Lemma 53.19.3. Let $k$ be a field. Let $(A, \mathfrak m, \kappa )$ be a Noetherian local $k$-algebra. The following are equivalent

1. $\kappa /k$ is separable, $A$ is reduced, $\dim _\kappa (\mathfrak m/\mathfrak m^2) = 2$, and there exists a nondegenerate $q \in \text{Sym}^2_\kappa (\mathfrak m/\mathfrak m^2)$ which maps to zero in $\mathfrak m^2/\mathfrak m^3$,

2. $\kappa /k$ is separable, $\text{depth}(A) = 1$, $\dim _\kappa (\mathfrak m/\mathfrak m^2) = 2$, and there exists a nondegenerate $q \in \text{Sym}^2_\kappa (\mathfrak m/\mathfrak m^2)$ which maps to zero in $\mathfrak m^2/\mathfrak m^3$,

3. $\kappa /k$ is separable, $A^\wedge \cong \kappa [[x, y]]/(ax^2 + bxy + cy^2)$ as a $k$-algebra where $ax^2 + bxy + cy^2$ is a nondegenerate quadratic form over $\kappa$.

Proof. Assume (3). Then $A^\wedge$ is reduced because $ax^2 + bxy + cy^2$ is either irreducible or a product of two nonassociated prime elements. Hence $A \subset A^\wedge$ is reduced. It follows that (1) is true.

Assume (1). Then $A$ cannot be Artinian, since it would not be reduced because $\mathfrak m \not= (0)$. Hence $\dim (A) \geq 1$, hence $\text{depth}(A) \geq 1$ by Algebra, Lemma 10.157.3. On the other hand $\dim (A) = 2$ implies $A$ is regular which contradicts the existence of $q$ by Algebra, Lemma 10.106.1. Thus $\dim (A) \leq 1$ and we conclude $\text{depth}(A) = 1$ by Algebra, Lemma 10.72.3. It follows that (2) is true.

Assume (2). Since the depth of $A$ is the same as the depth of $A^\wedge$ (More on Algebra, Lemma 15.43.2) and since the other conditions are insensitive to completion, we may assume that $A$ is complete. Choose $\kappa \to A$ as in More on Algebra, Lemma 15.38.3. Since $\dim _\kappa (\mathfrak m/\mathfrak m^2) = 2$ we can choose $x_0, y_0 \in \mathfrak m$ which map to a basis. We obtain a continuous $\kappa$-algebra map

$\kappa [[x, y]] \longrightarrow A$

by the rules $x \mapsto x_0$ and $y \mapsto y_0$. Let $q$ be the class of $ax_0^2 + bx_0y_0 + cy_0^2$ in $\text{Sym}^2_\kappa (\mathfrak m/\mathfrak m^2)$. Write $Q(x, y) = ax^2 + bxy + cy^2$ viewed as a polynomial in two variables. Then we see that

$Q(x_0, y_0) = ax_0^2 + bx_0y_0 + cy_0^2 = \sum \nolimits _{i + j = 3} a_{ij} x_0^ iy_0^ j$

for some $a_{ij}$ in $A$. We want to prove that we can increase the order of vanishing by changing our choice of $x_0$, $y_0$. Suppose that $x_1, y_1 \in \mathfrak m^2$. Then

$Q(x_0 + x_1, y_0 + y_1) = Q(x_0, y_0) + (2ax_0 + by_0)x_1 + (bx_0 + 2cy_0)y_1 \bmod \mathfrak m^4$

Nondegeneracy of $Q$ means exactly that $2ax_0 + by_0$ and $bx_0 + 2cy_0$ are a $\kappa$-basis for $\mathfrak m/\mathfrak m^2$, see discussion preceding the lemma. Hence we can certainly choose $x_1, y_1 \in \mathfrak m^2$ such that $Q(x_0 + x_1, y_0 + y_1) \in \mathfrak m^4$. Continuing in this fashion by induction we can find $x_ i, y_ i \in \mathfrak m^{i + 1}$ such that

$Q(x_0 + x_1 + \ldots + x_ n, y_0 + y_1 + \ldots + y_ n) \in \mathfrak m^{n + 3}$

Since $A$ is complete we can set $x_\infty = \sum x_ i$ and $y_\infty = \sum y_ i$ and we can consider the map $\kappa [[x, y]] \longrightarrow A$ sending $x$ to $x_\infty$ and $y$ to $y_\infty$. This map induces a surjection $\kappa [[x, y]]/(Q) \longrightarrow A$ by Algebra, Lemma 10.96.1. By Lemma 53.19.2 the kernel of $k[[x, y]] \to A$ is principal. But the kernel cannot contain a proper divisor of $Q$ as such a divisor would have degree $1$ in $x, y$ and this would contradict $\dim (\mathfrak m/\mathfrak m^2) = 2$. Hence $Q$ generates the kernel as desired. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).