Finding the completion of a coordinate ring Consider $A=\mathbb C[x,y]/(y^2-x(x+1))$, and consider the $\mathfrak m$-adic completion, where $\mathfrak m =(x,y)$. I want to show that this completion is isomorphic to $\mathbb C[[u,v]]/(uv)$, where the double brackets denote a ring of formal power series. 
I think I've figured out the trick, but I'm unable to complete the argument. I believe the completion of $A$ is just $\mathbb C[[x,y]]/(y^2-x^2(x+1))$. The "trick" is to note that we can factor the generator of the ideal in the power series ring as $(y-x\sqrt{x+1})(y+x\sqrt{x+1})$, because we can expand the square roots into power series. 
Consider the map $\mathbb C[[u,v]]\rightarrow \mathbb C[[x,y]]$ with 
$$\phi(u)=y+x\sqrt{x+1}$$
$$\phi(v)=y-x\sqrt{x+1}.$$
If we can show this is surjective, we can compose it with the quotient map to $\mathbb C[[x,y]]/(y^2-x^2(x+1))$. If we then show the kernel is $(uv)$, we are done by the standard ring isomorphism theorem. 
So my two questions are: How do we show surjectivity? And how do we compute the kernel after we compose with the quotient map? Instead of computing the kernel, we could note the map descends to a map on $\mathbb C[[u,v]]/(uv)$ to $\mathbb C[[x,y]]/(y^2-x^2(x+1))$ and show this is injective (possibly by finding an inverse?), but I don't see how to do this.
For surjectivity, we just need to show that $y$ and $x$ are in the image. We can get $y$ by $(u+v)/2$, but getting $x$ is going to require a power series that I don't see how to compute. 
I realize this has some geometric interpretation, but I'm looking for purely algebraic and elementary answers, as I'm new to this material. Thanks.
 A: Suppose you are given two power series without constant term $f_1,f_2 \in F[[x,y]]$ (for a field $F$). The map
$$ \sum a_{ij} x^i y^j \mapsto \sum a_{ij} f_1^i f_2^j$$ is then a well-defined $F$-algebra homomorphism (we need the constant terms to be zero to get formal convergence). It is an isomorphism if and only if the Jacobian determinant of $(f_1,f_2)$ is a unit in $F[[x,y]]$ (i.e., has non-zero constant term). Or, writing
$$f_1=a x+b y+\cdots, \quad f_2=cx+dy+\cdots$$ if and only if $ad-bc \neq 0$. 
In your case, the relevant expansions are
$$\phi(u)=x+y+\cdots \quad \text{and} \quad \phi(v)=-x+y+\cdots $$ so the Jacobian determinant at zero is $2$ and the map is an isomorphism.
[1. This is just a formal analytic version of the implicit function theorem; the unproved assertions are straightforward algebra. 
2. All of this is happening at the level of the formal power series rings without quotienting, but by your wise choice the element $uv$ corresponds to $y^2-x^2(x+1)$ so it descends to the appropriate quotients.]
