I now want to talk about some of the material in Hartshorne, II.8.  First, we need some preliminaries from commutative algebra.

Let ${A}$ be a commutative ring, ${B}$ an ${A}$-algebra, and ${M}$ a ${B}$-module. Then an ${A}$-derivation of ${B}$ in ${M}$ is a linear map ${D: B \rightarrow M}$ satisfying ${D(a)=0}$ for ${a \in A}$ and

$\displaystyle D(bb') = (Db) b' + b (Db').$

The set of all such derivations forms a ${B}$-module ${\mathrm{Der}_A(B,M)}$. If we regard this as a set, clearly, we have a contravariant functor

$\displaystyle \mathrm{Der}_A(B, -): B-\mathrm{mod} \rightarrow \mathrm{Set}$

because if ${B \rightarrow B'}$ is a homomorphism of ${A}$-algebras, we can pull back a derivation.

Before proceeding, I should say something about the canonical example. Let ${M}$ be a smooth manifold and ${O_x}$ the local ring (of germs of smooth functions) at ${x \in M}$. Then ${\mathbb{R}}$ becomes an ${O_x}$-module if the germ ${f}$ acts by multiplication by ${f(x)}$. More precisely, we have an exact sequence

$\displaystyle 0 \rightarrow m_x \rightarrow O_x \rightarrow \mathbb{R} \rightarrow 0$

for ${m_x \subset O_x}$ the maximal ideal of functions vanishing at ${x}$, and this is the way ${\mathbb{R}}$ is an ${O_x}$-module.

Anyway, an ${\mathbb{R}}$-derivation ${O_x \rightarrow \mathbb{R}}$ is just a tangent vector at ${x}$.

Now back to the algebraic theory. It turns out that the functor ${\mathrm{Der}_A}$ is representable. In other words, for each ${A}$-algebra ${B}$, there is a ${B}$-module ${\Omega_{B/A}}$ such that

$\displaystyle \hom_B( \Omega_{B/A}, M) \simeq \mathrm{Der}_A(B,M) ,$

the isomorphism being functorial. In addition, there must be a “universal” derivation ${d: B \rightarrow \Omega_{B/A}}$ (corresponding to the identity ${\Omega_{B/A} \rightarrow \Omega_{B/A}}$ in the above functorial isomorphism), that any derivation factors through.

The construction of ${\Omega_{B/A}}$ is straightforward. We define it as the ${B}$-module generated by symbols ${db, b\in B}$, modulo the relations ${da = 0}$ for ${a \in A}$, ${d(b+b') = db + db'}$, and ${d(bb') = b' db + b db'}$. It is now clear that we have a functorial isomorphism as above. Now, ${\Omega_{B/A}}$ is called the module of Kahler differentials of ${B}$ over ${A}$.

I will now prove some basic properties of Kahler differentials.

Tensor products

First, ${\Omega}$ linearizes tensor products. In detail, if ${B_1, B_2}$ are ${A}$-algebras, then

$\displaystyle \Omega_{B_1 \otimes_A B_2/A} \simeq B_1 \otimes \Omega_{B_2/A} \oplus B_2 \otimes \Omega_{B_1/A}.$

To see this, we will consider what a ${A}$-derivation ${D}$ of ${B_1 \otimes_A B_2}$ in a ${B_1 \otimes_A B_2}$-module ${M}$ looks like. It induces maps ${D_1: B_1 \rightarrow M, D_2: B_2 \rightarrow M}$ given by ${D_1(b_1) := D(b_1 \otimes 1), D_2(b_2) := D(1 \otimes b_2)}$, which are themselves ${A}$-derivations. Conversely, given a pair of such ${A}$-derivations ${D_1, D_2}$, we can define an ${A}$-derivation on ${B_1 \otimes_A B_2}$ by

$\displaystyle D(b_1 \otimes b_2) = D( (b_1 \otimes 1)(1 \otimes b_2)) := (b_1 \otimes 1) D_2(b_2) + (1 \otimes b_2)D_1(b_1).$

In particular, we find that for any ${B_1 \otimes_A B_2}$-module ${M}$,

$\displaystyle \mathrm{Der}_A(B_1 \otimes_A B_2, M) \simeq \mathrm{Der}_A(B_1, M) \times \mathrm{Der}_A(B_2, M).$

In particular, we have a functorial isomorphism:

$\displaystyle \hom_{B_1 \otimes_A B_2}( \Omega_{B_1 \otimes_A B_2/A}, M) \simeq \hom_{B_1}( \Omega_{B_1 /A}, M) \times \hom_{B_2}( \Omega_{B_2 /A}, M).$

Since ${M}$ is a ${B_1 \otimes_A B_2}$-module as well, the last functor is naturally isomorphic to

$\displaystyle \hom_{B_1 \otimes_A B_2}( B_2\otimes_A \Omega_{B_1 /A} \oplus B_1 \otimes_A \Omega_{B_2 /A} , M) ,$

and since ${ B_2\otimes_A \Omega_{B_1 /A} \oplus B_1 \otimes_A \Omega_{B_2 /A}}$ belongs to the category of ${B_1 \otimes_A B_2}$-modules, Yoneda’s lemma completes the proof.

First exact sequence: chains of rings

Differentials behave well with respect to changing rings:

Theorem 1

Let ${A \rightarrow B \rightarrow C}$ be a sequence of homomorphisms of commutative rings. Then there is an exact sequence of ${C}$-modules:$\displaystyle C \otimes_B \Omega_{B/A} \rightarrow \Omega_{C/A} \rightarrow \Omega_{C/B} \rightarrow 0.$

The first map sends ${c \otimes db}$ to ${c d f(b)}$, where ${f : B \rightarrow C}$ is the ring-homomorphism. The second sends ${dc \rightarrow dc}$.

Anyway, to prove this, we’ll use a categorical argument. Let ${M}$ be a ${C}$-module. I claim there is an exact sequence

$\displaystyle 0 \rightarrow \mathrm{Der}_B(C, M) \rightarrow \mathrm{Der}_A(C,M) \rightarrow \mathrm{Der}_A(B,M) .$

Here the maps are the obvious one: a ${B}$-derivation of ${C}$ is automatically an ${A}$-derivation, and an ${A}$-derivation of ${C}$ restricts to an ${A}$-derivation of ${B}$. The image of ${\mathrm{Der}_B(C, M)}$ consists of derivations that vanish on ${B}$, so it is the kernel of the last map, which establishes exactness. This is an exact sequence

$\displaystyle 0 \rightarrow \hom_C( \Omega_{C/B}, M) \rightarrow \hom_C( \Omega_{C/A}, M) \rightarrow \hom_B(\Omega_{B/A}, M )$

or equivalently, an exact sequence

$\displaystyle 0 \rightarrow \hom_C( \Omega_{C/B}, M) \rightarrow \hom_C( \Omega_{C/A}, M) \rightarrow \hom_C(C \otimes_B \Omega_{B/A}, M )$

which as one may check, comes from the sequence defined above

$\displaystyle C \otimes_B \Omega_{B/A} \rightarrow \Omega_{C/A} \rightarrow \Omega_{C/B} \rightarrow 0$

which is therefore exact, ${M}$ being arbitrary.

The second exact sequence: quotients

Now suppose ${I \subset B}$ is an ideal in the ${A}$-algebra ${B}$.

Theorem 2

There is an exact sequence of ${A/I}$ modules$\displaystyle I/I^2 \rightarrow \Omega_{B/A}/I\Omega_{B/A} \rightarrow \Omega_{(B/I)/A} \rightarrow 0.$

Here the first map is ${i \rightarrow d(i)}$.

Note that ${I/I^2}$ is a ${B/I}$-module. Again, we shall prove this by abstract nonsense. Let ${M}$ be an ${B/I}$-module. Then, I claim there is an exact sequence

$\displaystyle 0 \rightarrow \mathrm{Der}_{A}(B/I, M) \rightarrow \mathrm{Der}_A(B, M) \rightarrow \hom_B(I/I^2, M).$

The first map here, ${\mathrm{Der}_{A}(B/I, M) \rightarrow \mathrm{Der}_A(B, M)}$, is the obvious pull-back of a derivation. The second is slightly more subtle: note that any derivation ${D}$ of ${B}$ in ${M}$ vanishes on ${I^2}$ because ${M}$ is annihilated by ${I}$. Also, the restriction ${D: I/I^2 \rightarrow M}$ is actually a ${B}$-homomorphism because

$\displaystyle D(bi) = b D(i) + iD(b) = b D(i)$

for the same reason. So, the sequence is defined. We will now prove it exact.

Exactness at the first step is clear. Also, it is clear that the composition of the last two maps is zero. Finally, any derivation of ${B}$ that restricts to ${0}$ on ${I}$ is a derivation of ${B/I}$, evidently.

Now, writing ${C=B/I}$, we see that this becomes an exact sequence, functorial in ${M}$,

$\displaystyle 0 \rightarrow \hom_{C}( \Omega_{C/A}, M) \rightarrow \hom_{C}( \Omega_{B/A}/I \Omega_{B/A}, M) \rightarrow \hom_{C}(I/I^2, M)$

which is seen to be induced by the sequence

$\displaystyle I/I^2 \rightarrow \Omega_{B/A}/I\Omega_{B/A} \rightarrow \Omega_{(B/I)/A} \rightarrow 0$

which is therefore exact.

Next time, I’ll say something about how differentials deal with localization, maybe something on separability, and then onto sheaves of differentials.