# Inner automorphisms of algebraic groups

I'm confused about the precise definition of an inner automorphism of an algebraic group. Here is what Milne says in his book on algebraic groups:

Let $k$ be a field, let $\overline{k}$ be an algebraic closure, and let $G$ be an algebraic group over $k$. Let $Z$ (or $Z(G)$) denote the center of $G$. Then an automorphism of $G$ is said to be inner if it becomes of the form inn$(g)$ over $\overline{k}$, where $g \in G(\overline{k})$. The action of $G$ on itself by inner automorphisms, $(x,y) \mapsto xyx^{-1}$, is invariant under $Z(G) \times e$ acting by translation, and so it factors through $G/Z(G)$. The inner automorphisms of $G$ are exactly the automorphisms of $G$ defined by elements of $G/Z(k)$.

I don't understand why we don't simply say that the inner automorphisms of $G$ can be identified with $G/Z$. I think Milne is identifying $G/Z(k)$ with a subset of the underlying topological space $G/Z$.

• I don't have the print copy of Milne to hand, and I can't find this in the last edition of the online notes that he posted. The point is that $G/Z$ has points valued over all field extensions over $k$ (and more), and, if we want to regard these as inducing automorphisms of $G$ (regarded as a $k$-group), then they had better preserve $G(k)$. An arbitrary-field-valued point of $G/Z$ doesn't behave this way, but a $k$-valued point $(G/Z)(k)$, which may lift only to a point of $G(\overline k)$, does. Is that related to your question? – LSpice Sep 7 '18 at 1:28
• Inner automorphisms are exactly the $k$-points of $G/Z$, the definition in terms of points over $\bar k$ is just an explanation how the points of the quotient look like. – Victor Petrov Sep 7 '18 at 17:13
• When they make that identification, they are regarding $\mathrm{Inn}(G)$ as the identity component of the algebraic group $\mathrm{Aut}(G)$ of algebraic automorphisms—that is, in particular, as an algebraic group—whereas Milne is regarding the inner automorphism group as an abstract group (as @anon points out below). – LSpice Sep 9 '18 at 23:56
• In particular, Milne's inner automorphism group can change after base change, whereas Borel's can't; and, more precisely, we have $\mathrm{Inn}_{\text{Milne}}(G) = (\mathrm{Inn}_{\text{Borel}}(G))(k)$, or we could perhaps even write more explicitly $\mathrm{Inn}_{\text{Milne}}(G_{/k}) = (\mathrm{Inn}_{\text{Borel}}(G_{/k}))(k)$. – LSpice Sep 9 '18 at 23:57
• @LSpice Shouldn't $\mathrm{Aut}(\mathbb{G}_m^n)=\mathrm{GL}_n(\mathbb{Z})$? – Not a grad student Nov 9 '18 at 20:14

The inner automorphisms of $G$ form an abstract group, whereas $G/Z$ is an algebraic group (i.e., group scheme of finite type over the field $k$), so you can't say that one is equal to the other --- they are different types of objects. By $(G/Z)(k)$ Milne means the group of $k$-rational points of $G/Z$, which is an abstract group. Each element of $(G/Z)(k)$ defines an automorphism of $G$, and these are exactly the inner automorphisms
Here is the passage in Milne's book 3.51: Let $G$ be an algebraic group over $k$. An automorphism of $G$ is inner if it becomes of the form $inn(g)$ over $k^{\mathrm{a}}$. Later (17.63), we shall see that the group $\mathcal{I}(k)$ of inner automorphisms of $G$ is equal to $\bar{G}(k)$, where $\bar{G}$ is the quotient of $G$ by its centre. For example, for $t\in k^{\times}$, the automorphism $\left( \begin{smallmatrix} a & b\\ c & d \end{smallmatrix} \right) \mapsto\left( \begin{smallmatrix} a & tb\\ t^{-1}c & d \end{smallmatrix} \right)$ of $SL_{2}$ is inner because it becomes the inner automorphism defined by $diag(\sqrt{t},\sqrt{t}\,^{-1})$ over $k^{\mathrm{a}}$. It is also the inner automorphism defined by the element $diag(t,1)$ of $PGL_{2}(k)$.