[This was intended to be comment to Ben's reply but I exceeded the allowable limit for comments.] Actually it doesn't work over any ring. Just take any ring $R$ for which $GL(V)(R) \to PGL(V)(R)$ is not surjective. This exists if $R$ has a non-trivial rank $1$ projective $L$ such that $L^n$ is isomorphic to $R^n$. R^n$, $n:=\dim V$. Then $V\bigotimes L$ is an $End(V)\bigotimes R$ module which is not isomorphic to $V$ though they are both indecomposable projective $End(V)\bigotimes R$-modules (rarely irreducible though). The factor As $R$-module $V\bigotimes L \cong L^n\cong R^n\cong V\bigotimes R$ so that the sum we get a different action of $n$ copies of them are isomorphic gives End(V)\bigotimes R$, i.e., an automorphism of $End(V)\bigotimes R \to End(V)\bigotimes R$ (which can be taken as the definition of the $R$-points of $PGL(V)$) that does is not lift to an given by conjugation by some element of $GL(V\bigotimes R)$. As projective modules over For $R$ a local ring such an $L$ can not exists as all projective modules are free the $GL(V)(R) \to PGL(V)(R)$ is surjective when and there one does indeed get that every automorphism of $End(V)\bigotimes R$ is local which given by conjugation by an element of $GL(V)(R)$. This is enough to show that $GL(V) \to PGL(V)$ is surjective as a map the automorphism group scheme of algebraic groups.$End(V)$.
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[This was intended to be comment to Ben's reply but I exceeded the allowable limit for comments.] Actually it doesn't work over any ring. Just take any ring $R$ for which $GL(V)(R) \to PGL(V)(R)$ is not surjective. This exists if $R$ has a non-trivial rank $1$ projective $L$ such that $L^n$ is isomorphic to $R^n$. Then $V\bigotimes L$ is an $End(V)\bigotimes R$ module which is not isomorphic to $V$ though they are both indecomposable projective $End(V)\bigotimes R$-modules (rarely irreducible though). The factor that the sum of $n$ copies of them are isomorphic gives an automorphism of $End(V)\bigotimes R$ (which can be taken as the definition of the $R$-points of $PGL(V)$) that does not lift to an element of $GL(V\bigotimes R)$. As projective modules over a local ring are free the $GL(V)(R) \to PGL(V)(R)$ is surjective when $R$ is local which is enough to show that $GL(V) \to PGL(V)$ is surjective as a map of algebraic groups. |
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