Let $GL(n, \mathbb{R})$ and $GL(n, \mathbb{C})$ denote, respectively, the algebraic groups of real and complex isomorphisms. I know that $GL(n, \mathbb{R})$ can be seen as a subgroup of $GL(n, \mathbb{C})$ but, can it be regarded as a complex algebraic subgroup of $GL(n, \mathbb{C})$? I think that the answer is no...
As $GL(n, \mathbb{C})$ is obtained through base extension from $GL(n, \mathbb{R})$ I know that I have a morphism $GL(n, \mathbb{C}) \to GL(n, \mathbb{R})$.
No. One way to see this is that a complex algebraic subgroup necessarily has a Lie algebra which is a complex Lie subalgebra of $\mathfrak{gl}_n(\mathbb{C})$, but the Lie algebra $\mathfrak{gl}_n(\mathbb{R})$ is not closed under scalar multiplication by complex numbers.
There is no such surjective morphism.