Elements of $R=S^{-1} \mathbb{C}[ x,y ]$ I would like to understand how are the elements of $R=S^{-1} \mathbb{C}[ x,y ]$ where S is the multiplicative set generated by $\{x-a | a \in \mathbb{C} \} \cup \{y-a | a \in \mathbb{C} \}$.
I would like to understand which is the identity element and this kinds of things in order to try to prove if it is a field or not.
 A: We know $S=\langle \{x-a\hspace{0,1cm}|\hspace{0,1cm}a\in \mathbb{C}\} \cup \{y-a\hspace{0,1cm}|\hspace{0,1cm}a\in \mathbb{C}\} \rangle$, what means that if $s\in S$ then it can be written as $s=(x-a_1)\cdot ... \cdot (x-a_n)(y-b_1)\cdot ... \cdot (y-b_m)$ for $a_i, b_j \in \mathbb{C}$ and some $n,m$ positive integers.
Moreover, we know that $$R=S^{-1}\mathbb{C}[x,y] = \{ [f,s]\hspace{0,1cm}|\hspace{0,1cm} f\in \mathbb{C}[x,y] , s \in S \} / \sim$$
where $[f,s]\sim[f',s']$ if and only if $\exists s'' \in S$ such that $s''(s'f-sf')=0$. Since $s\in S$ implies $s\neq0$, we have that $[f,s]\sim[f',s']$ if and only if $(s'f-sf')=0$ if and only if $f/s =f'/s'$.
Knowing that, we need to see if all elements in $R$ have their inverse in $R$. This means $\forall [f,s] \in S^{-1} \mathbb{C}[x,y]$, $\exists [f',s'] \in S^{-1} \mathbb{C}[x,y]$ such that $[f,s]\cdot[f',s'] \sim  1_R $ where $1_R \sim [t,t]$ for any $t\in S$. This is because $S\subset \mathbb{C}[x,y]$ and we need to have the same element in order to have $t/t = 1_R$.
Now consider $[(x+1)(x+y),(x+1)] \in S^{-1} \mathbb{C}[x,y]$. This element is clearly in $R$ since $(x+1)$ has the form we have described before and $(x+1)(x+y) \in \mathbb{C}[x,y]$. The inverse of this element should be $[g,j]$ such that $g\in \mathbb{C}[x,y]$ and $j\in S$. It must satisfy the following expression: $$[(x+1)(x+y),(x+1)] \cdot [g,j] \sim 1_R$$
That means $(x+1)(x+y)gt=(x+1)jt$, $j=(x+y)g$ where $g\in \mathbb{C}[x,y]$. But $  \nexists j\in S$ such that $j=(x+y)g$ with $g\in  \mathbb{C}[x,y]$ since $j$ has the form described above.
The element $[(x+1)(x+y),(x+1)]$ has no inverse in $R$. Hence not all elements have their inverse in $R$. Hence $R$ is not a field.
