I am reading Toric Variety now and as a well documented example we know $\mathbb{C^s}$ is toric variety with torus $\mathbb(C^*)^s$. I am wondering how will the action of $\mathbb(C^*)^s$ will look like on $\mathbb C^s$. I think it will look like the multiplication map, that is F:$\mathbb(C^*)^s \times \mathbb{C^s} \rightarrow \mathbb C^s $, will be $F(t,s)=t.s$. Can anyone tell me if this intuition is right?
For proof I am thinking Let us assume that this statement is not true. since it's a morphism, then we can consider the map is $F=(F_1,F_2,....F_s)$, now for a fixed $t \in \mathbb (C^*)^s $, we will get a polynomial map from $\mathbb C^s \rightarrow C^s$. Now let this map is $F_1=(F_{11},F_{12},...F_{1s})$. Now, these $F_{1i}(x)$ will be $t_ix_i$, $\forall x \in \mathbb (C^*)^s$, where $t=(t_1,t_2,....t_s),x=(x_1,x_2,....x_s)$ then consider $F_{1i}-t_ix_i$, which will be a polynomial on $C[x_1,x_2,....x_s]$, which will be zero in $\mathbb (C^*)^s$ and non zero at the origin , which will make $\mathbb (C^*)^s$ a closed subset of $\mathbb {C^s}$. Which is a contradriction, so we can say that for any fixed $t$ $F_1(x)=t.x$, hence $\forall t \in \mathbb (C^*)^s$ and $\forall x \in \mathbb C^s$ , $F(t,x)=t.x$. Is this argument all right? It will be great if you can tell me. Any help will be appreciated.