The Stacks project

Proof. By Lemma 53.19.5 we may change our system of coordinates and the choice of generators during the proof.

If (1) holds, then we may change coordinates such that $x_1, \ldots , x_{n - 2}$ map to zero in $A/I$ and $A/I = k[[x_{n - 1}, x_ n]]/(a x_{n - 1}^2 + b x_{n - 1}x_ n + c x_ n^2)$ for some nondegenerate quadric $a x_{n - 1}^2 + b x_{n - 1}x_ n + c x_ n^2$. Then we can explicitly compute to show that both (2) and (3) are true.

Assume the $(n - 1) \times (n - 1)$ minors of the matrix $(\partial f_ j/\partial x_ i)$ generate $I + \mathfrak m_ A$. Suppose that for some $i$ and $j$ the partial derivative $\partial f_ j/\partial x_ i$ is a unit in $A$. Then we may use the system of parameters $f_ j, x_1, \ldots , x_{i - 1}, \hat x_ i, x_{i + 1}, \ldots , x_ n$ and the generators $f_ j, f_1, \ldots , f_{j - 1}, \hat f_ j, f_{j + 1}, \ldots , f_ m$ of $I$. Then we get a regular system of parameters $x_1, \ldots , x_ n$ and generators $x_1, f_2, \ldots , f_ m$ of $I$. Next, we look for an $i \geq 2$ and $j \geq 2$ such that $\partial f_ j/\partial x_ i$ is a unit in $A$. If such a pair exists, then we can make a replacement as above and assume that we have a regular system of parameters $x_1, \ldots , x_ n$ and generators $x_1, x_2, f_3, \ldots , f_ m$ of $I$. Continuing, in finitely many steps we reach the situation where we have a regular system of parameters $x_1, \ldots , x_ n$ and generators $x_1, \ldots , x_ t, f_{t + 1}, \ldots , f_ m$ of $I$ such that $\partial f_ j/\partial x_ i \in \mathfrak m_ A$ for all $i, j \geq t + 1$.

In this case the matrix of partial derivatives has the following block shape

Hence every $(n - 1) \times (n - 1)$-minor is in $\mathfrak m_ A^{n - 1 - t}$. Note that $I \not= \mathfrak m_ A$ otherwise the ideal of minors would contain $1$. It follows that $n - 1 - t \leq 1$ because there is an element of $\mathfrak m_ A \setminus \mathfrak m_ A^2 + I$ (otherwise $I = \mathfrak m_ A$ by Nakayama). Thus $t \geq n - 2$. We have seen that $t \not= n$ above and similarly if $t = n - 1$, then there is an invertible $(n - 1) \times (n - 1)$-minor which is disallowed as well. Hence $t = n - 2$. Then $A/I$ is a quotient of $k[[x_{n - 1}, x_ n]]$ and Lemma 53.19.2 implies in both cases (2) and (3) that $I$ is generated by $x_1, \ldots , x_{n - 2}, f$ for some $f = f(x_{n - 1}, x_ n)$. In this case the condition on the minors exactly says that the quadratic term in $f$ is nondegenerate, i.e., $A/I$ is as in Lemma 53.19.3. $\square$