Find all power series $f$ such that $\mathbb{C}[[x, y]]/(f(x, y))\cong \mathbb{C}[[x, y]]/(xy)$ Find all power series $f$ such that $\mathbb{C}[[x, y]]/(f(x, y)) \cong  \mathbb{C}[[x, y]]/(xy)$. The isomorphism $\phi$ should be identical on $\mathbb{C}$, i.e., $\forall c \in \mathbb{C} \subseteq \mathbb{C}[[x, y]]/(f(x, y))$, $\phi(c) = c$.
This problem is the third problem of the 2018 Alibaba math contest (final) —— Algebra track. It is not an ongoing contest.
My only attempt is to show the power series of the Right Hand Side (RHS) are in the form  $\sum\limits_{i = 0}a_ix^i + \sum\limits_{j = 0}b_iy^j (xy = 0)$ but I do not know how to proceed. Does this problem require some advanced theories or theorems? Is the complex field $\mathbb{C}$ anything special?
 A: $\mathbb{C}[[x, y]]$ is a local ring with unique maximal ideal $m = (x, y)$, hence so is any quotient of it, including both $\mathbb{C}[[x, y]]/(f)$ and $\mathbb{C}[[x, y]]/(xy)$. The quotient $m/m^2$ is a $2$-dimensional vector space with basis $\{ x, y \}$. Any isomorphism
$$\varphi : \mathbb{C}[[x, y]]/(xy) \to \mathbb{C}[[x, y]]/(f)$$
must send $m$ to $m$ and must induce an isomorphism $m/m^2 \cong m/m^2$; it follows that $\varphi(x)$ and $\varphi(y)$ must have the property that $\bmod m^2$ they form a basis of $m/m^2$. Up to a linear change of coordinates we can assume WLOG that in fact $\varphi(x) \equiv x \bmod m^2$ and $\varphi(y) \equiv y \bmod m^2$; write
$$\varphi(x) = x + a(x, y), \varphi(y) = y + b(x, y)$$
where $a, b$ have only quadratic or higher terms; we will be thinking of them as power series in $\mathbb{C}[[x, y]]$ as well as elements of the quotient $\mathbb{C}[[x, y]]/(f)$ (by picking arbitrary lifts). Now we come to the key technical argument:

Lemma: Any homomorphism $\phi : \mathbb{C}[[x, y]] \to \mathbb{C}[[x, y]]$ of the form $\phi(x) = x + a, \phi(y) = y + b$ where $a, b \in m^2$ is an isomorphism.

Proof. By construction $\phi$ sends $m^k$ to $m^k$ (the set of polynomials whose lowest degree term has degree at least $k$) and is the identity $\bmod m^{k+1}$ (since $\bmod m^{k+1}$ the contributions of $a$ and $b$ disappear); hence it induces an isomorphism on $m^k/m^{k+1}$ for all $k$, from which it follows that it induces an isomorphism $\bmod m^k$ for all $k$ and hence (because $\mathbb{C}[[x, y]]$ is its own $m$-adic completion) that $\phi$ is an isomorphism. $\Box$
It follows that $\phi$ as defined above, using any choice of lifts $a, b \in \mathbb{C}[[x, y]]$, is an automorphism of $\mathbb{C}[[x, y]]$ lifting $\varphi$. Hence it must send the ideal $(xy)$ to the ideal $(f)$, from which it follows that $f$ is equal to $(x + a)(y + b)$ times a unit.
Conversely, every $f$ of this form occurs, since given arbitrary $a, b \in m^2$ we can construct the isomorphism $\phi$ as above, and since units are just elements with nonzero constant term we can absorb the unit into either $x + a$ or $y + b$. Undoing the WLOG-change-of-coordinates above, we come to the following conclusion:

$\mathbb{C}[[x, y]]/(xy) \cong \mathbb{C}[[x, y]/(f)$ iff $f$ factors as a product $gh$ where $g, h \in m$ and $g, h$ are linearly independent in $m/m^2$.

There is a geometric picture here which explains the meaning of this condition. $\text{Spf } \mathbb{C}[[x, y]]/(xy)$ is a formal scheme describing a formal neighborhood of the origin in $\text{Spec } \mathbb{C}[x, y]/(xy)$ which is the union of the $x$ and $y$-axes in the affine plane $\mathbb{A}^2$. This formal neighborhood is sensitive to the nodal singularity caused by the crossing of the two axes; in particular it's sensitive to the fact that there are two tangent directions, one in the $x$-direction and one in the $y$-direction (that's what all the $m/m^2$ business was about; $m/m^2$ is the dual of the tangent space of the origin).
Similarly $\text{Spf } \mathbb{C}[[x, y]/(f)$ describes a formal neighborhood of the origin in the curve $\{ f(x, y) = 0 \}$. So we expect intuitively that this is isomorphic to $\text{Spf } \mathbb{C}[[x, y]]/(xy)$ iff $\{ f(x, y) = 0 \}$ has the same singularity behavior at the origin as $\{ xy = 0 \}$ does; namely, two curves meeting at the origin but not tangent there, each individually nonsingular at the origin. These two curves are cut out by the two factors $x + a, y + b$ or $g, h$ we constructed above.
