Proof. Suppose that $\mathfrak {M}$ is an $\mathfrak {S}_A$ -module that is projective as an $A[[u]]$ -module. Certainly $W(k') \otimes _{{\mathbf Z}_p} \mathfrak {M}$ is projective over $\mathfrak {S}_A$ , and we claim that it has $\mathfrak {M}$ as an $\mathfrak {S}_A$ -module direct summand. Indeed, this follows by rewriting $\mathfrak {M}$ as $W(k')\otimes _{W(k')} \mathfrak {M}$ and noting that $W(k')$ is a $W(k')$ -module direct summand of $W(k') \otimes _{{\mathbf Z}_p} W(k')$ .

Proof. This is [Reference Caraiani, Emerton, Gee and SavittCEGS20b, Lemma 2.1.4].

Proof. This is due to Fontaine [Reference FontaineFon90, Section 1.2], and can be proved in exactly the same way as [Reference KisinKis09, Lemma 1.2.7].

Proof. The extension $K_{\infty }/K$ is totally wildly ramified. Since the irreducible $\overline {{\mathbf F}}_p$ -representations of $G_K$ are induced from tamely ramified characters, we see that $\overline {r}|_{G_{K_{\infty }}}$ is irreducible if and only if $\overline {r}$ is irreducible, and if $\overline {r}$ or $\overline {r}'$ is irreducible, then we are done. In the reducible case, we see that $\overline {r}$ and $\overline {r}'$ are extensions of the same characters, and the result then follows from [Reference Gee, Liu and SavittGLS15, Lemma 5.4.2] and Lemma A.5 (2).

Proof. Since Z is an effective Cartier divisor if and only if its ideal sheaf ${\mathcal I}_Z$ is an invertible sheaf on X, this follows from the fact that the invertibility of a quasi-coherent sheaf is a local property in the fpqc topology.

Proof. It follows from [Sta13, Tag 0ANQ] that closed immersions $Z \to \operatorname {\mathrm {Spf}} A$ are necessarily of the form $\operatorname {\mathrm {Spf}} B \to \operatorname {\mathrm {Spf}} A$ , and correspond to continuous morphisms $A \to B$ , for some complete linearly topologised ring B, which are taut (in the sense of [Sta13, Tag 0AMX]), have closed kernel and dense image. Since A is adic, it admits a countable basis of neighbourhoods of the origin, and so it follows from [Sta13, Tag 0APT] (recalling also [Sta13, Tag 0AMV]) that $A\to B$ is surjective. Because any ideal of definition I of A is finitely generated, it follows from [Sta13, Tag 0APU] that B is endowed with the I-adic topology. Finally, since A is Noetherian, any ideal in A is I-adically closed. Thus closed immersions $\operatorname {\mathrm {Spf}} B \to \operatorname {\mathrm {Spf}} A$ are determined by giving the kernel of the corresponding morphism $A \to B$ , which can be arbitrary. The same is true of closed immersions $\operatorname {\mathrm {Spec}} B \to \operatorname {\mathrm {Spec}} A$ , and so the lemma follows.

Proof. The morphism $\operatorname {\mathrm {Spf}} B \to \operatorname {\mathrm {Spf}} A$ corresponds to an adic flat morphism $A \to B$ ([Sta13, Tag 0AN0] and [Reference EmertonEme, Lemma 8.18]) and hence is induced by a flat morphism $\operatorname {\mathrm {Spec}} B \to \operatorname {\mathrm {Spec}} A$ , which is furthermore faithfully flat if and only if $\operatorname {\mathrm {Spf}} B \to \operatorname {\mathrm {Spf}} A$ is surjective (again by [Reference EmertonEme, Lemma 8.18]). The present lemma thus follows from Lemma 2.4.2.

Proof. Essentially by definition (and taking into account [Reference EmertonEme, Lemma 8.18]), it suffices to prove this in the case when ${\mathcal X} = \operatorname {\mathrm {Spf}} B$ , where B is a flat Noetherian adic $W(k)$ -algebra admitting $(p)$ as an ideal of definition. In this case, the restriction map

$$\begin{align*}[\operatorname{\mathrm{Spec}} W(k)[X,Y]/(XY - p) / {\mathbf G}_m](\operatorname{\mathrm{Spec}} B)\to [\operatorname{\mathrm{Spec}} W(k)[X,Y]/(XY - p) / {\mathbf G}_m](\operatorname{\mathrm{Spf}} B)\end{align*}$$

is an equivalence of groupoids. Indeed, the essential surjectivity follows from the (standard and easily verified) fact that if $\{M_i\}$ is a compatible family of locally free $B/p^iB$ -modules of rank one, then $M := \varprojlim M_i$ is a locally free B-module of rank one, for which each of the natural morphisms $M/p^iM \to M_i$ is an isomorphism. The full faithfulness follows from the fact that a locally free B-module of rank one is p-adically complete, and so is recovered as the inverse limit of its compatible family of quotients $\{M/p^iM\}.$

We are therefore reduced to the same statement with ${\mathcal X} = \operatorname {\mathrm {Spec}} B$ . The composite morphism $\operatorname {\mathrm {Spec}} B \to [\mathbf A^1/{\mathbf G}_m]$ induced by (2.4.7) corresponds to giving a pair $({\mathcal D},X)$ , where ${\mathcal D}$ is a line bundle on $\operatorname {\mathrm {Spec}} B$ , and X is a global section of ${\mathcal D}^{-1}$ . Indeed, giving a morphism $\operatorname {\mathrm {Spec}} B \to [\mathbf A^1/{\mathbf G}_m]$ is equivalent to giving a ${\mathbf G}_m$ -torsor $P \to \operatorname {\mathrm {Spec}} B$ , together with a ${\mathbf G}_m$ -equivariant morphism $P \to \mathbf A^1$ . Giving a ${\mathbf G}_m$ -torsor P over $\operatorname {\mathrm {Spec}} B$ is equivalent to giving an invertible sheaf ${\mathcal D}$ on $\operatorname {\mathrm {Spec}} B$ (the associated ${\mathbf G}_m$ -torsor is then obtained by deleting the zero section from the line bundle $D\to X$ corresponding to ${\mathcal D}$ ), and giving a ${\mathbf G}_m$ -equivariant morphism $P \to \mathbf A^1$ is equivalent to giving a global section of ${\mathcal D}^{-1}$ .

It follows that giving a morphism $\operatorname {\mathrm {Spec}} B \to [\operatorname {\mathrm {Spec}} W(k)[X,Y]/(XY - p) / {\mathbf G}_m]$ corresponds to giving a line bundle ${\mathcal D}$ and sections $X \in {\mathcal D}^{-1}$ , $Y \in {\mathcal D}$ satisfying $X Y = p$ . To say that B is flat over $W(k)$ is just to say that p is a regular element on B, and so we see that X (respectively, Y) is a regular section of ${\mathcal D}^{-1}$ (respectively, ${\mathcal D}$ ). Again, since p is a regular element on B, we see that Y is uniquely determined by X and the equation $X Y = p$ , and so giving a morphism $\operatorname {\mathrm {Spec}} B\to [\operatorname {\mathrm {Spec}} W(k)[X,Y]/(XY - p) / {\mathbf G}_m]$ is equivalent to giving a line bundle ${\mathcal D}$ and a regular section X of ${\mathcal D}^{-1}$ , such that $pB \subset X\otimes _B {\mathcal D} \subset {\mathcal D}^{-1}\otimes _B {\mathcal D}\buildrel \sim \over \longrightarrow B$ ; this last condition guarantees the existence of the (then uniquely determined) Y.

Now, giving a line bundle ${\mathcal D}$ on $\operatorname {\mathrm {Spec}} B$ and a regular section $X\in {\mathcal D}^{-1}$ is the same as giving the zero locus D of X, which is a Cartier divisor on $\operatorname {\mathrm {Spec}} B$ . (There is a canonical isomorphism $({\mathcal D},X) \cong \bigl ({\mathcal I}_D,1\bigr )$ , where ${\mathcal I}_D$ denotes the ideal sheaf of D.) The condition that $pB \subset X\otimes _B {\mathcal D}$ is equivalent to the condition that $p \in {\mathcal I}_D$ , that is, that D be contained in $\operatorname {\mathrm {Spec}} B/pB$ , and we are done.

Proof. This follows immediately from Lemma 2.4.8, by the definition of ${\mathcal G}_{\eta }$ .

Proof. Note that since $\Phi _{\mathfrak {M}'}[1/E(u)]$ and $\Phi _{\mathfrak {M}"}[1/E(u)]$ are both isomorphisms by assumption, it follows from the snake lemma that $\Phi _{\mathfrak {M}}[1/E(u)]$ is an isomorphism. Similarly, we have a short exact sequence of $\mathfrak {S}_A$ -modules

$$\begin{align*}0\to\operatorname{\mathrm{coker}}\Phi_{\mathfrak{M}'}\to\operatorname{\mathrm{coker}}\Phi_{\mathfrak{M}}\to\operatorname{\mathrm{coker}}\Phi_{\mathfrak{M}"}\to 0.\end{align*}$$

The claims about the height and rank of $\mathfrak {M}$ follow immediately.

Proof. This is immediate.

Proof. We first describe the embedding (3.1.4) explicitly. Suppose that

$$ \begin{align*}0 \to \mathfrak{N} \to \mathfrak{E} \to \mathfrak{M} \to 0\end{align*} $$

is an extension in $\mathcal {K}(A)$ . Since $\mathfrak {M}$ is projective over $\mathfrak {S}_A$ , and since $\operatorname {\mathrm {Gal}}(K'/K)$ is of prime-to-p order, we split this short exact sequence over the twisted group ring $\mathfrak {S}_A[\operatorname {\mathrm {Gal}}(K'/K)],$ say via some element $\sigma \in \operatorname {\mathrm {Hom}}_{\mathfrak {S}_A[\operatorname {\mathrm {Gal}}(K'/K)]}(\mathfrak {M},\mathfrak {E}).$ This splitting is well-defined up to the addition of an element $\alpha \in \operatorname {\mathrm {Hom}}_{\mathfrak {S}_A[\operatorname {\mathrm {Gal}}(K'/K)]}(\mathfrak {M},\mathfrak {N}).$

This splitting is a homomorphism in $\mathcal {K}(A)$ if and only if the element

$$ \begin{align*}\Phi_{\mathfrak{E}}\circ \varphi^{*}\sigma - \sigma \circ \Phi_{\mathfrak{M}} \in \operatorname{\mathrm{Hom}}_{\mathfrak{S}_A[\operatorname{\mathrm{Gal}}(K'/K)]}(\varphi^{*}\mathfrak{M},\mathfrak{N})\end{align*} $$

vanishes. If we replace $\sigma $ by $\sigma +\alpha ,$ then this element is replaced by

$$ \begin{align*}(\Phi_{\mathfrak{E}}\circ \varphi^{*}\sigma - \sigma \circ \Phi_{\mathfrak{M}}) + (\Phi_{\mathfrak{N}} \circ \varphi^{*} \alpha - \alpha \circ \Phi_{\mathfrak{M}}).\end{align*} $$

Thus, the coset of $\Phi _{\mathfrak {E}}\circ \varphi ^{*}\sigma - \sigma \circ \Phi _{\mathfrak {M}}$ in $H^1\bigl (C^{\bullet }(\mathfrak {N})\bigr )$ is well-defined, independent of the choice of $\sigma ,$ and this coset is the image of the class of the extension $\mathfrak {E}$ under the embedding

(3.1.6) $$ \begin{align} \operatorname{\mathrm{Ext}}^1_{\mathcal{K}(A)}(\mathfrak{M},\mathfrak{N}) \hookrightarrow H^1\bigl(C^{\bullet}(\mathfrak{N})\bigr) \end{align} $$

(up to a possible overall sign, which we ignore, since it doesn’t affect the claim of the lemma).

Now, given any element $\nu \in \operatorname {\mathrm {Hom}}_{\mathfrak {S}_A[\operatorname {\mathrm {Gal}}(K'/K)]}(\varphi ^{*}\mathfrak {M},\mathfrak {N})$ , we may give the $\mathfrak {S}_A[\operatorname {\mathrm {Gal}}(K'/K)]$ -module $\mathfrak {E} := \mathfrak {N} \oplus \mathfrak {M}$ the structure of a $\mathfrak {S}_A[F,\operatorname {\mathrm {Gal}}(K'/K)]$ -module as follows: we need to define a $\varphi $ -linear morphism $\mathfrak {E}\to \mathfrak {E}$ , or equivalently, a linear morphism $\Phi _{\mathfrak {E}}:\varphi ^{*}\mathfrak {E}\to \mathfrak {E}$ . We do this by setting

$$ \begin{align*}\Phi_{\mathfrak{E}} := \begin{pmatrix} \Phi_{\mathfrak{N}} & \nu \\ 0 & \Phi_{\mathfrak{M}} \end{pmatrix}.\end{align*} $$

Then, $\mathfrak {E}$ is an extension of $\mathfrak {M}$ by $\mathfrak {N}$ , and if we let $\sigma $ denote the obvious embedding of $\mathfrak {M}$ into $\mathfrak {E}$ , then one computes that

$$ \begin{align*}\nu = \Phi_{\mathfrak{E}}\circ \varphi^{*}\sigma - \sigma \circ \Phi_{\mathfrak{M}} .\end{align*} $$

This shows that (3.1.6) is an isomorphism, as claimed.

Proof. Indeed, there is a natural isomorphism

$$ \begin{align*}\operatorname{\mathrm{Hom}}_{\mathfrak{S}_A[\operatorname{\mathrm{Gal}}(K'/K)]}(\mathfrak{M}, \mathfrak{N}\, \widehat{\otimes}_A Q) \cong \operatorname{\mathrm{Hom}}_{\mathfrak{S}_B[\operatorname{\mathrm{Gal}}(K'/K)]}(\mathfrak{M}\, \widehat{\otimes}_A B, \mathfrak{N}\, \widehat{\otimes}_A Q),\end{align*} $$

and similarly with $\varphi ^{*}\mathfrak {M}$ in place of $\mathfrak {M}$ .

Proof of Lemma 3.1.10

Since $p^a=0$ in A, there exists $H(u)\in \mathfrak {S}_A$ with $u^{e'ah}=E(u)^hH(u)$ in $\mathfrak {S}_A$ . Thus, the image of $\Phi _{\mathfrak {M}}$ contains $u^{e'ah} \mathfrak {M}=v^{eah}\mathfrak {M}$ , and there exists a map $\Upsilon : \mathfrak {M} \to \varphi ^{*} \mathfrak {M}$ , such that $\Phi _{\mathfrak {M}} \circ \Upsilon $ is multiplication by $v^{eah}$ .

We begin with (1). Suppose that $f\in \ker (\delta \otimes \operatorname {id}_Q)\cap v^MC^0\otimes _AQ$ . Since $C^0\otimes _AQ$ is v-adically separated, it is enough, applying induction on M, to show that $f\in v^{M+1}C^0\otimes _AQ$ . Since $f\in \ker (\delta \otimes \operatorname {id}_Q)$ , we have $f\circ \Phi _{\mathfrak {M}}=\Phi _{\mathfrak {N}}\circ \varphi ^{*}f$ . Since $f\in v^MC^0\otimes _AQ$ , we have $f \circ \Phi _{\mathfrak {M}} = \Phi _{\mathfrak {N}} \circ \varphi ^{*}f \in v^{pM} C^1 \otimes _A Q$ . Precomposing with $\Upsilon $ gives $v^{eah} f \in v^{pM} C^0 \otimes _A Q$ . Since $C^0 \otimes _A Q$ is v-torsion free, it follows that $f\in v^{pM-eah}C^0\otimes _AQ\subseteq u^{M+1}C^0\otimes _AQ$ , as required.

We now move on to (2). Set $M = N - eah$ . By precomposing with $\Upsilon $ , we see that $\alpha \circ \Phi _{\mathfrak {M}} \in v^N C^1\otimes _A Q$ implies $\alpha \in v^M C^0\otimes _A Q$ ; from this, together with the inequality $pM \ge N$ , it is straightforward to check that

$$\begin{align*}(\Phi_{\mathfrak{M}}^{*})^{-1}(v^N C^1\otimes_AQ) = (\delta \otimes \operatorname{id}_Q)^{-1}(v^N C^1\otimes_AQ)\cap v^M C^0\otimes_AQ.\end{align*}$$

Note that M satisfies the condition in (1). To complete the proof, we will show that for any M as in (1) and any $N \ge M + eah$ , the map $\delta $ induces an isomorphism

$$\begin{align*}(\delta \otimes \operatorname{id}_Q)^{-1}(v^N C^1\otimes_AQ)\cap v^M C^0\otimes_AQ \stackrel{\sim}{\longrightarrow} v^N C^1\otimes_AQ.\end{align*}$$

By (1), $\delta \otimes \operatorname {id}_Q$ induces an injection $(\delta \otimes \operatorname {id}_Q)^{-1}(v^N C^1\otimes _AQ)\cap v^M C^0\otimes _AQ\hookrightarrow v^N C^1\otimes _AQ$ , so it is enough to show that $(\delta \otimes \operatorname {id}_Q)(v^MC^0\otimes _AQ)\supseteq v^N C^1\otimes _AQ$ . Equivalently, we need to show that

$$ \begin{align*}v^N C^1 \otimes_A Q \to (C^1\otimes_A Q) / (\delta\otimes\operatorname{id}_Q)\big(v^M C^0\otimes_A Q) \end{align*} $$

is identically zero. Since the formation of cokernels is compatible with tensor products, we see that this morphism is obtained by tensoring the corresponding morphism

$$ \begin{align*}v^N C^1 \to C^1/ \delta\bigl(v^M C^0\bigr) \end{align*} $$

with Q over A, so we are reduced to the case $Q=A$ . (Recall from Remark 3.1.11(1) that the hypotheses of the Lemma are satisfied in this case, and that $C^1$ is v-adically separated.)

We claim that for any $g\in v^NC^1$ , we can find an $f\in v^{N-eah}C^0$ , such that $\delta (f)-g\in v^{p(N-eah)}C^1$ . Admitting the claim, given any $g\in v^N C^1$ , we may find $h\in v^MC^0$ with $\delta (h)=g$ by successive approximation in the following way: Set $h_0=f$ for f as in the claim; then $h_0\in v^{N-eah}C^0\subseteq v^MC^0$ , and $\delta (h_0)-g\in v^{p(N-eah)}C^1\subseteq v^{N+1}C^1$ . Applying the claim again, with N replaced by $N+1$ , and g replaced by $g-\delta (h_0)$ , we find $f\in v^{N+1-eah}C^0\subseteq v^{M+1}C^0$ with $\delta (f)-g+\delta (h_0)\in v^{p(N+1-eah)}C^1\subseteq v^{N+1}C^1$ . Setting $h_1=h_0+f$ , and proceeding inductively, we obtain a Cauchy sequence converging (in the v-adically complete $A[[v]]$ -module $C^0$ ) to the required element h. It remains to prove the claim. Since $\delta (f)=\Phi _{\mathfrak {N}}\circ \varphi ^{*}f-f\circ \Phi _{\mathfrak {M}} $ , and since if $f\in v^{N-eah}C^0$ , then $\Phi _{\mathfrak {N}}\circ \varphi ^{*}f\in v^{p(N-eah)}C^1$ , it is enough to show that we can find an $f\in v^{N-eah}C^0$ with $f\circ \Phi _{\mathfrak {M}}=-g$ . Since $\Phi _{\mathfrak {M}}$ is injective, the map $\Upsilon \circ \Phi _{\mathfrak {M}}$ is also multiplication by $v^{eah}$ , and so it suffices to take f with $v^{eah} f = -g \circ \Upsilon \in v^N C^0$ .

Proof. By Remarks 3.1.11 and 2.1.6(2), we can apply Lemma 3.1.10 to both $C^i(\mathfrak {N}\, \widehat {\otimes }_AQ)$ and $C^i(\mathfrak {N})\otimes _AQ$ , and we see that it is enough to show that the natural morphism of complexes

is a quasi-isomorphism. In fact, it is even an isomorphism.

Proof. The statements for $\mathfrak {N}/u^i\mathfrak {N}$ follow easily from those for $\mathfrak {N}$ , by considering the short exact sequence $0 \to u^i\mathfrak {N} \to \mathfrak {N} \to \mathfrak {N}/u^i\mathfrak {N} \to 0$ in $\mathcal {K}(A)$ and applying Corollary 3.1.7. By Corollary 3.1.8, it is enough to consider the cohomology of the complex $C^{\bullet }$ . By Lemma 3.1.10 with $Q=A$ , the cohomology of $C^{\bullet }$ agrees with the cohomology of the induced complex

$$\begin{align*}C^0/\big((\Phi_{\mathfrak{M}}^{*})^{-1}(v^N C^1) )\to C^1/ v^N C^1,\end{align*}$$

for an appropriately chosen value of N. It follows that for an appropriately chosen value of N, $\operatorname {\mathrm {Ext}}^1_{\mathcal {K}(A)}(\mathfrak {M},\mathfrak {N})$ can be computed as the cokernel of the induced morphism $C^0/v^N C^0 \to C^1/ v^N C^1$ .

Under our hypothesis on $\mathfrak {N}$ , $C^0/v^N C^0$ and $C^1/v^NC^1$ are finitely generated projective A-modules, and thus finitely presented. It follows that $\operatorname {\mathrm {Ext}}^1_{\mathcal {K}(A)}(\mathfrak {M},\mathfrak {N})$ is finitely presented.

In the case that A is furthermore assumed to be Noetherian, it is enough to note that since $v^NC^0\subseteq (\Phi _{\mathfrak {M}}^{*})^{-1}(v^N C^1)$ , the quotient $C^0/\bigl ((\Phi _{\mathfrak {M}}^{*})^{-1}(v^N C^1) \bigr )$ is a finitely generated A-module.

Proof. By Lemmas 3.1.3 and 3.1.9 (the latter applied with $Q=B$ ), we can and do think of $f_B$ as being an element of the kernel of $\delta :C^0(\mathfrak {N}\, \widehat {\otimes }_A B)\to C^1(\mathfrak {N}\, \widehat {\otimes }_A B)$ , the complex $C^{\bullet }$ here and throughout this proof denoting $C^{\bullet }_{\mathfrak {M}}$ as usual.

Fix N as in Lemma 3.1.10, and write $\overline {f}_B$ for the corresponding element of $C^0(\mathfrak {N}\, \widehat {\otimes }_A B)/v^N=(C^0(\mathfrak {N})/v^N)\otimes _A B$ (this equality following easily from the assumption that $\mathfrak {M}$ and $\mathfrak {N}$ are projective $\mathfrak {S}_A$ -modules of finite rank). Since $C^0(\mathfrak {N})/v^N$ is a projective A-module of finite rank, it follows that for some finite type A-subalgebra $B'$ of B, there is an element $\overline {f}_{B'}\in (C^0(\mathfrak {N})/v^N)\otimes _A B'=C^0(\mathfrak {N}\, \widehat {\otimes }_A B')/v^N$ , such that $\overline {f}_{B'}\otimes _{B'}B=\overline {f}_B$ . Denote also by $\overline {f}_{B'} $ the induced element of

$$\begin{align*}C^0(\mathfrak{N}\, \widehat{\otimes} _A B')/\big(\Phi_{\mathfrak{M}}^{*})^{-1}(v^N C^1(\mathfrak{N}\, \widehat{\otimes} _A B')).\end{align*}$$

By Lemma 3.1.10 (and Lemma 3.1.3), we have a commutative diagram with exact rows

in which the vertical arrows are induced by $\, \widehat {\otimes }_{B'}B$ . By a diagram chase, we only need to show that $\delta (\overline {f}_{B'})=0$ . Since $\delta (f_B)=0$ , it is enough to show that the right-hand vertical arrow is an injection. This arrow can be rewritten as the tensor product of the injection of A-algebras $B'\hookrightarrow B$ with the flat (even projective of finite rank) A-module $C^1(\mathfrak {N})/v^N$ , so the result follows.

Proof. We first prove the lemma in the case of an A-module Q. It follows from Lemmas 3.1.5 and 3.1.10 that we may compute $\operatorname {\mathrm {Ext}}^1_{\mathcal {K}(A)}(\mathfrak {M},\mathfrak {N})$ as the cokernel of the morphism

$$ \begin{align*}C^0(\mathfrak{N})/v^N C^0(\mathfrak{N}) \buildrel \delta \over \longrightarrow C^1(\mathfrak{N})/v^N C^1(\mathfrak{N}),\end{align*} $$

for some sufficiently large value of N (not depending on $\mathfrak {N}$ ), and hence that we may compute $\operatorname {\mathrm {Ext}}^1_{\mathcal {K}(A)}(\mathfrak {M},\mathfrak {N})\otimes _A Q$ as the cokernel of the morphism

$$ \begin{align*}\bigl(C^0(\mathfrak{N})/v^N C^0(\mathfrak{N})\bigr) \otimes_A Q \buildrel \delta \over \longrightarrow \bigl(C^1(\mathfrak{N})/v^N C^1(\mathfrak{N})\bigr) \otimes_A Q.\end{align*} $$

We may similarly compute $\operatorname {\mathrm {Ext}}^1_{\mathcal {K}(A)}(\mathfrak {M},\mathfrak {N}\, \widehat {\otimes }_A Q)$ as the cokernel of the morphism

$$ \begin{align*}C^0(\mathfrak{N}\, \widehat{\otimes}_A Q)/v^N C^0(\mathfrak{N}\, \widehat{\otimes}_A Q) \buildrel \delta \over \longrightarrow C^1(\mathfrak{N}\, \widehat{\otimes}_A Q)/v^N C^1(\mathfrak{N}\, \widehat{\otimes}_A Q)\end{align*} $$

(Remark 2.1.6 (2) shows that $\mathfrak {N}\, \widehat {\otimes }_A Q$ satisfies the necessary hypotheses for Lemma 3.1.10 to apply.) Once we note that the natural morphism

$$ \begin{align*}\bigl( C^i(\mathfrak{N})/v^N C^i(\mathfrak{N})\bigr)\otimes_A Q \to C^i(\mathfrak{N}\, \widehat{\otimes}_A Q)/v^N C^i(\mathfrak{N}\, \widehat{\otimes}_A Q)\end{align*} $$

is an isomorphism for $i = 0$ and $1$ (because $\mathfrak {M}$ is a finitely generated projective $\mathfrak {S}_A$ -module), we obtain the desired isomorphism

$$ \begin{align*}\operatorname{\mathrm{Ext}}^1_{\mathcal{K}(A)}(\mathfrak{M},\mathfrak{N})\otimes_A Q \buildrel \sim \over \longrightarrow \operatorname{\mathrm{Ext}}^1_{\mathcal{K}(A)}(\mathfrak{M},\mathfrak{N}\, \widehat{\otimes}_A Q).\end{align*} $$

If B is an A-algebra, and Q is a B-module, then by Lemma 3.1.9, there is a natural isomorphism

$$ \begin{align*}\operatorname{\mathrm{Ext}}^1_{\mathcal{K}(A)}(\mathfrak{M}, \mathfrak{N}\, \widehat{\otimes}_A Q) \buildrel \sim \over \longrightarrow \operatorname{\mathrm{Ext}}^1_{\mathcal{K}(B)}(\mathfrak{M}\, \widehat{\otimes}_A B, \mathfrak{N}\, \widehat{\otimes}_A Q);\end{align*} $$

combined with the preceding base-change result, this yields one of our claimed isomorphisms, namely

$$\begin{align*}\operatorname{\mathrm{Ext}}^1_{\mathcal{K}(A)}(\mathfrak{M},\mathfrak{N})\otimes_A Q \buildrel \sim \over \longrightarrow \operatorname{\mathrm{Ext}}^1_{\mathcal{K}(B)}(\mathfrak{M}\, \widehat{\otimes}_A B, \mathfrak{N}\, \widehat{\otimes}_A Q).\end{align*}$$

Taking Q to be B itself, we then obtain an isomorphism

$$\begin{align*}\operatorname{\mathrm{Ext}}^1_{\mathcal{K}(A)}(\mathfrak{M},\mathfrak{N})\otimes_A B \buildrel \sim \over \longrightarrow \operatorname{\mathrm{Ext}}^1_{\mathcal{K}(B)}(\mathfrak{M}\, \widehat{\otimes}_A B, \mathfrak{N}\, \widehat{\otimes}_A B).\end{align*}$$

This allows us to identify $\operatorname {\mathrm {Ext}}^1_{\mathcal {K}(A)}(\mathfrak {M},\mathfrak {N})\otimes _A Q$ , which is naturally isomorphic to $\bigl (\operatorname {\mathrm {Ext}}^1_{\mathcal {K}(A)}(\mathfrak {M},\mathfrak {N})\otimes _A B\bigr ) \otimes _B Q$ , with $\operatorname {\mathrm {Ext}}^1_{\mathcal {K}(B)}(\mathfrak {M}\, \widehat {\otimes }_A B, \mathfrak {N}\, \widehat {\otimes }_A B)\otimes _B Q$ , yielding the second claimed isomorphism.

Proof. If $\mathfrak m$ is a maximal ideal of A, then $\kappa (\mathfrak m)$ is certainly a finite type A-algebra, and so evidently (1) implies (2). It is even a finitely generated A-module, and so also (2) follows from (3). We next prove that (2) implies (3). To this end, recall that if A is any ring, and M is any A-module, then M injects into the product of its localisations at all maximal ideals. If A is Noetherian, and M is finitely generated, then, by combining this fact with the Artin–Rees Lemma, we see that M embeds into the product of its completions at all maximal ideals. Another way to express this is that, if I runs over all cofinite length ideals in A (i.e. all ideals for which $A/I$ is finite length), then M embeds into the projective limit of the quotients $M/IM$ (the point being that this projective limit is the same as the product over all $\mathfrak m$ -adic completions). We are going to apply this observation with A replaced by $\mathfrak {S}_A$ , and with M taken to be $\mathfrak {N}\otimes _A Q$ for some finitely generated A-module Q.

In $A[[u]]$ , one sees that u lies in the Jacobson radical (because $1 + fu$ is invertible in $A[[u]]$ for every $f \in A[[u]]$ ), and thus in every maximal ideal, and so the maximal ideals of $A[[u]]$ are of the form $(\mathfrak m, u)$ , where $\mathfrak m$ runs over the maximal ideals of A. Thus, the ideals of the form $(I,u^n)$ , where I is a cofinite length ideal in A, are cofinal in all cofinite length ideals in $A[[u]]$ . Since $\mathfrak {S}_A$ is finite over $A[[u]]$ , we see that the ideals $(I,u^n)$ in $\mathfrak {S}_A$ are also cofinal in all cofinite length ideals in $A[[u]]$ . Since $A[[u]]$ , and hence $\mathfrak {S}_A$ , is furthermore Noetherian when A is, we see that if Q is a finitely generated A-module, and $\mathfrak {N}$ is a finitely generated $\mathfrak {S}_A$ -module, then $\mathfrak {N}\otimes _A (Q/IQ)$ is u-adically complete, for any cofinite length ideal I in A, and hence equal to the limit over n of $\mathfrak {N} \otimes _A Q/(I,u^n)$ . Putting this together with the observation of the preceding paragraph, we see that the natural morphism

$$ \begin{align*}\mathfrak{N}\otimes_A Q \to \varprojlim_I \mathfrak{N}\otimes_A (Q/IQ)\end{align*} $$

(where I runs over all cofinite length ideals of A) is an embedding. The induced morphism

$$ \begin{align*}\operatorname{\mathrm{Hom}}_{\mathcal{K}(A)}(\mathfrak{M},\mathfrak{N}\otimes_A Q) \to \varprojlim_I \operatorname{\mathrm{Hom}}_{\mathcal{K}(A)}(\mathfrak{M},\mathfrak{N}\otimes_A (Q/IQ))\end{align*} $$

is then evidently also an embedding.

Thus, to conclude that $ \operatorname {\mathrm {Hom}}_{\mathcal {K}(A)}(\mathfrak {M},\mathfrak {N}\otimes _A Q)$ vanishes, it suffices to show that $\operatorname {\mathrm {Hom}}_{\mathcal {K}(A)}(\mathfrak {M},\mathfrak {N}\otimes _A (Q/IQ))$ vanishes for each cofinite length ideal I in A. An easy induction (using the flatness of $\mathfrak {N}$ ) on the length of $A/I$ reduces this to showing that $\operatorname {\mathrm {Hom}}_{\mathcal {K}(A)}\bigl (\mathfrak {M},\mathfrak {N}\otimes _A \kappa (\mathfrak m)\bigr ),$ or, equivalently, $\operatorname {\mathrm {Hom}}_{\mathcal {K}(\kappa (\mathfrak {m}))}\bigl (\mathfrak {M}\otimes _A \kappa (\mathfrak m), \mathfrak {N}\otimes _A \kappa (\mathfrak m)\bigr ),$ vanishes for each maximal ideal $\mathfrak m$ . Since this is the hypothesis of (2), we see that indeed (2) implies (3).

It remains to show that (3) implies (1) when A is Jacobson. Applying the result ‘(2) implies (3)’ (with A replaced by B, and taking Q in (3) to be B itself as a B-module) to $\mathfrak {M}\, \widehat {\otimes }_A B$ and $\mathfrak {N}\, \widehat {\otimes }_A B$ , we see that it suffices to prove the vanishing of

$$ \begin{align*}\operatorname{\mathrm{Hom}}_{\mathcal{K}(B)}\bigl( (\mathfrak{M}\, \widehat{\otimes}_A B)\otimes_B \kappa(\mathfrak n), (\mathfrak{N}\, \widehat{\otimes}_A B)\otimes_B \kappa(\mathfrak n) \bigr) = \operatorname{\mathrm{Hom}}_{\mathcal{K}(A)}\bigl( \mathfrak{M}, \mathfrak{N}\, \widehat{\otimes}_A \kappa(\mathfrak n) \bigr) \end{align*} $$

for each maximal ideal $\mathfrak n$ of B. Since A is Jacobson, the field $\kappa (\mathfrak n)$ is, in fact, a finitely generated A-module, hence $\mathfrak {N}\, \widehat {\otimes }\kappa (\mathfrak n) = \mathfrak {N}\otimes _A \kappa (\mathfrak n)$ , and so the desired vanishing is a special case of (3).