The Stacks project

Proof. Proof of (1). We view $\text{Der}_ k(P, P)$ as a $P$-module. If it has finite dimension over $k$, then it has finite length as a $P$-module, hence it is supported in finitely many closed points of $\mathop{\mathrm{Spec}}(P)$ (Algebra, Lemma 10.52.11). Since $\text{Der}_ k(P, P) = \mathop{\mathrm{Hom}}\nolimits _ P(\Omega _{P/k}, P)$ we see that $\text{Der}_ k(P, P)_\mathfrak p = \text{Der}_ k(P_\mathfrak p, P_\mathfrak p)$ for any prime $\mathfrak p \subset P$ (this uses Algebra, Lemmas 10.131.8, 10.131.15, and 10.10.2). Let $\mathfrak p$ be a minimal prime ideal of $P$ corresponding to an irreducible component of dimension $d > 0$. Then $P_\mathfrak p$ is an Artinian local ring essentially of finite type over $k$ with residue field and $\Omega _{P_\mathfrak p/k}$ is nonzero for example by Algebra, Lemma 10.140.3. Any nonzero finite module over an Artinian local ring has both a sub and a quotient module isomorphic to the residue field. Thus we find that $\text{Der}_ k(P_\mathfrak p, P_\mathfrak p) = \mathop{\mathrm{Hom}}\nolimits _{P_\mathfrak p}(\Omega _{P_\mathfrak p/k}, P_\mathfrak p)$ is nonzero too. Combining all of the above we find that (1) is true.

Proof of (2). For a prime $\mathfrak p$ of $P$ we will use that $\mathop{N\! L}\nolimits _{P_\mathfrak p/k} = (\mathop{N\! L}\nolimits _{P/k})_\mathfrak p$ (Algebra, Lemma 10.134.13) and we will use that $\text{Ext}_ P^1(\mathop{N\! L}\nolimits _{P/k}, P)_\mathfrak p = \text{Ext}_{P_\mathfrak p}^1(\mathop{N\! L}\nolimits _{P_\mathfrak p/k}, P_\mathfrak p)$ (More on Algebra, Lemma 15.65.4). Given a prime $\mathfrak p \subset P$ then $k \to P$ is smooth at $\mathfrak p$ if and only if $(\mathop{N\! L}\nolimits _{P/k})_\mathfrak p$ is quasi-isomorphic to a finite projective module placed in degree $0$ (this follows immediately from the definition of a smooth ring map but it also follows from the stronger Algebra, Lemma 10.137.12).

Assume that $P$ is smooth over $k$ at all but finitely many primes. Then these “bad” primes are maximal ideals $\mathfrak m_1, \ldots , \mathfrak m_ n \subset P$ by Algebra, Lemma 10.61.3 and the fact that the “bad” primes form a closed subset of $\mathop{\mathrm{Spec}}(P)$. For $\mathfrak p \not\in \{ \mathfrak m_1, \ldots , \mathfrak m_ n\}$ we have $\text{Ext}^1_ P(\mathop{N\! L}\nolimits _{P/k}, P)_\mathfrak p = 0$ by the results above. Thus $\text{Ext}^1_ P(\mathop{N\! L}\nolimits _{P/k}, P)$ is a finite $P$-module whose support is contained in $\{ \mathfrak m_1, \ldots , \mathfrak m_ r\}$. By Algebra, Proposition 10.63.6 for example, we find that the dimension over $k$ of $\text{Ext}^1_ P(\mathop{N\! L}\nolimits _{P/k}, P)$ is a finite integer combination of $\dim _ k \kappa (\mathfrak m_ i)$ and hence finite by the Hilbert Nullstellensatz (Algebra, Theorem 10.34.1). $\square$