$PGL(n)$ is a closed subgroup of $GL(n^2)$

Question

Assume the base field is complex numbers. Let $PGL(n)$ be the projective linear group. I heard it is a closed algebraic subgroup of $GL(n^2)$, but I am having trouble proving it.

What I tried is embedding a matrix in $PGL(n)$ using the ratios of its entries, but this works bad when there is $0$ in the matrix.

Similar question has been asked here.How to realise $\mathrm{PGL}_2$ as a closed subgroup of some $\mathrm{GL}_n$ explicitly?

Thanks for the help.

Answer 1

This works over every field $k$.

Lemma: the natural map $PGL_n\to Aut(M_n)$ given by $A\mapsto (M\mapsto AMA^{-1})$ is a group isomorphism (we are considering the automorphism of $M_n$ as a $k$-algebra, not just a vector space of course).

Proof: Skolem-Noether theorem. If you want I might be able to prove this more directly.

Now consider the inclusion $Aut(M_n)\subseteq GL_{n^2}$ given by writing every automorphism in matrix form, using the standard basis $E_{ij}$ of $M_n$. This makes $Aut{(M_n)}$ into a closed subset, since the conditions of multiplicativity are given by polynomial equations.