**Proof.**
First assume that $k$ is algebraically closed. In this case the equivalence of (1) and (3) is trivial. The equivalence of (1) and (3) with (2) holds because the only nondegenerate quadric in two variables is $xy$ up to change in coordinates. The equivalence of (1) and (6) is Lemma 53.16.1. After replacing $X$ by an affine neighbourhood of $x$, we may assume there is a closed immersion $X \to \mathbf{A}^ n_ k$ mapping $x$ to $0$. Let $f_1, \ldots , f_ m \in k[x_1, \ldots , x_ n]$ be generators for the ideal $I$ of $X$ in $\mathbf{A}^ n_ k$. Then $\Omega _{X/k}$ corresponds to the $R = k[x_1, \ldots , x_ n]/I$-module $\Omega _{R/k}$ which has a presentation

\[ R^{\oplus m} \xrightarrow {(\partial f_ j/\partial x_ i)} R^{\oplus n} \to \Omega _{R/k} \to 0 \]

(See Algebra, Sections 10.131 and 10.134.) The first Fitting ideal of $\Omega _{R/k}$ is thus the ideal generated by the $(n - 1) \times (n - 1)$-minors of the matrix $(\partial f_ j/\partial x_ i)$. Hence (2), (4), (5) are equivalent by Lemma 53.19.6 applied to the completion of $k[x_1, \ldots , x_ n] \to R$ at the maximal ideal $(x_1, \ldots , x_ n)$.

Now assume $k$ is an arbitrary field. In cases (2), (4), (5), (6) the residue field $\kappa (x)$ is separable over $k$. Let us show this holds as well in cases (1) and (3). Namely, let $Z \subset X$ be the closed subscheme of $X$ defined by the first Fitting ideal of $\Omega _{X/k}$. The formation of $Z$ commutes with field extension (Divisors, Lemma 31.10.1). If (1) or (3) is true, then there exists a point $\overline{x}$ of $X_{\overline{k}}$ such that $\overline{x}$ is an isolated point of multiplicity $1$ of $Z_{\overline{k}}$ (as we have the equivalence of the conditions of the lemma over $\overline{k}$). In particular $Z_{\overline{x}}$ is geometrically reduced at $\overline{x}$ (because $\overline{k}$ is algebraically closed). Hence $Z$ is geometrically reduced at $x$ (Varieties, Lemma 33.6.6). In particular, $Z$ is reduced at $x$, hence $Z = \mathop{\mathrm{Spec}}(\kappa (x))$ in a neighbourhood of $x$ and $\kappa (x)$ is geometrically reduced over $k$. This means that $\kappa (x)/k$ is separable (Algebra, Lemma 10.44.1).

The argument of the previous paragraph shows that if (1) or (3) holds, then the first Fitting ideal of $\Omega _{X/k}$ generates $\mathfrak m_ x$. Since $\mathcal{O}_{X, x} \to \mathcal{O}_{X_{\overline{k}}, \overline{x}}$ is flat and since $\mathcal{O}_{X_{\overline{k}}, \overline{x}}$ is reduced and has depth $1$, we see that (4) and (5) hold (use Algebra, Lemmas 10.163.2 and 10.162.2). Conversely, (4) implies (5) by Algebra, Lemma 10.156.3. If (5) holds, then $Z$ is geometrically reduced at $x$ (because $\kappa (x)/k$ separable and $Z$ is $x$ in a neighbourhood). Hence $Z_{\overline{k}}$ is reduced at any point $\overline{x}$ of $X_{\overline{k}}$ lying over $x$. In other words, the first fitting ideal of $\Omega _{X_{\overline{k}}/\overline{k}}$ generates $\mathfrak m_{\overline{x}}$ in $\mathcal{O}_{X_{\overline{k}, \overline{x}}}$. Moreover, since $\mathcal{O}_{X, x} \to \mathcal{O}_{X_{\overline{k}}, \overline{x}}$ is flat we see that $\text{depth}(\mathcal{O}_{X_{\overline{k}}, \overline{x}}) = 1$ (see reference above). Hence (5) holds for $\overline{x} \in X_{\overline{k}}$ and we conclude that (3) holds (because of the equivalence over algebraically closed fields). In this way we see that (1), (3), (4), (5) are equivalent.

The equivalence of (2) and (6) follows from Lemma 53.19.4.

Finally, we prove the equivalence of (2) $=$ (6) with (1) $=$ (3) $=$ (4) $=$ (5). First we note that the geometric number of branches of $X$ at $x$ and the geometric number of branches of $X_{\overline{k}}$ at $\overline{x}$ are equal by Varieties, Lemma 33.39.2. We conclude from the information available to us at this point that in all cases this number is equal to $2$. On the other hand, in case (1) it is clear that $X$ is geometrically reduced at $x$, and hence

\[ \delta \text{-invariant of }X\text{ at }x \leq \delta \text{-invariant of }X_{\overline{k}}\text{ at }\overline{x} \]

by Varieties, Lemma 33.38.8. Since in case (1) the right hand side is $1$, this forces the $\delta $-invariant of $X$ at $x$ to be $1$ (because if it were zero, then $\mathcal{O}_{X, x}$ would be a discrete valuation ring by Varieties, Lemma 33.38.4 which is unibranch, a contradiction). Thus (5) holds. Conversely, if (2) $=$ (5) is true, then assumptions (a), (b), (c) of Varieties, Lemma 33.27.5 hold for $x \in X$ by Lemma 53.19.4. Thus Varieties, Lemma 33.38.9 applies and shows that we have equality in the above displayed inequality. We conclude that (5) holds for $\overline{x} \in X_{\overline{k}}$ and we are back in case (1) by the equivalence of the conditions over an algebraically closed field.
$\square$

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