Proof.
Proof of (1). Consider the derivations $\partial /\partial x_ i$ of $k[[x_1, \ldots , x_ n]]$ over $k$. Write $f_ i = \partial f/\partial x_ i$. The derivation
\[ \theta = \sum h_ i \partial /\partial x_ i \]
of $k[[x_1, \ldots , x_ n]]$ induces a derivation of $P = k[[x_1, \ldots , x_ n]]/(f)$ if and only if $\sum h_ i f_ i \in (f)$. Moreover, the induced derivation of $P$ is zero if and only if $h_ i \in (f)$ for $i = 1, \ldots , n$. Thus we find
\[ \mathop{\mathrm{Ker}}((f_1, \ldots , f_ n) : P^{\oplus n} \longrightarrow P) \subset \text{Der}_ k(P, P) \]
The left hand side is a finite dimensional $k$-vector space only if $n = 1$; we omit the proof. We also leave it to the reader to see that the right hand side has finite dimension if $n = 1$. This proves (1).
Proof of (2). Let $Q$ be a flat deformation of $P$ over $k[\epsilon ]$ as in the proof of Lemma 93.8.3. Choose lifts $q_ i \in Q$ of the image of $x_ i$ in $P$. Then $Q$ is a complete local ring with maximal ideal generated by $q_1, \ldots , q_ n$ and $\epsilon $ (small argument omitted). Thus we get a surjection
\[ k[\epsilon ][[x_1, \ldots , x_ n]] \longrightarrow Q,\quad x_ i \longmapsto q_ i \]
Choose an element of the form $f + \epsilon g \in k[\epsilon ][[x_1, \ldots , x_ n]]$ mapping to zero in $Q$. Observe that $g$ is well defined modulo $(f)$. Since $Q$ is flat over $k[\epsilon ]$ we get
\[ Q = k[\epsilon ][[x_1, \ldots , x_ n]]/(f + \epsilon g) \]
Finally, if we changing the choice of $q_ i$ amounts to changing the coordinates $x_ i$ into $x_ i + \epsilon h_ i$ for some $h_ i \in k[[x_1, \ldots , x_ n]]$. Then $f + \epsilon g$ changes into $f + \epsilon (g + \sum h_ i f_ i)$ where $f_ i = \partial f/\partial x_ i$. Thus we see that the isomorphism class of the deformation $Q$ is determined by an element of
\[ k[[x_1, \ldots , x_ n]]/ (f, \partial f/\partial x_1, \ldots , \partial f/\partial x_ n) \]
This has finite dimension over $k$ if and only if its support is the closed point of $k[[x_1, \ldots , x_ n]]$ if and only if $\sqrt{(f, \partial f/\partial x_1, \ldots , \partial f/\partial x_ n)} = (x_1, \ldots , x_ n)$.
$\square$
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