an injection of finitely generated R-modules. Since B is finitely
generated, we can find an R-module surjection Rn ։ B for some n ∈ N, so that we
can write B ∼
= Rn /K for some submodule K ⊆ Rn . Then A ⊆ B is of the form
A∼
= L/K for some L ⊆ Rn .
It is enough to show that M ⊗ L → M ⊗ Rn is injective whenever L is a submodule
of Rn , because in that case, we can represent the above situation by the diagram
0
K
L
A
0
0
K
Rn
B
0
with exact rows, and then tensoring with M gives us
(0
)M ⊗ K
M ⊗L
M ⊗A
0
0
M ⊗K
M ⊗ Rn
M ⊗B
0
The bottom row is still exact because by assumption M ⊗ · preserves the injectivity
of K ֒→ Rn . (The top row then must still be exact since M ⊗ · also preserves the
injectivity of the middle vertical arrow L ֒→ Rn , but that fact will not be used.) Then
a diagram chase shows that M ⊗ A → M ⊗ B must also be injective: if x ∈ M ⊗ A
is sent to 0 ∈ M ⊗ B, then letting y ∈ M ⊗ L map to x ∈ M ⊗ A, we find that y (as
an element of M ⊗ Rn ) is sent to 0 in M ⊗ N ; thus y ∈ M ⊗ K. But then y is in the
kernel of M ⊗ L → M ⊗ N ′ , so its image x must be 0 after all.
Now we will prove that M ⊗ · preserves injections of the form L ֒→ Rn by induction
on n. The base case n = 1 is the hypothesis, because then L ⊆ R is an ideal of R.
Now let L ֒→ Rn with n arbitrary, and set I = L ∩ R = {r : (r, 0, . . . , 0) ∈ L}. Then
I is the kernel of L ⊆ Rn → Rn−1 , so L/I ֒→ Rn−1 . Then we have
0
I
L
L/I
0
0
R
Rn
Rn−1
0.
Tensoring with M preserves the first and third vertical injections by the base case
and the induction hypothesis, respectively, so by another diagram chase it preserves
the center injection as well.
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• As a corollary, we can characterize flat modules over a PID:
Corollary 6. Let R be a PID and M an R-module. Then M is flat if and only if M
is torsion-free.
Proof. M is torsion-free if and only if r· : M → M is injective for every r 6= 0 in
R. Another way to put this is to say that the map (r) ⊗R M → M is injective for
every r 6= 0 in R. Since 0 ⊗R M → M is always injective, and all the ideals of R are
principal, we find that M is torsion free if and only if I ⊗R M → M is injective for
every ideal I ⊆ R. This is equivalent to M being flat by the previous lemma.
• In particular, if R is a PID and M is a finitely generated flat R-module, then M is
free. (This absolutely fails if M is not finitely generated: for example, take M = Q
over R = Z!)
— End of bonus topic —
• There’s a much more general characterization of finitely-generated flat modules in
terms of freeness:
Theorem 7. Let R be a ring and M a finitely generated R-module. Then M is flat if
and only if for every prime ideal p ⊆ R, the localization Mp is free as an Rp -module.
• We’ll prove this theorem by first showing that flatness is a local property, and then
showing that over a local ring, the finite flat modules are free.
Lemma 8. Let R be a ring and M an R-module. Then the following are equivalent:
1. M is flat.
2. Mp is flat as an Rp -module for each prime ideal p ⊆ R.
Proof of 1⇒2. This is a special case of the assertion that if M is a flat R-module and
A is any R-algebra, then A ⊗R M is a flat A-module. This is easy to check, and is an
optional homework exercise.
Proof of 2⇒1. Let A ֒→ B be an injective R-module homomorphism, and suppose
that Mp is a flat Rp -module for each prime p. Let K be the kernel of M ⊗R A →
M ⊗R B; we wish to show that K = 0. We will do so by showing that Kp = 0 for
each prime p.
Let p be a prime ideal of R. Since the sequences 0 → A → B and 0 → K →
M ⊗R A → M ⊗R B are exact, and localization preserves exact sequences, we get
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exact sequences 0 → Ap → Bp and 0 → Kp → (M ⊗R A)p → (M ⊗R B)p . We may
tensor the first sequence with the flat Rp -module Mp to obtain an exact sequence
0 → Mp ⊗Rp Ap → Mp ⊗Rp Bp .
On the other hand, we know that (M ⊗R A)p ∼
= Mp ⊗Rp Ap and (M ⊗R B)p ∼
= Mp ⊗Rp Bp
from Homework 3, so we find that
0 → Kp → Mp ⊗Rp Ap → Mp ⊗Rp Bp
is exact. Therefore Kp is the kernel of an injective homomorphism, and thus is zero.
Hence the support of K is empty, and K = 0.
• All that remains, then is to prove the following lemma:
Lemma 9. Let R be a local ring and M a finitely generated R-module. Then M is
flat if and only if M is free.
The idea will be start with a basis for M/mM as an R/m-vector space (where m is
the maximal ideal of R), and lift it to a basis of M as an R-module.
• The first step is Nakayama’s lemma, which removes flatness from Lemma 9 and weakens a basis to a generating set. We’ll prove it in stages using this version of the
Cayley-Hamilton theorem:
Theorem 10 (Cayley-Hamilton). Let R be a ring, I ⊆ R an ideal, M a finitelygenerated R-module, and ϕ : M → M an R-module homomorphism with ϕ(M ) ⊆ IM .
Then ϕ satisfies a monic polynomial with coefficients in I.
P
Proof. Suppose {m1 , . . . , mn } generates M , and for each i write ϕ(mi ) = nj=1 aij mj
n
with each aij ∈ I. Then det(ϕ · 1n×n − aij i,j=1 ) acts by 0 on each mi , hence must
be the zero endomorphism of M . Expanding the determinant, we obtain a monic
polynomial equation satisfied by ϕ:
ϕn − b1 ϕn−1 − b2 ϕn−2 − · · · − bn = 0
with each bi ∈ I.
Lemma 11 (Nakayama, baby version). Let R be a ring, I an ideal, and M a finitelygenerated R-module with M = IM . Then there exists b ∈ I with (1 − b)M = 0.
Proof. Apply the Cayley-Hamilton theorem to the identity map M → M . Then we
obtain
1 − b1 − b2 − · · · − b n = 0
as an endomorphism of M , so (1 − b)M = 0 with b = b1 + · · · + bn ∈ I.
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Lemma 12 (Nakayama, child version). Let R be a ring, I an ideal contained in
every maximal ideal of R, and M a finitely-generated R-module with M = IM . Then
M = 0.
Proof. Let (1 − b)M = 0 with b ∈ I. Since b is in every maximal ideal, (1 − b) is in
none, hence is a unit and M = 0.
Lemma 13 (Nakayama, teenage version). Let R be a ring, I an ideal contained in
every maximal ideal of R, M a finitely-generated R-module, and N ⊆ M a submodule
with M = IM + N . Then M = N .
Proof. If M = IM + N , then I(M/N ) = M/N . Hence M/N = 0.
Lemma 14 (Nakayama, grown-up version). Let R be a ring, I an ideal contained in
every maximal ideal of R, M a finitely-generated R-module, and let m1 , . . . , mn ∈ M
generate M/IM . Then m1 , . . . , mn generate M .
Proof. Let N = Rhm1 , . . . , mn i ⊆ M . If N → M/IM is surjective, then M =
IM + N , so M = N .
• Generally speaking, for an ideal I to be contained in every maximal ideal is a very
strong condition. But if R is a local ring, all it means is that I is a proper ideal. So
Nakayama’s lemma is usually invoked in the following way: if M is a finitely generated
over a local ring R (with maximal ideal m) and m1 , . . . , mn generate the R/m-vector
space M/mM , then m1 , . . . , mn generate M .
• All that’s left is to show that if M is flat, then passing to M/mM also reflects linear
independence:
Lemma 15. Let R be a local ring with maximal ideal m. Let M be a flat Rmodule, and suppose that m1 , . . . , mn are R/m-linearly independent in M/mM . Then
m1 , . . . , mn are R-linearly independent in M .
Proof (following Matsumura). We use induction on n. The base case n = 0 is trivial:
the empty set is always linearly independent.
P
Now suppose n ≥ 1, and let nj=1P
aj mj = 0 in M . Let I be the ideal (a1 , . . . , an ).
Then I ⊗ M → M is injective, so nj=1 aj ⊗ mj = 0 in I ⊗ M . Now I is generated
by n elements, so it has a surjection Rn → I sending the jth basis element ej to aj .
Let the kernel of this surjection be K ⊆ Rn . Then we have an exact sequence
(0
)K ⊗ M
Rn ⊗ M
P
: nj=1 ej ⊗ mj
8
I ⊗M
0
0
P
P
Thus there exists an element i ki ⊗yi ∈PK ⊗M whose image in Rn ⊗M is nj=1 ej ⊗
mj . Writing each ki ∈ K ⊆ Rn as ki = nj=1 bij ej , we find that
X
bij ej ⊗ yi 7→
i,j
X
ej ⊗ mj ,
j
P
so mj = i bij yi forP
each j ∈ {1, . . . , n}. Furthermore, since each ki is in the kernel
n
of R → I, we have nj=1 bij aj = 0 for each i.
P
Now m1 67→ 0 in M/mM , so m1 = P
/ mM and at least one bi1 ∈
/ m, i.e. is a
i bi1 yi ∈
n
unit in R. Thus from the equation j=1 bP
ij aj = 0 for this i, we find that a1 is an
R-linear combination of the other ai : a1 = nj=2 cj aj for some cj ∈ R. Then we can
write the equation a1 m1 + a2 m2 + · · · + an mn = 0 as
a2 (m2 + c2 m1 ) + · · · + an (mn + cn m1 ) = 0.
Now assuming that the images of the {mj : 1 ≤ j ≤ n} in M/mM are R/m-linearly
independent, so are the images of the {mj + cj m1 : 2 ≤ j ≤ n}, since a nontrivial
linear relation among the latter would give a nontrivial linear relation among the
former. Hence by the induction hypothesis applied to m2 + c2 m1 , . . . , mn + cn m1 ,
we find that each of a2 , . . . , an must vanish in R. (And so must a1 , being a linear
combination of the a2 , . . . , an ). Thus m1 , . . . , mn are linearly independent in M as
well.
• Putting it all together, let R be a local ring with maximal ideal m and M a finitely
generated flat R-module. Then a basis for M/mM as an R/m-vector space lifts to
a generating set for M by Nakayama’s lemma (since M is finitely generated), and
this generating set for M is R-linearly independent because its images in M/mM are
R/m-linearly independent (since M is flat). Thus we have found an R-basis for M ,
and M is free.
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