The Stacks project

Proof. Let $A_1 \to A$ and $A_2 \to A$ be morphisms of $\mathcal{C}_\Lambda $. Assume $A_2 \to A$ is surjective. According to Formal Deformation Theory, Lemma 90.16.4 it suffices to show that the functor $\mathcal{F}(A_1 \times _ A A_2) \to \mathcal{F}(A_1) \times _{\mathcal{F}(A)} \mathcal{F}(A_2)$ is an equivalence of categories. This is a special case of More on Algebra, Lemma 15.7.7. $\square$

Proof. Recall that $\text{Inf}(\mathcal{D}\! \mathit{ef}_ P)$ is the set of automorphisms of the trivial deformation $P[\epsilon ] = P \otimes _ k k[\epsilon ]$ of $P$ to $k[\epsilon ]$ equal to the identity modulo $\epsilon $. By Deformation Theory, Lemma 91.2.1 this is equal to $\mathop{\mathrm{Hom}}\nolimits _ P(\Omega _{P/k}, P)$ which in turn is equal to $\text{Der}_ k(P, P)$ by Algebra, Lemma 10.131.3.

Recall that $T\mathcal{D}\! \mathit{ef}_ P$ is the set of isomorphism classes of flat deformations $Q$ of $P$ to $k[\epsilon ]$, more precisely, the set of isomorphism classes of $\mathcal{D}\! \mathit{ef}_ P(k[\epsilon ])$. Recall that a $k[\epsilon ]$-algebra $Q$ with $Q/\epsilon Q = P$ is flat over $k[\epsilon ]$ if and only if

is exact. This is proven in More on Morphisms, Lemma 37.10.1 and more generally in Deformation Theory, Lemma 91.5.2. Thus we may apply Deformation Theory, Lemma 91.2.2 to see that the set of isomorphism classes of such deformations is equal to $\text{Ext}^1_ P(\mathop{N\! L}\nolimits _{P/k}, P)$. $\square$

Proof. By Lemma 93.8.3 we have to show $\text{Ext}^1_ P(\mathop{N\! L}\nolimits _{P/k}, P) = (0)$. Since $k \to P$ is smooth $\mathop{N\! L}\nolimits _{P/k}$ is quasi-isomorphic to the complex consisting of a finite projective $P$-module placed in degree $0$. $\square$

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). $\square$

Proof. This follows immediately from Lemmas 93.8.2 and 93.8.5 and Formal Deformation Theory, Lemma 90.16.6 and Remark 90.15.7. $\square$

Proof. Given a deformation of $P$ we can take the localization of it to get a deformation of the localization; this is clear and we encourage the reader to skip the proof. More precisely, let $(A, Q) \to (k, P)$ be a morphism in $\mathcal{F}$, i.e., an object of $\mathcal{D}\! \mathit{ef}_ P$. Let $S_ Q \subset Q$ be the inverse image of $S$. Then Hence $(A, S_ Q^{-1}Q) \to (k, S^{-1}P)$ is the desired object of $\mathcal{D}\! \mathit{ef}_{S^{-1}P}$. $\square$

Proof. Given a deformation of $P$ we can take the henselization of it to get a deformation of the henselization; this is clear and we encourage the reader to skip the proof. More precisely, let $(A, Q) \to (k, P)$ be a morphism in $\mathcal{F}$, i.e., an object of $\mathcal{D}\! \mathit{ef}_ P$. Denote $J_ Q \subset Q$ the inverse image of $J$ in $Q$. Let $(Q^ h, J_ Q^ h)$ be the henselization of the pair $(Q, J_ Q)$. Recall that $Q \to Q^ h$ is flat (More on Algebra, Lemma 15.12.2) and hence $Q^ h$ is flat over $A$. By More on Algebra, Lemma 15.12.7 we see that the map $Q^ h \to P^ h$ induces an isomorphism $Q^ h \otimes _ A k = Q^ h \otimes _ Q P = P^ h$. Hence $(A, Q^ h) \to (k, P^ h)$ is the desired object of $\mathcal{D}\! \mathit{ef}_{P^ h}$. $\square$

Proof. Given a deformation of $P$ we can take the strict henselization of it to get a deformation of the strict henselization; this is clear and we encourage the reader to skip the proof. More precisely, let $(A, Q) \to (k, P)$ be a morphism in $\mathcal{F}$, i.e., an object of $\mathcal{D}\! \mathit{ef}_ P$. Since the kernel of the surjection $Q \to P$ is nilpotent, we find that $Q$ is a local ring with the same residue field as $P$. Let $Q^{sh}$ be the strict henselization of $Q$. Recall that $Q \to Q^{sh}$ is flat (More on Algebra, Lemma 15.45.1) and hence $Q^{sh}$ is flat over $A$. By Algebra, Lemma 10.156.4 we see that the map $Q^{sh} \to P^{sh}$ induces an isomorphism $Q^{sh} \otimes _ A k = Q^{sh} \otimes _ Q P = P^{sh}$. Hence $(A, Q^{sh}) \to (k, P^{sh})$ is the desired object of $\mathcal{D}\! \mathit{ef}_{P^{sh}}$. $\square$

Proof. Given a deformation of $P$ we can take the completion of it to get a deformation of the completion; this is clear and we encourage the reader to skip the proof. More precisely, let $(A, Q) \to (k, P)$ be a morphism in $\mathcal{F}$, i.e., an object of $\mathcal{D}\! \mathit{ef}_ P$. Observe that $Q$ is a Noetherian ring: the kernel of the surjective ring map $Q \to P$ is nilpotent and finitely generated and $P$ is Noetherian; apply Algebra, Lemma 10.97.5. Denote $J_ Q \subset Q$ the inverse image of $J$ in $Q$. Let $Q^\wedge $ be the $J_ Q$-adic completion of $Q$. Recall that $Q \to Q^\wedge $ is flat (Algebra, Lemma 10.97.2) and hence $Q^\wedge $ is flat over $A$. The induced map $Q^\wedge \to P^\wedge $ induces an isomorphism $Q^\wedge \otimes _ A k = Q^\wedge \otimes _ Q P = P^\wedge $ by Algebra, Lemma 10.97.1 for example. Hence $(A, Q^\wedge ) \to (k, P^\wedge )$ is the desired object of $\mathcal{D}\! \mathit{ef}_{P^\wedge }$. $\square$

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

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

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

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

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

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 $\sqrt{(f, \partial f/\partial x_1, \ldots , \partial f/\partial x_ n)} = (x_1, \ldots , x_ n)$. $\square$