Let $k$ be a field. The ring $M_n(k)$ of $n\times n$ matrices over $k$ has some automorphisms, given by conjugation by elements of $\GL_n(k)$. These are inner automorphisms, and this action happens to be the adjoint action of $\GL_n$ on its Lie algebra. Are there any other automorphisms? The answer is no, and the reason fits in the framework of automorphisms of central separable algebras. In this post we'll sketch how this works.

Let $R$ be a commutative ring. An $R$ algebra $A$ is called **separable** if $A$ is projective as an $A\otimes_R A^{\rm op}$-module and $A$ is called **central separable** if it is separable and the center of $A$ coincides with $R$.

A typical example is a matrix ring $M_n(R)$ over $R$. In fact, whenever $M$ is a finitely generated projective and faithful module then ${\rm Hom}_R(M,M)$ is a central separable $R$-algebra. So even if you don't have much intuition for central separable algebras in general, everything we say about them will certainly apply to algebras of the form ${\rm Hom}_R(M,M)$.

For a commutative ring, let's call a finitely generated projective module of constant rank one a **line bundle**. The **Picard group** of a commutative ring $R$ is the group of isomorphism classes of line bundles. The group operation is the tensor product. If $A$ is a central separable algebra, then $A^A = \{ b\in A : ab = ba \forall a\in A\} = R$ is a trivial example of a line bundle.

More generally, $(A_\sigma)^A = \{ b\in A : ab = b\sigma(a)\forall a\in A\}$ is also a line bundle. This gives a well-defined map

$$

{\rm Aut}_R(A)\to {\rm Pic}(R)

$$

that happens to fit into an exact sequence

$$

1\to {\rm Inn}_R(A)\to {\rm Aut}_R(A)\to {\rm Pic}(R).

$$

Consequently, if ${\rm Pic}(R)$ is trivial, then every $R$-algebra automorphism of $A$ must in fact be inner. For example, this works for fields and principal ideal domains, and hence answers our query that we started with: why is every algebra automorphism of a full matrix ring inner? The construction of the exact sequence, which I will not explain at this point, actually gives a somewhat constructive answer: if $\sigma$ is a given automorphism then $(A_\sigma)^A$ is a free $R$-module on a generator $v$ that is a unit in $A$, and $\sigma(x) = v^{-1}xv$.