Derivations and differentials
Johan Commelin
April 24, 2012
In the following text all rings are commutative with 1, unless otherwise
specified.
1 Modules of derivations
Let A be a ring, α : A → B an A algebra, and M a B-module.
1.1 Definition. An A-derivation is an A-linear map d : B → M, satisfying the
Leibniz rule, i.e., d( f g) = f d( g) + gd( f ) for all f , g ∈ B.
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Observe that the set of derivations Der A ( B, M ) is an B-module in a natural
way. We write Der A ( B) for Der A ( B, B).
We are used to the fact that the derivative of a constant is 0. An analogues
statement is true in this setting.
1.2 Lemma. For all c ∈ α( A) and d ∈ Der A ( B, M ) we have d(c) = 0.
Proof.
d(c) = cd(1) = cd(1) + 1 · d(c) − d(c) = d(1 · c) − d(c) = 0
Other facts that we are used to are also true.
1.3 Exercise.
1. Prove that, for d ∈ Der A ( B, M ) we have
d( an ) = nan−1 d( a)
and therefore, in a ring of characteristic n, we have d( an ) = 0.
2. Prove that, for d ∈ Der A ( B) we have a Leibniz formula for powers of d,
n
n i
d ( a ) · d n −i ( b );
d ( ab) = ∑
i
i =0
n
and if A is of characteristic n, this reduces to dn ( ab) = adn (b) + bdn ( a).
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Johan Commelin (s0849065, Universiteit Leiden)
April 24, 2012
Let A be a ring, and let a commutative diagram
C/I
B
β
q
C
of A-algebras be given, where I is an ideal of C, and q is the reduction map.
A map γ : B → C is called a lift of β if the obvious triangle commutes.
1.4 Lemma. Suppose I 2 = 0. Let S be the set of lifts B → C. Let γ be a fixed element of
S. Then S − γ = Der A ( B, I ).
Proof. Let γ0 ∈ S be given. Clearly γ0 − γ maps into I, since q ◦ (γ0 − γ) =
β − β = 0.
Also γ0 − γ is clearly A-linear. It satisfies the Leibniz rule, since
0 = (γ0 − γ)( f )(γ0 − γ)( g) = γ0 ( f g) + γ( f g) − γ( f )γ0 ( g) − γ0 ( f )γ( g),
and therefore
(γ0 − γ)( f g) = γ( f )γ0 ( g) + γ( g)γ0 ( f ) − 2γ( f g) =
γ( f )(γ0 − γ)( g) + γ( g)(γ0 − γ)( f ).
Finally observe that I inherits a B-module structure via γ, and that this does
not depend on the choice of γ, since I 2 = 0.
This shows that S − γ ⊂ Der A ( B, I ). It is left as an easy exercise to verify
that for d ∈ Der A ( B, I ) the map d + γ is a lift of β.
1.5 Lemma. The functor M 7→ Der A ( B, M ) is representable.
Proof. We want to show that there exists a B-module Ω B/A together with an
A-derivation d B/A : B → Ω B/A , such that Der A ( B, M ) ∼
= HomB (Ω B/A , M) in a
functorial way.
Define µ : B ⊗ A B → B by f ⊗ g 7→ f g. Put I = ker µ, Ω B/A = I/I 2 and
C = ( B ⊗ A B)/I 2 . Then µ induces a map µ0 : C → B, and
µ0
0 → Ω B/A → C → B → 0
is an exact sequence of B-modules. For i ∈ {1, 2} define λi : B → C by λ1 ( f ) =
f ⊗ 1 and λ2 ( f ) = 1 ⊗ f . Observe that these are splittings of the exact sequence.
Observe that as (Ω B/A )2 = 0 as an ideal of C. Define d = d B/A = λ1 − λ2 .
Indeed, d is a derivation, by lemma 1.4.
We claim that said isomorphism is given by
Der A ( B, M ) ∼
= HomB (Ω B/A , M)
δ 7→ ( f ⊗ g 7→ f δg)
φ ◦ d B/A ←[ φ.
It is left as an exercise to prove that the maps are each others inverses. 2
Johan Commelin (s0849065, Universiteit Leiden)
April 24, 2012
The module Ω B/A is called the module of Kähler differentials.
We also give another construction of Ω B/A . For all b ∈ B, we denote with db
an abstract symbol. Write Ω0 for the free A-module generated by {db : b ∈ B}.
We define Ω B/A to be the quotient of Ω0 by the submodule generated by
d(c f + c0 g) − cd f + c0 dg
∀c, c0 ∈ A,
∀ f , g ∈ B.
d( f g) − f dg − gd f
f, g ∈ B
Finally, we define d B/A : B → Ω B/A by f 7→ d f .
By definition this module represents the functor Der A ( B, _), and therefore
it is isomorphic to the previous construction in lemma 1.5.
1.6 Lemma. If the map A → B is surjective, then Ω B/A = 0.
Proof. Immediate from the construction of Ω B/A , and the result of lemma 1.2.
1.7 Exercise. Prove that, if B is generated by S ⊂ B as an A-algebra, then Ω B/A is
generated by d(S), using the Leibniz rule repeatedly.
In particular, if B is finitely generated as A algebra, then Ω B/A is finitely
generated as B-module.
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1.8 Exercise. If B = A[ x1 , . . . , xn ], then Ω B/A = ⊕in=1 Adxi .
Use this to show that if B = A[ x1 , . . . , xn ]/I with I = ( f 1 , . . . , f m ) then
Ω B/A = coker(d : I/I 2 → ⊕in=1 Bdxi . If we precompose d with ⊕im=1 Sei →
I/I 2 : ei 7→ f i , then Ω B/A is the cokernel of the jacobian matrix
 ∂f
∂f 
1
. . . ∂xm1
∂x1
 .
.. 
 .

...
. .
 .
∂f
∂ fn
. . . ∂xmn
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∂x
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2 Formally smooth, unramified, etale
We return to the situation before, about lifts of ring maps. Let A be a ring, and
let a commutative diagram
C/I
β
q
B
α
C
A
of A-algebras be given, where I is an ideal of C, and q is the reduction map.
Let S be the set of lifts B → C of β.
2.1 Definition. If for all pairs (C, I ), where I ⊂ C is an ideal satisfying I 2 = 0, we
have
#S ≥ 1 we say that α is formally smooth;
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Johan Commelin (s0849065, Universiteit Leiden)
April 24, 2012
#S ≤ 1 we say that α is formally unramified.
If α is both formally smooth and formally unramified, then we say that α is
formally etale.
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2.2 Lemma. The map α is formally unramified if and only if Ω B/A = 0.
Proof. The implication to the left is clear from the definition and lemma 1.4.
For the other implication take consider the construction of Ω B/A via C =
( B ⊗ A B)/I 2 and Ω B/A = I/I 2 . Then (Ω B/A )2 = 0. Also we had two lifts λ1
and λ2 . If α is formally unramified, we have λ1 = λ2 , and hence d = λ1 − λ2 =
0. Since d( B) generates Ω B/A we conclude that Ω B/A = 0.
2.3 Exercise. Let A be a ring, and S ⊂ A a set. Prove that the localization A →
S−1 A is formally etale.
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3 Two exact sequences
Let a commutative diagram
B
φ
B0
α0
α
A
ψ
A0
of ring maps be given.
3.1 Exercise. Show that an A0 -derivation d : B → M induces an A-derivation
d ◦ ψ.
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By the universal property of Ω B/A we see that φ induces a B-linear map
Ω B/A → Ω B0 /A0 . Explicitly the map is given by f dg 7→ φ( f )dφ( g).
3.2 Lemma. If φ is surjective, so is the induced map Ω B/A → Ω B0 /A0 , and its kernel is
generated by {d f |φ( f ) ∈ A0 }.
Proof. Note that Ω B/A is generated by d( B), while Ω B0 /A0 is generated by d( B0 ).
Clearly d( B0 ) = d(φ( B)), and hence the image of the induced map generates
Ω B0 /A0 . But then it is surjective, since it is B-linear.
If f dg is mapped to 0, then φ( f )dφ( g) = 0. This shows that the kernel is
generated by elements of the form id f with i ∈ ker φ and f ∈ B, together with
elements of the form d f satisfying φ( f ) ∈ A0 . But since di f = id f + f di, we see
that id f = di f − f di. Hence the elements of the second form suffice.
All these preperations lead to two important exact sequences.
3.3 Proposition. Let A → B → C be ring maps. Then there is a canonical exact sequence
of C-modules
Ω B/A ⊗ B C → ΩC/A → ΩC/B → 0.
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Johan Commelin (s0849065, Universiteit Leiden)
April 24, 2012
Proof. Observe that this sequence is exact if for every C-module M the sequence
HomC (Ω B/A ⊗ B C, M ) ← HomC (ΩC/A , M) ← HomC (ΩC/B , M) ← 0
is exact. But that sequence actually is
Der A ( B, M ) ← Der A (C, M) ← DerB (C, M) ← 0.
Now the exactness is clear.
3.4 Exercise. In the above situation, if B → C is formally smooth, then the sequence actually is short exact (i.e., the first map is injective) and split.
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We now consider the case where the second map is surjective. In the above
sequence we would have ΩC/B = 0, by lemma 1.6. Therefore we are interested
in the kernel of the first map of the sequence.
3.5 Proposition. Let A → B → C be ring maps, where the second map is surjective,
with kernel I ⊂ B. Then there exists a canonical exact sequence of C-modules
I/I 2 → Ω B/A ⊗ B C → ΩC/A → 0,
where the first map is given by f 7→ d f ⊗ 1.
Proof. We leave it as an exercise to the reader to show that the first map is
well-defined. By proposition 3.3 it is clear that we only need to show exactness
at the second term.
Since I maps to 0 in C, it follows that for f ∈ I, d f maps to 0 in ΩC/A .
Therefore the composition of the first two maps is zero.
Again let M be an arbitrary C-module. Then
HomC ( I/I 2 , M ) ← Der A ( B, M ) ← Der A (C, M )
is exact, since if δ ∈ Der A ( B, M ) maps to 0, this means that δ( I ) = 0. But then
δ comes from a derivation C = B/I → M. This proves the exactness of the
sequence.
3.6 Exercise. In the above situation, if the compostion A → C is formally smooth,
then the sequence actually is short exact (i.e., the first map is injective) and
split.
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4 Colimits
Let A be ring.
4.1 Lemma. Let B, A0 be A-algebras. Put B0 = B ⊗ A A0 . Then Ω B0 /A0 ∼
= Ω B/A ⊗ A A0 .
I.e., formation of differentials commutes with arbitrary change of base.
Proof. Observe that d ⊗ 1 : B0 → Ω B/A ⊗ A A0 is an A0 -derivation, which gives
a map φ : Ω B0 /A0 → Ω B/A ⊗ A A0 , by the universal property.
On the other hand, the composite map
d
B = B ⊗ A A → B ⊗ A A0 → ΩB0 /A0
is an A-derivation, which induces a map ψ : Ω B/A → Ω B0 /A0 . Since the target
is a B0 -module, we get another induced map, which is the inverse of φ.
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Johan Commelin (s0849065, Universiteit Leiden)
April 24, 2012
The following fact we shall not prove. It is an analogue to the fact that for
two manifolds X and Y we have the identity
T( x,y) X × Y = Tx X ⊕ Ty Y.
Observe that in the following lemma, instead of taking a product of algebras,
we take a coproduct, since moving between geometric categories (manifolds,
varieties, schemes) and the category of rings is contravariant.
4.2 Lemma. Let Bi be A-algebras, for i ∈ I. Let T denote the coproduct ⊗ A Bi . Then
Ω T/B ∼
= ⊕i (Ω Bi /A ⊗ Bi T ).
Proof. See Eisenbud, 394.
4.3 Remark. In Eisenbud, 397, is proven that formation of differentials commutes
with arbitrary colimits.
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As a consequence we state the following lemma, that we also do not prove.
4.4 Lemma. Let A → B be a map of rings. Let S ⊂ B be a subset. Then
ΩS−1 B/A ∼
= Ω B/A ⊗ B S−1 B.
Proof. See Eisenbud, 397.
Finally, we prove that formation of differentials commutes with finite products.
4.5 Lemma. Let B1 and B2 be A-algebras. Write B = B1 × B2 . Then
Ω B/A = Ω B1 /A × Ω B2 /A .
Proof. Let e1 denote (1, 0) ∈ B and e2 = (0, 1). Let M be an arbitrary B-module,
and δ : B → M an A-derivation. Since ei is idempotent, δei = 0. Hence by the
Leibniz rule δ(ei f ) = ei δ f . But then δ maps Bi = ei B to ei M. Thus δ corresponds
to a map Ω Bi /A → ei M. It follows that Ω B1 /A × Ω B2 /A satisfies the universal
property for Kähler differentials.
4.6 Exercise. Let A → B be a map of rings. Let δ : B → M be an A-derivation. Let
e ∈ B be an idempotent. Prove that δe = 0.
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April 24, 2012
Johan Commelin (s0849065, Universiteit Leiden)
5 Differential forms
Let B be a ring, and M an B-module.
5.1 Definition. Let k be an integer. Let M⊗k denote the k-fold tensor product of
M over B. Let N denote the submodule generated by
m1 ⊗ m2 ⊗ . . . ⊗ m k ,
∃i, j : i 6= j, mi = m j .
The k-th exterior power of M is defined as M⊗k /N, and is denoted Λk M. An
element of Λk is written as a wedge product: m1 ∧ . . . ∧ mk with mi ∈ M.
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Observe that Λ0 M = B.
Let α : A → B be an A-algebra. Recall that Ω B/A is a B-module. We write
ΩkB/A for Λk Ω B/A . We define maps
+1
di : ΩiB/A → ΩiB/A
f · ω1 ∧ . . . ∧ ω i 7 → d f ∧ ω1 . . . ∧ ω i .
5.2 Exercise. Prove that the maps di are well-defined.
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Observe that di+1 di = 0. Therefore we have an (algebraic) de Rham complex
associated to α : A → B
d
di
d1
+1
Ω•B/A : 0 → Ω0B/A → Ω1B/A → · · · → ΩiB/A → . . . ΩiB/A
→ ...
We can therefore define de Rham cohomology groups
i
HdR
( B/A) = ker di / im di−1
associated to B/A.
5.3 Remark. It can be proven (see e.g., Hartshorne) that if A = C and B is the ring
i coincides with
of global sections of a smooth affine variety X over C, then HdR
sing
the usual singular homology group Hi
( X, C).
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6 Tangent spaces
In view of the previous it makes sense to view Der A ( B) as the tangent space,
since it is the dual module to Ω B/A .
6.1 Exercise. The map Der A ( B) × Der A ( B) → Der A ( B) defined by [d1 , d2 ] = d1 ◦
d2 − d2 ◦ d1 is a Lie bracket. In particular if A is a field, this turns Der A ( B) into
a Lie algebra.
Proof. We verify that the map is well-defined. Clearly [d1 , d2 ] is an A-linear
map B → B. Also
[d1 , d2 ]( f g) = d1 d2 ( f g) − d2 d1 ( f g)
= d1 ( f d2 ( g) + gd2 ( f )) − d2 ( f d1 ( g) + gd1 ( f ))
= f d1 d2 ( g) + d1 ( f )d2 ( g) + gd1 d2 ( f ) + d1 ( g)d2 ( f )
− f d2 d1 ( g) − d2 ( f )d1 ( g) − gd2 d1 ( f ) − d2 ( g)d1 ( f )
= f [d1 , d2 ]( g) + g[d1 , d2 ]( f )
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Johan Commelin (s0849065, Universiteit Leiden)
April 24, 2012
shows that [d1 , d2 ] is a derivation.
Clearly [_, _] is anti-symmetric. To show that [_, _] is a Lie bracket, we must
prove that it satisfies the Jacobi identity.
[d, [e, f ]] + [ f , [d, e]] + [e, [ f , d]] = de f − d f e − e f d + f ed+
f de − ed f − de f + d f e
e f d − f ed − f de + ed f
=0
Consequently, Der A ( B) is a Lie algebra.
Let ( A, m) be a local ring containing a field k isomorphic to its residue field
A/m.
6.2 Lemma. The map
d : m/m2 → Ω A/k ⊗ A k
is an isomorphism.
Proof. We have a sequence k → A → k, where the second map is surjective.
Therefore, by proposition 3.5 we see that
m/m2 → Ω A/k ⊗ A k → Ωk/k → 0
is an exact sequence. The surjectivity follows immediately.
(Since k → k is formally smooth, we also get injectivity from this sequence.
We will give another proof below.)
To prove the injectivity we pas to the dual modules, and prove surjectivity
there.
d∗ : Homk (Ω A/k ⊗ A k, k) → Homk (m/m2 , k)
Observe that the left hand side is isomorphic to Hom A (Ω A/k , k ) ∼
= Derk ( A, k).
Note that for any derivation δ : A → k we have δ(m2 ) = 0 by the Leibniz rule.
Let h ∈ Homk (m/m2 , k) be given. Note that f ∈ A can be uniquely written
as c + f 0 , with c ∈ k and f 0 ∈ m. Define δ f = h( f 0 ). We claim that δ is a kderivation, and that d∗ (δ) = h. Clearly, δ is k-linear. Also, if we write f = c + f 0
and g = c0 + g0 , then
δ( f g) = h(c0 f 0 + cg0 ) = c0 h( f 0 ) + ch( g0 ) = gδ f + f δg.
Thus δ is indeed a k-derivation. Also, since δ(m2 ) = 0, we see that the restriction
of δ to m yields h, i.e., d∗ (δ) = h. Hence d∗ is surjective, and therefore d is
injective.
As a consequence of the provious lemma we see that the notions of tangent
space in algebraic geometry and differential geometry coincide.
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