Isomorphism of formal power series factorrings over polynomials This problem is taken from the Hartshorne's book Algebraic Geometry, Chapter 1, Section 5, Problem 14(a). 
Two polynomials $f(x,y)$ and $g(x,y)$ are written in the form
$$f(x,y) = f_{r}(x,y) + f_{r+1}(x,y) + \cdots + f_{m}(x,y)$$
$$g(x,y) = g_{s}(x,y) + g_{k+1}(x,y) + \cdots + g_{n}(x,y)$$
such that $f_{i}(x,y)$ and $g_{j}(x,y)$, $r \leq i \leq m$, $s \leq j \leq n$, are homogeneous polynomials of degree $i$ and $j$ respectively, and $f_{r}, g_{s} \neq 0$. So $f(x,y)$ (resp. $g(x,y)$) contains no monomials of the degree smaller than $r$ (resp. $s$). Let $k[[x,y]]$ denote the ring of formal power series of variables $x,y$, char $k = 0$. 

Prove that if the factor rings $k[[x,y]]/(f)$ and $k[[x,y]]/(g)$ are isomorphic as $k$-algebras, then $r=s$.

Please give me some hint or link, where there is a proof of this statement.
Thank you in advance!
 A: Obviously if $f$ is a unit, $g$ is a unit, and vice versa.  Also, $r=1$ iff the embedding dimension of $k[[x,y]]/(f)$ is $1$ iff the embedding dimension of $k[[x,y]]/(g)$ is $1$ iff $s=1$.  
So, we may assume both $r$ and $s$ are greater than $2$, that is, $f,g \in \mathfrak{m}^2$, where $\mathfrak{m}$ denotes the maximal ideal $(x,y)$. Since $f,g \in \mathfrak{m}^2$, we may lift this isomorphism to a $k$-algebra automorphism $\phi$ of $k[[x,y]]$ mapping $(f)$ to $(g)$.  It isn't obvious that this is possible, but it is: it follows from the theorem on page 8 of these notes from Hochster on complete local rings.
To be precise, say $\psi$ is a $k$-algebra isomorphism between $k[[x,y]]/(f)$ and $k[[x,y]]/(g)$.  $\psi(x)$ and $\psi(y)$ generate the maximal ideal of $k[[x,y]]/(g)$.  Thus, since $g \in \mathfrak{m}^2$, we have $\psi(x)$ and $\psi(y)$ generate $\mathfrak{m}/(g) / \mathfrak{m}^2/(g) \cong  \mathfrak{m}/\mathfrak{m}^2$.  This means $\psi(x)$ and $\psi(y)$ lift to generators of $\mathfrak{m}$ in $k[[x,y]]$ (by NAK).
Applying parts (b) and (c) of the Theorem on page 8 of Hochster's notes, it follows that $\psi$ lifts to a $k$-algebra automorphism $\phi$ of $k[[x,y]]$.  We know $\phi$ maps $f$ to a generator of $(g)$, since $\phi((f))=(g)$.  So, after multiplying $\phi$ by a unit, we may assume $\phi(f)=g$.  
$\phi$ is determined by $\phi(x)$ and $\phi(y)$; let's use the Krull Intersection Theorem to see this.  The claim is that if $x \mapsto p$ and $y \mapsto q$ under $\phi$, then any power series $h(x,y)$ in $k[[x,y]]$ must map to $h(p, q)$.  First of all, note that this composition is well-defined, since $p$ and $q$ have $0$ constant term.
To prove the claim, let $h_i(x,y)$ denote the truncation of $h$ to the $i^{th}$ degree; that is, $h_i$ is the sum of all the terms of $h$ which have degree at most $i$.  It is clearly the case that $\phi(h_i(x,y))=h_i(p,q)$ for all $i$.  Thus, for each $i \ge 1$, $\phi(h(x,y))=h_i(p,q) + \text{some element of } \mathfrak{m}^{i+1}$.  This means that, for each $i$, 
$$\phi(h(x,y)) - h(p,q)= h_i(p,q) + \text{some element of } \mathfrak{m}^{i+1} - h_i(p,q) + \text{some other element of } \mathfrak{m}^{i+1}$$
$$\in \mathfrak{m}^{i+1}.$$
By the Krull Intersection Theorem, this implies that $\phi(h(x,y)) = h(p,q)$.
In particular, we have $g(x,y)=f(\phi(x), \phi(y))$.  We know $\phi(x)$ and $\phi(y)$ are in $\mathfrak{m}$, so the lowest degree "piece" of $g(x,y)$ can't have smaller degree than the lowest degree "piece" of $f(x,y)$. Thus, $s \ge r$.
But we can apply the same argument using $\phi^{-1}$ to show that $s \le r$.  Thus, $r=s$.
