Let $G$ be a linear algebraic group and let $C[G] = C[x_1,...,x_n] / I(G)$ denote the coordinate ring of $G$.
(Note that $I(G)$ is the ideal containing all those polynomials in $C[x_1,...,x_n]$ of which elements of $G$ are the common zeroes.)
Now, we the action of $C$- points of $G$ on $C[G]$ is defined by : $$ G_C \times C[G] \to C[G]$$ $$(x,f(y)) \mapsto f(x^{-1}y) $$
I am not able to see the condition $x_1.(x_2.f) = (x_1.x_2).f$ being satisfied by this. Please help me in realising that this indeed is a group action. Thank you !
When evaluating $x_1\cdot(x_2\cdot f)(y)$, it's tempting to write $x_1\cdot(f(x_2^{-1}y))$ which I think is where the confusion comes from, but this doesn't make sense as $f(x_2^{-1}y)$ is a scalar. Instead, we have $x_2\cdot f\in C[G]$, so letting $x_1$ act on this function, $$ (x_1\cdot(x_2\cdot f))(y)=(x_2\cdot f)(x_1^{-1}y)=f(x_2^{-1}x_1^{-1}y)=f((x_1x_2)^{-1}y)=((x_1x_2)\cdot f)(y) $$ hence $x_1\cdot(x_2\cdot f)=(x_1x_2)\cdot f$.