What is going on with this Bézout-like identity?

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So consider the ring $C_f := \overline{k}[x,y]/(f)$ where $f \in C_f$ an irreducible polynomial, and consider the localization of this ring $(C_f)_M$ where $M$ is the maximal generated by the images of $x-a$ and $y-b$.

Let's assume the maximal ideal of $(C_f)_M$ is generated by a single element, $z$. Since $M$ is generated by the images of $x$ and $y$ (translating $(a, b) \mapsto (0,0)$), we can define

$ux + vy = z$ for $u,v, \in (C_f)_M$

$x = zs$ for $s \in (C_f)_M$

$y = zr$ for $r \in (C_f)_M$.

Since $(C_f)_M$ a domain we can write $us + vr = 1$.

Now, how is it the case that, given this identity, either $s$ or $r$ is a unit in $(C_f)_M$ ?

Furthermore, how is it the case that, since we can write $rx -sy = 0$ in $(C_f)_M$, we can find a polynomial $s(x,y) \in \overline{k}[x,y]$ such that it has non-trivial constant term , and how is that related, if at all, to the first question?

EDIT: I did not make it clear, but also you need to select, without loss of generality, $s$ as the unit above.