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 ${A}$. (more…)

I learned the material in this post from the book by Humphreys on Lie algebras and representation theory.

Recall that if ${A}$ is any algebra (not necessarily associative), then the derivations of ${A}$ form a Lie algebra ${Der(A)}$, and that if ${A}$ is actually a Lie algebra, then there is a homomorphism ${\mathrm{ad}: A \rightarrow Der(A)}$. In this case, the image of ${\mathrm{ad}}$ is said to consist of inner derivations.

Theorem 1 Any derivation of a semisimple Lie algebra ${\mathfrak{g}}$ is inner.

To see this, consider ${\mathrm{ad}: \mathfrak{g} \rightarrow D :=Der(\mathfrak{g})}$; by semisimplicity this is an injection. Let the image be ${D_i}$, the inner derivations. Next, I claim that ${[D, D_i] \subset D_i}$. Indeed, if ${\delta \in D}$ and ${\mathrm{ad} x \in D_i}$, we have

$\displaystyle [\delta, \mathrm{ad} x] y = \delta( [x,y]) - [x, \delta(y)] = [\delta(x),y] = (\mathrm{ad}(\delta(x)))y.$

In other words, ${[\delta, \mathrm{ad} x] = \mathrm{ad}(\delta(x))}$. This proves the claim.

Consider the Killing form ${B_D}$ on ${D}$ and the Killing form ${B_{D_i}}$ on ${D_i}$. The above claim and the definition as a trace shows that ${B_D|_{D_i \times D_i} = B_{D_i}}$. (more…)