I'm trying to find the toric variety associated to the cone $\sigma_0$ which is the region in the real plane with $x\geq 0$ and $y-x\geq 0.$ I found that it's dual cone is $\check{\sigma_0}$ the region in the real plane with $y+x\geq 0$ and $y\geq 0.$ So we have a commutative semi group $S_{\sigma_0} = \check{\sigma_0} \cap \mathbb{Z}^2$, with generators $(-1,1) $ and $(0,1).$ But there doesn't seem to be any $\mathbb{N}$ linear relations between these generators. The variety associated to the cone is $\operatorname{Spec} (\mathbb{C}[S_{\sigma_0}])$ and in Fulton's Toric Varieties book on page 6, it states that
$$ \operatorname{Spec} (\mathbb{C}[S_{\sigma_0}]) = \operatorname{Spec}( \mathbb{C}[X,X^{-1}Y] ) = \mathbb{C}^2. $$
How does he obtain both of these inequalities? I can't even see the second one. I find the notation $\mathbb{C}[X,X^{-1}Y]$ to be confusing, for example I would prefer that instead of $\mathbb{C}[X,X^{-1}]$ we write instead $\mathbb{C}[X,Y]/(XY-1)$ because it makes it clearer to make that its Spec is the set of points $(x,y)\in\mathbb{C}^2$ such that $xy=1.$ So how do we do the same thing for $\mathbb{C}[X,X^{-1}Y]$?
First, you've made a small mistake with your second generator for $S_{\sigma_0};$ it should be $(1,0)$ rather than $(0,1).$ Thus the monoid algebra generated by this monoid is $\mathbb C[\chi^{(1,0)},\chi^{(-1,1)}] =\mathbb C[X,X^{-1}Y].$
We can think of the ring $\mathbb C[X,X^{-1}Y]$ as the subring of the Laurent polynomials $\mathbb C[X,X^{-1},Y,Y^{-1}]$ generated by $X$ and $X^{-1}Y,$ and show that $\mathbb C[X,X^{-1}Y]$ is isomorphic to the polynomial ring $\mathbb C[U,V]$ by the obvious map $U\mapsto X,V\mapsto X^{-1}Y.$
(In general, we begin with a cone contained in a lattice $\mathbb Z^n,$ so we always think of these monoid algebras as subrings of some $\mathbb C[X_1,X_1^{-1},\ldots,X_n,X_n^{-1}].$ It is not always trivial to see what the relations amongst generators will be!)
Once you are comfortable with this isomorphism (which interestingly describes one chart of the blowup of $\mathbb A^2$ at the origin), the equality $\mathbb A^2 = \operatorname{Spec}(\mathbb C[X,X^{-1}Y])$ is trivial.