Let me define a linear algebraic group first of all:
A linear algebraic group is an affine variety (zeroes of a set of polynomials in $n$ variables with coefficients in a field $C$) such that the maps $$\phi : G \times G \to G $$ $$(x,y) \mapsto xy$$
and $$\psi : G \to G$$ $$x \mapsto x^{-1}$$
are morphisms of varieties.
As I understand (please correct if I am wrong), $G$ contains points in $C^n$ as well as $K^n$ where $K$ is a field extension of $C$.
Now, G acts on iteslf by right multiplication : Let $g \in G$ be a $C$-point ,i.e, $g \in C^n$ and consider the map
$$\rho_{g} : G \to G $$ $$h \mapsto hg$$
This map is ofcourse a bijection. I do not see that it as a homomorphism though. (please correct if I am wrong).
Now, I have read in a few books that this map $\rho_{g}$ induces an $C$-algebra automorphism :
$$\rho^{*}_g : C[G] \to C[G]$$ $$ f \mapsto f.\rho_g$$
Note that $C[G]$ is the coordinate ring of $G$, i.e, $C[G]= C[x_1,...,x_n]/I(G) $ where $I(G)$ is the ideal consisting of all polynomials satisfied by $G$.
Now, I have few doubts which I state below:
Since $g$ is a $C$-point, I can realise that $f.\rho_g$ indeed is in $C[G]$ but what if $g$ is not a $C$-point. What happens then ?
I have checked that $\rho^{*}_g$ is an additive homorphism and a bijection. But I am failing to realise it as a ring homomorphism. Please help me regarding this.
Thank you !
If $g$ is a $K$-point then it gives an automorphism of $K[G]$.
This is equivalent to the claim that the corresponding map $G \to G$ is a map of varieties, which is true because the multiplication map is a map of varieties.