Proof. By replacing $(f_1,\ldots ,f_d)$ by $(f_1^2,\ldots ,f_d^2)$ , we may assume that $(f_1,\ldots ,f_d)\subset \operatorname {\mathrm {ker}}\lambda ^2$ ; see [Reference MatsumuraMat80, 15.A, Theorem 26]. Write $S[1/\varpi ]$ for the localization of S at $\varpi $ and $\widehat {S[1/\varpi ]}$ for the completion of the latter at the point corresponding to $\lambda $ . By our hypothesis, the ring $\widehat {S[1/\varpi ]}$ is a power series ring over E in $n\ge d$ indeterminates. Let $\widehat I$ denote its maximal ideal. Choose $g_1,\ldots ,g_d$ in $\ker \lambda $ whose images in $\widehat I/\widehat I^2$ are linearly independent over E. Then $(h_1,\ldots ,h_d)$ with $h_i=f_i+\varpi g_i$ has all properties required.

Proof. Let $\Pi \colon S={\mathcal O}[[z_1,\ldots ,z_n]]\to B$ be a surjective ring homomorphism. Let ${{\mathfrak p}}_\lambda \subset {\mathfrak m}_B$ be the prime ideal $\ker \lambda $ , and denote by ${\mathfrak q}_\lambda \subset {\mathfrak m}_S$ its inverse image under $\Pi $ , that is, ${\mathfrak q}_\lambda =\ker \lambda \circ \Pi $ . Let $m=n-d\ge 0$ .

By hypothesis $B/\varpi $ has dimension d. Because S is $\varpi $ -torsion free and $S/\varpi $ is regular, we can find a regular sequence $(f_1,\ldots ,f_m)$ in $\ker \Pi \subset S$ such that $(f_1,\ldots ,f_m,\varpi )$ is a regular sequence. Because $S[1/\varpi ]$ is regular of dimension n and ${\mathfrak q}_\lambda [1/\varpi ]$ is a maximal ideal of that ring, the ring $S[1/\varpi ]$ is formally smooth at ${\mathfrak q}_\lambda [1/\varpi ]$ of dimension n.

It follows from Lemma 3.5 that there exist $h_1,\ldots ,h_m\in \operatorname {\mathrm {ker}} \Pi +\varpi S$ such that $h_1,\ldots ,h_m,\varpi $ is a regular sequence in S and such that for $A=S/(h_1,\ldots ,h_m)$ and induced augmentation $\lambda _A\colon A\to {\mathcal O}$ the ring $A[1/\varpi ]$ is formally smooth at $\lambda _A$ of dimension $n-m=d$ . It follows that one has an induced surjection $A\to B$ , where A is a local complete, complete intersection ${\mathcal O}$ -algebra, flat over ${\mathcal O}$ of relative dimension d and that the induced surjection $A[1/\varpi ]\to B[1/\varpi ]$ becomes an isomorphism after completion at ${\mathfrak q}_\lambda [1/\varpi ]$ .

Proof of Proposition 3.2.

Because R is Cohen–Macaulay and flat over ${\mathcal O}$ of relative dimension d, we can find a regular sequence $\varpi ,f_1,\ldots ,f_d$ in R. If we replace each $f_i$ by an element in $f_i+\varpi R$ the resulting sequence is again regular. Now, using that $\operatorname {\mathrm {ker}} \lambda $ together with $\varpi $ generate the maximal ideal of R, we may assume that $f_1,\ldots ,f_d$ lie in $\operatorname {\mathrm {ker}} \lambda $ . Again by hypothesis $R[1/\varpi ]$ is Cohen–Macaulay of dimension d and formally smooth at $\lambda $ , and hence it is formally smooth at $\lambda $ of dimension d.

Then by Lemma 3.5, there exist $h_1,\ldots ,h_d\in \operatorname {\mathrm {ker}} \lambda $ such that $h_1,\ldots ,h_d,\varpi $ is a regular sequence in R and such that for $B=R/(h_1,\ldots ,h_d)$ and the induced augmentation $\lambda _B\colon B\to {\mathcal O}$ the ring $B[1/\varpi ]$ is formally smooth at $\lambda _B$ of dimension $0$ . It follows that the continuous ${\mathcal O}$ -algebra map $\theta \colon S={\mathcal O}[[y_1,\ldots ,y_d]]\to R$ with $y_i\mapsto f_i$ makes R into a finite free S-module such that in the notation of (P), we have $B=R_\theta $ and $\lambda _B=\lambda _\theta $ , and moreover $R_\theta $ is finite free over ${\mathcal O}$ . Hence, $R_\theta [1/\varpi ]$ is a product of Artin E-algebras, and the smoothness at $\lambda _\theta $ shows that the component corresponding to $\lambda _\theta $ is equal to E. From this, it follows that $\Phi _{\lambda _\theta }(R_\theta ) =\operatorname {\mathrm {ker}} \lambda _\theta /(\operatorname {\mathrm {ker}}\lambda _\theta )^2$ is of finite ${\mathcal O}$ -length, as it is finitely generated over ${\mathcal O}$ and ${\mathcal O}$ -torsion.

Proof. Pick any lift $\widetilde {x}_0\in {\mathfrak m}_A$ of x. Let the set of minimal primes of A be $\{P_1,\ldots ,P_n\}$ , labeled so that there is some $0\le a\le n$ for which $\widetilde {x}_0\not \in P_1,P_2,\ldots ,P_a$ , and $\widetilde {x}_0\in P_{a+1},\ldots ,P_n$ .

Now, fix any $i>a$ , so that $\widetilde {x}_0\in P_i$ . Note that if $I\subseteq P_i$ then $P_i/I$ would be a minimal prime of B containing x, contradicting our assumption. Hence, $I\not \subseteq P_i$ , and so there is some $r_i\in I\smallsetminus P_i$ .

Also, for any $j\ne i$ , $P_j\not \subseteq P_i$ , and so there is some $s_{ij}\in P_j\smallsetminus P_i$ . Now, define

$$\begin{align*}y_i:= r_i\prod_{j\ne i} s_{ij}\end{align*}$$

so that $y_i\in I$ , $y_i\in P_j$ for $j\ne i$ and $y_i\not \in P_i$ . Finally, let

$$\begin{align*}\widetilde{x} = \widetilde{x}_0+y_{a+1}+y_{b+2}+\cdots+y_{n}.\end{align*}$$

Then we have $\widetilde {x}\equiv \widetilde {x}_0\equiv x{\operatorname {mod}}{I}$ , $\widetilde {x}\equiv \widetilde {x}_0\not \equiv 0\, {\operatorname {mod}}\, {P_i}$ for $i\le a$ and $\widetilde {x}\equiv y_i\not \equiv 0\, {\operatorname {mod}}\, {P_i}$ for $i>a$ . So $\widetilde {x}$ is our desired lift.

Proof of Lemma 3.7.

Identifying S with its image in R, we get that $(y_1,\ldots ,y_d,\varpi )$ , and thus $(\varpi ,y_1,\ldots ,y_d)$ , is a regular sequence for R. We claim that we can inductively construct a sequence $\widetilde {y}_1,\widetilde {y}_2,\ldots ,\widetilde {y}_d\in \widetilde {R}$ such that $\varphi _\infty (\widetilde {y}_i) = y_i$ for all i and $\dim \widetilde {R}/(\varpi ,\widetilde {y}_1,\ldots ,\widetilde {y}_j) = d-j = \dim R/(\varpi ,y_1,\ldots ,y_j)$ for all $0\le j\le d$ .

As $\widetilde {R}$ and R are both flat over ${\mathcal O}$ of relative dimension d, we have $\dim \widetilde {R}/(\varpi ) = d = \dim R/(\varpi )$ . Now, assume that $\widetilde {y}_1,\ldots ,\widetilde {y}_j$ have been constructed for some $j<d$ . Let $A_j = \widetilde {R}/(\varpi ,\widetilde {y}_1,\ldots ,\widetilde {y}_j)$ and $B_j = R/(\varpi ,y_1,\ldots ,y_j)$ so that $\varphi _\infty :\widetilde {R}\to R$ induces a map $\varphi _j:A_j\to B_j$ . As $(\varpi ,y_1,\ldots ,y_d)$ is a regular sequence for R, $y_{j+1}$ is by definition not a zero divisor in $B_j$ , and so in particular cannot be contained in any minimal primes of $B_j$ . By Lemma 3.8 it follows there is some $y_{j+1}'\in A_j$ with $\varphi _j(y_{j+1}') = y_{j+1}$ which is not contained in any minimal prime of $A_j$ . Let $\widetilde {y}_{j+1}\in \widetilde {R}$ be any lift of $y_{j+1}'$ . But now

$$\begin{align*}\widetilde{R}/(\varpi,\widetilde{y}_1,\ldots,\widetilde{y}_j,\widetilde{y}_{j+1})\cong A_j/(y_{j+1}')\end{align*}$$

which has dimension $\dim A_j - 1 = d-(j+1)$ , by the assumption that $y_{j+1}'$ is not contained in any minimal prime of $A_j$ . This completes the induction.

Now, $(\varpi ,\widetilde {y}_1,\ldots ,\widetilde {y}_d)$ is a system of parameters for $\widetilde {R}$ . As $\widetilde {R}$ is a complete intersection and thus Cohen–Macaulay, it follows that $(\varpi ,\widetilde {y}_1,\ldots ,\widetilde {y}_d)$ , and thus $(y_1,\ldots ,y_d,\varpi )$ , is a regular sequence for $\widetilde {R}$ .

So now defining ${\widetilde {\theta }}:S\to \widetilde {R}$ by ${\widetilde {\theta }}(y_i)=\widetilde {y}_i$ makes $\widetilde {R}$ into a finite free S module, as desired.

The fact that $\widetilde {R}_\theta $ is a complete intersection of dimension $1$ , and finite free over ${\mathcal O}$ , now follows immediately from the fact that $\widetilde {R}$ is a complete intersection. For the last assertion, the proof of [Reference Böckle, Khare and ManningBKM21, Theorem 7.16] gives rise to a commutative diagram with exact rows:

where $\Phi _{\lambda }(R) = (\ker \lambda )/(\ker \lambda )^2 = {\widehat {\Omega }}_{R/{\mathcal O}}\otimes _\lambda {\mathcal O}$ and $\Phi _{\widetilde {\lambda }}(\widetilde {R}) = (\ker \widetilde {\lambda })/(\ker \widetilde {\lambda })^2 = {\widehat {\Omega }}_{\widetilde {R}/{\mathcal O}}\otimes _{\widetilde {\lambda }}{\mathcal O}$ , and the maps $\Theta $ and ${\widetilde {\Theta }}$ are given in terms of differentials by $e_i\mapsto dy_i$ .

Now, as in [Reference Böckle, Khare and ManningBKM21, Theorem 7.16], the fact that $\operatorname {\mathrm {Spec}} R[1/\varpi ]$ and $\operatorname {\mathrm {Spec}}\widetilde {R}[1/\varpi ]$ are both equidimensional of dimension d and $\lambda $ and $\widetilde {\lambda }$ , respectively, correspond to formally smooth points on these schemes, implies that $\Phi _{\lambda }(R)$ and $\Phi _{\widetilde {\lambda }}(\widetilde {R})$ both have rank d as ${\mathcal O}$ -modules.

But now the fact that $\Phi _{\lambda _\theta }(R_\theta )$ is finite implies that $\Theta $ must be injective. By commutativity, this implies that ${\widetilde {\Theta }}$ is also injective, which in turn implies that $\Phi _{\widetilde {\lambda }_\theta }(\widetilde {R}_\theta )$ is also finite.

Proof of Theorem 3.9(1).

Clearly, we have $\pi _\theta (\widetilde {R}[I]) \subseteq \widetilde {R}_\theta [I_\theta ]$ (since $I_\theta = \pi _\theta (I)$ and so $\widetilde {R}_\theta [I_\theta ] = \widetilde {R}[I]$ ), so it suffices to prove that $\pi _\theta |_{\widetilde {R}[I]}:\widetilde {R}[I]\to \widetilde {R}_\theta [I_\theta ]$ is surjective.

We first note that as $\widetilde {R}$ and $\widetilde {R}_\theta $ are complete intersections, and thus are Gorenstein, we get the following:

Lemma 3.10. There are isomorphisms $\Psi :\widetilde {R}\xrightarrow {\sim } \operatorname {\mathrm {Hom}}_{S}(\widetilde {R},S)$ and $\Psi _\theta :\widetilde {R}_\theta \xrightarrow {\sim } \operatorname {\mathrm {Hom}}_{{\mathcal O}}(\widetilde {R}_\theta ,{\mathcal O})$ of $\widetilde {R}$ -modules, fitting into a commutative diagram:

where the vertical map $\sigma :\operatorname {\mathrm {Hom}}_{S}(\widetilde {R},S)\to \operatorname {\mathrm {Hom}}_{S}(\widetilde {R},{\mathcal O}) = \operatorname {\mathrm {Hom}}_{{\mathcal O}}(\widetilde {R}_\theta ,{\mathcal O})$ is just composition with the map $S \to S/(y_1,\ldots ,y_d)={\mathcal O}$ .

Proof. As $\widetilde {R}$ is Cohen–Macaulay and free of finite rank over S, we have $\omega _{\widetilde {R}} \cong \operatorname {\mathrm {Hom}}_{S}(\widetilde {R},S)$ . But as $\widetilde {R}$ is a complete intersection, it is Gorenstein, and so $\omega _{\widetilde {R}}\cong \widetilde {R}$ . Composing these isomorphisms gives the desired isomorphism $\Psi :\widetilde {R} \xrightarrow {\sim } \operatorname {\mathrm {Hom}}_{S}(\widetilde {R},S)$ .

Now, note that (as $\widetilde {R}$ is a free S-module):

$$ \begin{align*} \Psi(\ker \pi_\theta) &= \Psi(y_1\widetilde{R}+\cdots+y_d\widetilde{R}) = y_1\Psi(\widetilde{R})+\cdots+y_d\Psi(\widetilde{R})\\ &= y_1\operatorname{\mathrm{Hom}}_{S}(\widetilde{R},S)+\cdots+y_d\operatorname{\mathrm{Hom}}_{S}(\widetilde{R},S) = \operatorname{\mathrm{Hom}}_{S}(\widetilde{R},y_1S+\cdots +y_dS)\\ &= \ker \sigma, \end{align*} $$

which implies that there is an injection $\Psi _\theta :\widetilde {R}\xrightarrow {\sim } \operatorname {\mathrm {Hom}}_{{\mathcal O}}(\widetilde {R},{\mathcal O})$ making the above diagram commute. As $\sigma $ is clearly surjective (since $\widetilde {R}$ is a projective S-module), it follows that $\Psi _\theta $ is also surjective.

Lemma 3.11. We have

$$\begin{align*}\Psi(\widetilde{R}[I]) = \{f:\widetilde{R} \to S|f(I) = 0\} = \operatorname{\mathrm{Hom}}_{S}(\widetilde{R}/I,S)\end{align*}$$

and

$$\begin{align*}\Psi_\theta(\widetilde{R}_\theta[I_\theta]) = \{f:\widetilde{R}_\theta\to {\mathcal O}|f(I_\theta) = 0\} = \operatorname{\mathrm{Hom}}_{{\mathcal O}}(\widetilde{R}_\theta/I_\theta,{\mathcal O}) = \operatorname{\mathrm{Hom}}_{S}(\widetilde{R}/I,{\mathcal O}).\end{align*}$$

Proof. As $\Psi $ is an isomorphism of $\widetilde {R}$ -modules, we have $\Psi (\widetilde {R}[I]) = \operatorname {\mathrm {Hom}}_{S}(\widetilde {R},S)[I]$ and thus

$$ \begin{align*} \Psi(\widetilde{R}[I]) &= \{f:\widetilde{R}\to S|rf=0 \text{ for all }r\in I\}\\ &= \{f:\widetilde{R}\to S|(rf)(x)=0 \text{ for all }r\in I\text{ and } x\in \widetilde{R}\}\\ &= \{f:\widetilde{R}\to S|f(rx)=0 \text{ for all }r\in I\text{ and } x\in \widetilde{R}\}\\ &= \{f:\widetilde{R}\to S|f(I) = 0\}\\ &=\operatorname{\mathrm{Hom}}_{S}(\widetilde{R}/I,S). \end{align*} $$

The proof for $\Psi _\theta (\widetilde {R}_\theta [I_\theta ])$ is identical.

Now, since $\widetilde {R}/I\cong R$ is a projective S-module, $\sigma $ induces a surjective map $\operatorname {\mathrm {Hom}}_{S}(\widetilde {R}/I,S)\to \operatorname {\mathrm {Hom}}_{S}(\widetilde {R}/I,{\mathcal O})$ . By Lemma 3.11, this is a surjective map $\sigma |_{\Psi (\widetilde {R}[I])}:\Psi (\widetilde {R}[I])\to \Psi _\theta (\widetilde {R}_\theta [I_\theta ])$ , so the commutative diagram from Lemma 3.10 gives that $\pi _\theta |_{\widetilde {R}[I]}:\widetilde {R}[I]\to \widetilde {R}_\theta [I_\theta ]$ is surjective. This completes the proof of (1).

Proof of Theorem 3.9(2).

By the definition of I, we have a short exact sequence of S-modules

$$\begin{align*}0\to I \to \widetilde{R}\xrightarrow{\varphi} R \to 0.\end{align*}$$

Applying $-\otimes _{S}{\mathcal O}$ to this gives an exact sequence

$$\begin{align*}\operatorname{\mathrm{Tor}}_1^{S}(R,{\mathcal O})\to I\otimes_{S}{\mathcal O}\to \widetilde{R}_\theta\xrightarrow{\varphi} R_\theta\to 0.\end{align*}$$

and so as $I_\theta = \ker \varphi _\theta $ , this gives as exact sequence

$$\begin{align*}\operatorname{\mathrm{Tor}}_1^{S}(R,{\mathcal O})\to I\otimes_{S}{\mathcal O}\to I_\theta\to 0.\end{align*}$$

But now as R is a finite free S-module, $\operatorname {\mathrm {Tor}}_1^{S}(R,{\mathcal O}) = 0$ and so we have an isomorphism $I\otimes _{S}{\mathcal O}\cong I_\theta $ of $\widetilde {R}_\theta $ -modules.

Now, by [Sta19, Lemma 07ZA] we indeed have:

$$\begin{align*}\pi_\theta(\operatorname{\mathrm{Fitt}}(I)) = \operatorname{\mathrm{Fitt}}(I\otimes_{S}{\mathcal O}) = \operatorname{\mathrm{Fitt}}(I_\theta),\end{align*}$$

as desired. This completes the proof of (2) and hence of Theorem 3.9.

Proof. This follows from Theorem 3.9 which shows that

$$\begin{align*}C_{1,\lambda_\theta}(R_\theta) = C_{1,\widetilde{\lambda}}(\widetilde{R}),\end{align*}$$

and the results of Appendix A which show that $C_{1,\lambda _\theta }(R_\theta )$ is well defined and independent of $\widetilde {R}_\theta $ .

Proof. As R is Cohen–Macaulay and $\widetilde {R}$ is Gorenstein, we have that $\omega _{R}\cong \operatorname {\mathrm {Hom}}_{S}(R,S)$ and $\widetilde {R}\cong \operatorname {\mathrm {Hom}}_{S}(\widetilde {R},S)$ as $\widetilde {R}$ -modules. Now, by [Sta19, Lemma 08YP]:

$$ \begin{align*} \widetilde{R}[I] &\cong \operatorname{\mathrm{Hom}}_{\widetilde{R}}(R,\widetilde{R}) \cong \operatorname{\mathrm{Hom}}_{\widetilde{R}}(R,\operatorname{\mathrm{Hom}}_{S}(\widetilde{R},S)) \cong \operatorname{\mathrm{Hom}}_{S}(R,S) \cong \omega_{R} \end{align*} $$

as R-modules.

Proof. This follows from the distinguished triangle

$$\begin{align*}C\otimes^{\mathbf{L}}_B L_{B/A}^{\operatorname{\mathrm{an}}}\to L_{C/A}^{\operatorname{\mathrm{an}}}\to L_{C/B}^{\operatorname{\mathrm{an}}}\to C\otimes^{\mathbf{L}}_B L_{B/A}^{\operatorname{\mathrm{an}}}[1] \end{align*}$$

from [Reference Gabber and RameroGR03, Theorem 7.1.33].

Proof. As the map $A\to B$ is finite, it is finite type (and not merely topologically finite type). By [Reference IyengarIye07, 6.11], $L_{B/A}$ is quasi-isomorphic to a bounded above complex of finite free B-modules ${\mathcal L}^\bullet $ . Using ${\mathcal L}^\bullet $ to compute $(L_{B/A})^\wedge $ , we get

$$\begin{align*}L_{B/A}^{\operatorname{\mathrm{an}}} = (L_{B/A})^{\wedge} \cong ({\mathcal L}^{\bullet})^{\wedge} = {\mathcal L}^{\bullet} \cong L_{B/A}, \end{align*}$$

as finitely generated B-modules are already ${\mathfrak m}_B$ -adically complete. The last claim now follows from the definition of $\widehat {\operatorname {\mathrm {Der}}}^i_A(B,M)$ and $\operatorname {\mathrm {Der}}^i_A(B,M)$ .

Proof. For the first claim, we argue as in [Reference Böckle, Khare and ManningBKM21, Lemma 7.1] (and note that the assumption that ${\mathcal R}$ is finitely generated over A in that lemma was used only in the last step, to conclude that $\Omega _{{\mathcal R}/A}$ was finitely generated over A). Specifically, for any $n>k$ we have $\Omega _{B/A}/{\mathfrak m}_B^k\Omega _{B/A} = \Omega _{B/A}\otimes _B B/{\mathfrak m}_B^k\cong \Omega _{(B/{\mathfrak m}_B^n)/A}\otimes B/{\mathfrak m}_B^k$ and so taking inverse limits gives

$$\begin{align*}\Omega_{B/A}/{\mathfrak m}_B^k\Omega_{B/A} \cong\varprojlim_n\left(\Omega_{(B/{\mathfrak m}_B^n)/A}\otimes_B B/{\mathfrak m}_B^k\right) \cong\varprojlim_n\left(\Omega_{(B/{\mathfrak m}_B^n)/A}\right)\otimes_B B/{\mathfrak m}_B^k = {\widehat{\Omega}}_{B/A}\otimes_B B/{\mathfrak m}_B^k. \end{align*}$$

Taking inverse limits again and using the fact that ${\widehat {\Omega }}_{B/A}$ is finite over B, and hence ${\mathfrak m}_B$ -adically complete gives

$$\begin{align*}{\widehat{\Omega}}_{B/A}\cong \varprojlim{\widehat{\Omega}}_{B/A}\otimes_B B/{\mathfrak m}_B^k\cong \varprojlim_k\Omega_{B/A}/{\mathfrak m}_B^k\Omega_{B/A}\end{align*}$$

as desired.

In particular, this shows that the module ${\widehat {\Omega }}_{B/A}$ is simply the module $\Omega _{B/A}^{\operatorname {\mathrm {an}}} = (\Omega _{B/A})^{\wedge }$ from [Reference Gabber and RameroGR03], and so the second claim follows from [Reference Gabber and RameroGR03, Lemma 7.1.27(iii)] and the definition of $\widehat {\operatorname {\mathrm {Der}}}^i_A(B,M)$ .

Proof. By [Reference Gabber and RameroGR03, Proposition 7.1.29], we have $L^{\operatorname {\mathrm {an}}}_{{\mathcal O}[[x_1,\ldots ,x_n]]/{\mathcal O}} = {\widehat {\Omega }}_{{\mathcal O}[[x_1,\ldots ,x_n]]/{\mathcal O}}[0] = {\mathcal O}[[x_1,\ldots ,x_n]]^n[0]$ and so

$$\begin{align*}\operatorname{\mathrm{RHom}}_{{\mathcal O}[[x_1,\ldots,x_n]]}(L^{\operatorname{\mathrm{an}}}_{{\mathcal O}[[x_1,\ldots,x_n]]/{\mathcal O}},M) =\operatorname{\mathrm{RHom}}_{{\mathcal O}[[x_1,\ldots,x_n]]}({\mathcal O}[[x_1,\ldots,x_n]]^n[0],M) = M^n[0]\end{align*}$$

so the claim follows.

Proof. As $B=A/I$ is clearly finite over A, Proposition 3.16 gives $\widehat {\operatorname {\mathrm {Der}}}^i_A(B,M) = \operatorname {\mathrm {Der}}^i(B,M)$ for all $i\ge 0$ and all M. The claim now follows from [Reference IyengarIye07, 6.12].

Proof. The first and last isomorphisms follow from Proposition 3.16, as R is finite over S and $R_\theta $ is finite over ${\mathcal O}$ .

For the second isomorphism, first note that as R is a finite free S-module, it is a projective resolution for itself in $D(S)$ , and so we have $R\otimes ^{\mathbf {L}}_{S}{\mathcal O} = R\otimes _{S}{\mathcal O} \cong R_\theta $ . By [Sta19, Lemma 08QQ], this implies that $L_{R/S}\otimes ^{\mathbf {L}}_{R}R_\theta \cong L_{R_\theta /{\mathcal O}}$ . But now [Sta19, Lemma 0E1W] gives that

$$\begin{align*}\operatorname{\mathrm{RHom}}_{R}(L_{R/S},M) = \operatorname{\mathrm{RHom}}_{R_\theta}(L_{R/S}\otimes^{\mathbf{L}}_{R}R_\theta,M)\cong\operatorname{\mathrm{RHom}}_{R_\theta}(L_{R_\theta/{\mathcal O}},M)\end{align*}$$

so the claim follows by definition.

Proof. Applying Proposition 3.15 to the ring maps ${\mathcal O}\to S\to R$ gives an exact sequence:

$$ \begin{align*} 0 &\to \widehat{\operatorname{\mathrm{Der}}}^0_{S}(R,E/{\mathcal O})\to \widehat{\operatorname{\mathrm{Der}}}^0_{\mathcal O}(R,E/{\mathcal O})\to \widehat{\operatorname{\mathrm{Der}}}^0_{\mathcal O}(S,E/{\mathcal O})\\ &\to \widehat{\operatorname{\mathrm{Der}}}^1_{S}(R,E/{\mathcal O})\to \widehat{\operatorname{\mathrm{Der}}}^1_{\mathcal O}(R,E/{\mathcal O})\to \widehat{\operatorname{\mathrm{Der}}}^1_{\mathcal O}(S,E/{\mathcal O}). \end{align*} $$

By Lemma 3.18, $\widehat {\operatorname {\mathrm {Der}}}^0_{\mathcal O}(S,E/{\mathcal O}) = (E/{\mathcal O})^d$ and $\widehat {\operatorname {\mathrm {Der}}}^1_{\mathcal O}(S,E/{\mathcal O}) = 0$ .

But now by the assumption that $\lambda :R \to {\mathcal O}$ represents a smooth point of $\operatorname {\mathrm {Spec}} R[1/\varpi ]$ we get that ${\widehat {\Omega }}_{R/{\mathcal O}}\otimes _\lambda {\mathcal O}$ has rank d as an ${\mathcal O}$ -module (as in [Reference Böckle, Khare and ManningBKM21, Theorem 7.16]), and so

$$\begin{align*}\widehat{\operatorname{\mathrm{Der}}}^0_{\mathcal O}(R,E/{\mathcal O}) = \operatorname{\mathrm{Hom}}_{R}({\widehat{\Omega}}_{R/{\mathcal O}},E/{\mathcal O}) = \operatorname{\mathrm{Hom}}_{{\mathcal O}}({\widehat{\Omega}}_{R/{\mathcal O}}\otimes_\lambda{\mathcal O},E/{\mathcal O}) = (E/{\mathcal O})^d\oplus G,\end{align*}$$

for some finite group G. Also as $\Phi _{\lambda _\theta }(R_\theta )={\widehat {\Omega }}_{R^\theta /{\mathcal O}}\otimes _{\lambda _\theta }{\mathcal O}$ is finite (as $\theta $ satisfies (P)),

$$\begin{align*}\widehat{\operatorname{\mathrm{Der}}}^0_{S}(R,E/{\mathcal O}) \cong \widehat{\operatorname{\mathrm{Der}}}^0_{\mathcal O}(R_\theta,E/{\mathcal O}) = \operatorname{\mathrm{Hom}}_{{\mathcal O}}({\widehat{\Omega}}_{R_\theta/{\mathcal O}},E/{\mathcal O}) = \operatorname{\mathrm{Hom}}_{{\mathcal O}}(\Phi_{\lambda_\theta}(R_\theta),E/{\mathcal O})\end{align*}$$

is finite as well. Now, the exact sequence simplifies to

$$\begin{align*}0\to \widehat{\operatorname{\mathrm{Der}}}^0_{S}(R,E/{\mathcal O}) \to (E/{\mathcal O})^d\oplus G\to (E/{\mathcal O})^d\to \widehat{\operatorname{\mathrm{Der}}}^1_{S}(R,E/{\mathcal O})\to \widehat{\operatorname{\mathrm{Der}}}^1_{\mathcal O}(R,E/{\mathcal O})\to 0.\end{align*}$$

But comparing coranks in the sequence $0\to \widehat {\operatorname {\mathrm {Der}}}^0_{S}(R,E/{\mathcal O}) \to (E/{\mathcal O})^d\oplus G\to (E/{\mathcal O})^d$ implies that $(E/{\mathcal O})^d\oplus G\to (E/{\mathcal O})^d$ has finite cokernel and hence must be surjective, as $E/{\mathcal O}$ does not have any nontrivial finite quotients. This implies that the map $\widehat {\operatorname {\mathrm {Der}}}^1_{S}(R, E/{\mathcal O})\to \widehat {\operatorname {\mathrm {Der}}}^1_{\mathcal O}(R,E/{\mathcal O})$ is indeed an isomorphism. This completes the proof of Theorem 3.20.

Proof. Applying Proposition 3.15 to the ring maps ${\mathcal O}\to \widetilde {R} \to R$ gives an exact sequence:

$$ \begin{align*} 0 &\to \widehat{\operatorname{\mathrm{Der}}}^0_{\widetilde{R}}(R,E/{\mathcal O})\to \widehat{\operatorname{\mathrm{Der}}}^0_{\mathcal O}(R,E/{\mathcal O})\to \widehat{\operatorname{\mathrm{Der}}}^0_{\mathcal O}(\widetilde{R},E/{\mathcal O})\\ &\to \widehat{\operatorname{\mathrm{Der}}}^1_{\widetilde{R}}(R,E/{\mathcal O})\to \widehat{\operatorname{\mathrm{Der}}}^1_{\mathcal O}(R,E/{\mathcal O})\to \widehat{\operatorname{\mathrm{Der}}}^1_{{\mathcal O}}(\widetilde{R},E/{\mathcal O}), \end{align*} $$

and Lemma 3.19 implies that $\widehat {\operatorname {\mathrm {Der}}}^0_{\widetilde {R}}(R,E/{\mathcal O}) = 0$ and $\widehat {\operatorname {\mathrm {Der}}}^1_{\widetilde {R}}(R,E/{\mathcal O}) = \operatorname {\mathrm {Hom}}_{R}(I/I^2,E/{\mathcal O})$ , so it’s enough to prove that $ \widehat {\operatorname {\mathrm {Der}}}^1_{\mathcal O}(\widetilde {R},E/{\mathcal O}) = 0$ (since by Proposition 3.17, $\widehat {\operatorname {\mathrm {Der}}}^0_{\mathcal O}(R,E/{\mathcal O})=\operatorname {\mathrm {Hom}}_{R}({\widehat {\Omega }}_{R},E/{\mathcal O})$ and $\widehat {\operatorname {\mathrm {Der}}}^0_{\mathcal O}(\widetilde {R},E/{\mathcal O})=\operatorname {\mathrm {Hom}}_{\widetilde {R}}({\widehat {\Omega }}_{\widetilde {R}},E/{\mathcal O})$ ).

Since $\widetilde {R}$ is a complete intersection, we can write $\widetilde {R} = P/J$ , where $P = {\mathcal O}[[x_1,\ldots ,x_{d+n}]]$ and $J=(f_1,\ldots ,f_n)$ is generated by a regular sequence. Applying Proposition 3.15 to the ring maps ${\mathcal O}\to P\to \widetilde {R}$ gives an exact sequence:

$$ \begin{align*} 0 &\to \widehat{\operatorname{\mathrm{Der}}}^0_{P}(\widetilde{R},E/{\mathcal O})\to \widehat{\operatorname{\mathrm{Der}}}^0_{\mathcal O}(\widetilde{R},E/{\mathcal O})\to \widehat{\operatorname{\mathrm{Der}}}^0_{\mathcal O}(P,E/{\mathcal O})\\ &\to \widehat{\operatorname{\mathrm{Der}}}^1_{P}(\widetilde{R},E/{\mathcal O})\to \widehat{\operatorname{\mathrm{Der}}}^1_{\mathcal O}(\widetilde{R},E/{\mathcal O})\to \widehat{\operatorname{\mathrm{Der}}}^1_{\mathcal O}(P,E/{\mathcal O}). \end{align*} $$

Now, Lemma 3.18 gives the identification $\widehat {\operatorname {\mathrm {Der}}}^0_{\mathcal O}(P,E/{\mathcal O}) = (E/{\mathcal O})^{d+n}$ and $\widehat {\operatorname {\mathrm {Der}}}^1_{\mathcal O}(P,E/{\mathcal O}) = 0$ and Lemma 3.19 gives $\widehat {\operatorname {\mathrm {Der}}}^0_{P}(\widetilde {R},E/{\mathcal O}) = 0$ and $\widehat {\operatorname {\mathrm {Der}}}^1_{P}(\widetilde {R},E/{\mathcal O}) = \operatorname {\mathrm {Hom}}_{\widetilde {R}}(J/J^2,E/{\mathcal O})$ . Moreover, as J is generated by a regular sequence of length n, it follows that $J/J^2\cong (\widetilde {R})^n$ as $\widetilde {R}$ -modules, and so $\widehat {\operatorname {\mathrm {Der}}}^1_{P}(\widetilde {R},E/{\mathcal O}) = \operatorname {\mathrm {Hom}}_{\widetilde {R}}(J/J^2,E/{\mathcal O})\cong (E/{\mathcal O})^n$ . Thus, the above exact sequence simplifies to

$$\begin{align*}0 \to \widehat{\operatorname{\mathrm{Der}}}^0_{\mathcal O}(\widetilde{R},E/{\mathcal O})\to (E/{\mathcal O})^{n+d} \to (E/{\mathcal O})^n\to \widehat{\operatorname{\mathrm{Der}}}^1_{\mathcal O}(\widetilde{R},E/{\mathcal O})\to 0. \end{align*}$$

But now, just as in the proof of Proposition 3.22 above, the fact that $\operatorname {\mathrm {Spec}} \widetilde {R}[1/\varpi ]$ is smooth of dimension d at $\widetilde {\lambda }$ implies that $\widehat {\operatorname {\mathrm {Der}}}^0_{\mathcal O}(\widetilde {R},E/{\mathcal O}) \cong (E/{\mathcal O})^d\oplus H$ for some finite group H, and so comparing ranks gives that $(E/{\mathcal O})^{n+d} \to (E/{\mathcal O})^n$ has finite cokernel, and hence is surjective. Thus, $\widehat {\operatorname {\mathrm {Der}}}^1_{\mathcal O}(\widetilde {R},E/{\mathcal O})=0$ , and so the claim follows.

Proof. The proofs of the independence statements for $C_{1,\lambda _\theta }(R_\theta )$ and $\operatorname {\mathrm {Der}}_{\mathcal O}^1(R_\theta ,E/{\mathcal O})$ follow from Theorems 3.9 and 3.20, respectively. The assertion for the Wiles defect $\delta _{\lambda _\theta }(R_\theta )$ is then immediate from Theorem 2.2.

Proof. This is a consequence of Theorem 3.25, Theorem 3.9, Corollary 3.12 and Theorem 3.20, combined with Remark A.7 which confirms the finiteness of length of the terms involved in the one-dimensional case.

The final claim about changing the coefficient ring is easy to verify in the case when R is finite free over ${\mathcal O}$ (this fact was already noted in [Reference Böckle, Khare and ManningBKM21, Section 3] for $\delta _{\lambda }(R)$ ), and the general claim follows from this.

Proof. This follows from Proposition 3.16 and Proposition A.6 of Appendix A (cf. Theorem 2.2).

Proof. If R is a complete intersection, then $\widehat {\operatorname {\mathrm {Der}}}^1_{\mathcal O}(R,E/{\mathcal O})=0$ by the argument given in the proof of Theorem 3.23. Further $C_{1,\lambda }(R)=0$ (as we can take the CI cover $\widetilde R=R$ ). This gives that $\delta _\lambda (R)=0$ .

Conversely, assume $\delta _\lambda (R)=0$ . Then by our results we have a quotient $(R_\theta ,\lambda _\theta ) \in C_{\mathcal O}$ of $(R,\lambda ) \in C_{\mathcal O}$ by a regular sequence $(y_1,\ldots , y_d)$ , namely $R_\theta =R/(y_1,\ldots , y_d)$ and $\lambda _\theta : R \to R_\theta \to {\mathcal O}$ (the last map being $\lambda $ ) with $R_\theta $ of dimension one. Further, $\delta _{\lambda _\theta }(R_\theta )=\delta _\lambda (R)=0$ . Thus, by Lemma 2.1, $R_\theta $ is a complete intersection, which implies that R is a complete intersection.

Proof. By definition, (3) will follow from (1) and (2).

For (1), we will first reduce to dimension $1$ . Let $S_1 = {\mathcal O}[[x_1,\ldots ,x_{d_1}]]$ and $S_2 = {\mathcal O}[[y_1,\ldots ,y_{d_2}]]$ . By Proposition 3.2, we may find maps $\theta _1:S_1\hookrightarrow R_1$ and $\theta _2:S_2\hookrightarrow R_2$ satisfying (P). Then the map $\theta = \theta _1{\widehat {\otimes }}_{{\mathcal O}}\theta _2:S_1{\widehat {\otimes }}_{{\mathcal O}}S_2\hookrightarrow R$ satisfies (P) as well. So consider the rings

$$ \begin{align*} R_{1,\theta_1} &= R_1\otimes_{S_1}{\mathcal O},& R_{2,\theta_1} &= R_1\otimes_{S_1}{\mathcal O},& &\text{and}& R_\theta &= R\otimes_{S_1{\widehat{\otimes}}_{{\mathcal O}}S_2}{\mathcal O} = R_{1,\theta_1}\otimes_{\mathcal O} R_{2,\theta_2} \end{align*} $$

and note that these are all finite free over ${\mathcal O}$ .

By Theorem 3.20, we now have that

$$ \begin{align*} \widehat{\operatorname{\mathrm{Der}}}^1_{{\mathcal O}}(R_1,E/{\mathcal O}) &= \operatorname{\mathrm{Der}}^1_{{\mathcal O}}(R_{1,\theta_1},E/{\mathcal O}),\\ \widehat{\operatorname{\mathrm{Der}}}^1_{{\mathcal O}}(R_2,E/{\mathcal O}) &= \operatorname{\mathrm{Der}}^1_{{\mathcal O}}(R_{2,\theta_2},E/{\mathcal O}),\\ \widehat{\operatorname{\mathrm{Der}}}^1_{{\mathcal O}}(R,E/{\mathcal O}) &= \operatorname{\mathrm{Der}}^1_{{\mathcal O}}(R_{\theta},E/{\mathcal O}). \end{align*} $$

But now by [Sta19, Lemma 09DA], as $R_1$ and $R_2$ are both free over ${\mathcal O}$ , and hence Tor-independent, we have

$$ \begin{align*} L_{R_\theta/{\mathcal O}} &\cong L_{R_{1,\theta_1}\otimes_{\mathcal O} R_{2,\theta_2}/{\mathcal O}} \cong L_{R_{1,\theta_1}/{\mathcal O}}\otimes^{\mathbf{L}}_{R_{1,\theta_1}}R_\theta\oplus L_{R_{2,\theta_2}/{\mathcal O}}\otimes^{\mathbf{L}}_{R_{2,\theta_1}}R_\theta. \end{align*} $$

Thus,

$$ \begin{align*} &\widehat{\operatorname{\mathrm{Der}}}^1_{{\mathcal O}}(R,E/{\mathcal O})\\& \quad \cong \operatorname{\mathrm{Der}}^1_{{\mathcal O}}(R_{\theta},E/{\mathcal O}) = H^1(\operatorname{\mathrm{RHom}}_{R_\theta}(L_{R/{\mathcal O}},E/{\mathcal O}))\\& \quad \cong H^1(\operatorname{\mathrm{RHom}}_{R_\theta}(L_{R_{1,\theta_1}/{\mathcal O}}\otimes^{\mathbf{L}}_{R_{1,\theta_1}}R_\theta\oplus L_{R_{2,\theta_2}/{\mathcal O}}\otimes^{\mathbf{L}}_{R_{2,\theta_1}}R_\theta,E/{\mathcal O}))\\& \quad \cong H^1(\operatorname{\mathrm{RHom}}_{R_\theta}(L_{R_{1,\theta_1}/{\mathcal O}}\otimes^{\mathbf{L}}_{R_{1,\theta_1}}R_\theta,E/{\mathcal O}))\oplus H^1(\operatorname{\mathrm{RHom}}_{R_\theta}(L_{R_{2,\theta_2}/{\mathcal O}}\otimes^{\mathbf{L}}_{R_{2,\theta_1}}R_\theta,E/{\mathcal O}))\\& \quad \cong H^1(\operatorname{\mathrm{RHom}}_{R_{1,\theta_1}}(L_{R_{1,\theta_1}/{\mathcal O}},E/{\mathcal O}))\oplus H^1(\operatorname{\mathrm{RHom}}_{R_{2,\theta_2}}(L_{R_{2,\theta_2}/{\mathcal O}},E/{\mathcal O}))\\& \quad = \operatorname{\mathrm{Der}}^1_{{\mathcal O}}(R_{1,\theta_1},E/{\mathcal O})\oplus \operatorname{\mathrm{Der}}^1_{{\mathcal O}}(R_{2,\theta_2},E/{\mathcal O}) = \widehat{\operatorname{\mathrm{Der}}}^1_{{\mathcal O}}(R_1,E/{\mathcal O})\oplus \widehat{\operatorname{\mathrm{Der}}}^1_{{\mathcal O}}(R_2,E/{\mathcal O}) \end{align*} $$

and so (1) follows.

It remains to prove (2). Consider triples $(\widetilde {R}_{1},I_{1},\varphi _{1})$ and $(\widetilde {R}_{2},I_{2},\varphi _{2})$ satisfying (CI) (with $(R_1,\lambda _1)$ and $(R_{2},\lambda _2)$ , respectively, in place of $(R,\lambda )$ ).

Define $\widetilde {R}= \widetilde {R}_{1}{\widehat {\otimes }}_{{\mathcal O}}\widetilde {R}_{2}$ , and note that $I_{1}{\widehat {\otimes }}_{{\mathcal O}} \widetilde {R}_{2}$ and $\widetilde {R}_{1}{\widehat {\otimes }}_{{\mathcal O}}I_{2}$ are both ideals of $\widetilde {R}$ . Let $\varphi =\varphi _{1}\otimes \varphi _{2}:\widetilde {R}= \widetilde {R}_{1}{\widehat {\otimes }}_{{\mathcal O}}\widetilde {R}_{2}\twoheadrightarrow R_{1}{\widehat {\otimes }}_{{\mathcal O}} R_{2} = R$ , and note that $\ker \varphi = \left (I_{1}{\widehat {\otimes }}_{{\mathcal O}} \widetilde {R}_{2}\right )+\left (\widetilde {R}_{1}{\widehat {\otimes }}_{{\mathcal O}}I_{2}\right )$ . Denoting this ideal $I \subseteq \widetilde {R}$ , the triple $(\widetilde {R},I,\varphi )$ satisfies (CI). So by the definition of $c_{1,\lambda }$ ,

$$ \begin{align*} c_{1,\lambda_1}(R_{1})\log|{\mathcal O}/p| &= \log\left|\lambda_1(\widetilde{R}_{1}[I_{1}])/\lambda_1(\operatorname{\mathrm{Fitt}}(I_{1}))\right|\\ c_{1,\lambda_2}(R_{1})\log|{\mathcal O}/p| &= \log\left|\lambda_2(\widetilde{R}_{2}[I_{2}])/\lambda_2(\operatorname{\mathrm{Fitt}}(I_{2}))\right|\\ c_{1,\lambda}(R)\log|{\mathcal O}/p| &= \log\left|\lambda(\widetilde{R}[I_{}])/\lambda(\operatorname{\mathrm{Fitt}}(I))\right|. \end{align*} $$

The desired equality will now follow from Lemma 3.33.

Proof. For the first claim, standard properties of annihilators imply that

$$ \begin{align*} \widetilde{R}[I] & = \widetilde{R}\left[\left(I_{1}{\widehat{\otimes}}_{{\mathcal O}} \widetilde{R}_{2}\right)+\left(\widetilde{R}_{1}{\widehat{\otimes}}_{{\mathcal O}}I_{2}\right)\right] = \widetilde{R}_{}\left[\left(I_{1}{\widehat{\otimes}}_{{\mathcal O}} \widetilde{R}_{2}\right)\right]\cap \widetilde{R}_{}\left[\left(\widetilde{R}_{1}{\widehat{\otimes}}_{{\mathcal O}}I_{2}\right)\right]\\ &= \left(\widetilde{R}_{1}\left[I_{1}\right]{\widehat{\otimes}}_{{\mathcal O}}\widetilde{R}_{2}\right)\cap \left(\widetilde{R}_{1}{\widehat{\otimes}}_{{\mathcal O}}\widetilde{R}_{2}\left[I_{2}\right]\right) = \widetilde{R}_{1}\left[I_{1}\right]{\widehat{\otimes}}_{{\mathcal O}}\widetilde{R}_{2}\left[I_{2}\right] \end{align*} $$

(where we’ve used that fact that $\left (A{\widehat {\otimes }}_{{\mathcal O}}\widetilde {R}_{2}\right )\cap \left (\widetilde {R}_{1}{\widehat {\otimes }}_{{\mathcal O}}B\right ) = \left (A{\widehat {\otimes }}_{{\mathcal O}}\widetilde {R}_{2}\right ) \left (\widetilde {R}_{1}{\widehat {\otimes }}_{{\mathcal O}}B\right ) = A{\widehat {\otimes }}_{{\mathcal O}}B$ for any ideals $A\subseteq \widetilde {R}_{1}$ and $B\subseteq \widetilde {R}_{2}$ ). Thus,

$$\begin{align*}\lambda\left(\widetilde{R}_{}[I_{}]\right) = (\lambda_1\otimes \lambda_2)\left(\widetilde{R}_{1}\left[I_{1}\right]{\widehat{\otimes}}_{{\mathcal O}}\widetilde{R}_{2}\left[I_{2}\right]\right) = \lambda_1\left(\widetilde{R}_{1}\left[I_{1}\right]\right) \lambda_2 \left(\widetilde{R}_{2}\left[I_{2}\right]\right). \end{align*}$$

For the statement about fitting ideals, fix presentations

$$ \begin{align*} 0&\to K_1\to \widetilde{R}_{1}^{m} \xrightarrow{A} I_{1}\to 0\\ 0&\to K_2\to \widetilde{R}_{2}^{n} \xrightarrow{B} I_{2}\to 0, \end{align*} $$

where $K_i$ is a finitely generated $\widetilde {R}_{i}$ -module. Then A and B induce surjective maps $A\otimes \operatorname {\mathrm {Id}}:\widetilde {R}_{}^{m} = \widetilde {R}_{1}^{m}{\widehat {\otimes }}_{{\mathcal O}}\widetilde {R}_{2}\to I_{1}{\widehat {\otimes }}_{{\mathcal O}}\widetilde {R}_{2}$ and $\operatorname {\mathrm {Id}}\otimes B:\widetilde {R}_{}^{n} = \widetilde {R}_{1}{\widehat {\otimes }}_{{\mathcal O}}\widetilde {R}_{2}^n\to \widetilde {R}_{1}{\widehat {\otimes }}_{{\mathcal O}}I_{2}$ , and so we may combine them to produce a surjective map

$$\begin{align*}C = (A\otimes\operatorname{\mathrm{Id}})-(\operatorname{\mathrm{Id}}\otimes B) : \widetilde{R}_{}^{m+n}= \widetilde{R}_{}^m\oplus \widetilde{R}_{}^n\to \left(I_{1}{\widehat{\otimes}}_{{\mathcal O}} \widetilde{R}_{2}\right)+\left(\widetilde{R}_{1}{\widehat{\otimes}}_{{\mathcal O}}I_{2}\right) = I.\end{align*}$$

Write $K \subseteq \widetilde {R}_{}^{m+n}$ for the kernel of C.

By definition: $\operatorname {\mathrm {Fitt}}(I_{1})$ is the ideal of $\widetilde {R}_{1}$ generated by all elements of the form $\det \left (u_1,\ldots ,u_m\right )\in \widetilde {R}_{1}$ for $u_1,\ldots ,u_m\in K_1\subseteq \widetilde {R}_{1}^m$ ; $\operatorname {\mathrm {Fitt}}(I_{2})$ is the ideal of $\widetilde {R}_{2}$ generated by all elements of the form $\det \left (v_1,\ldots ,v_n\right )\in \widetilde {R}_{1}$ for $v_1,\ldots ,v_n\in K_2\subseteq \widetilde {R}_{2}^n$ ; and lastly $\operatorname {\mathrm {Fitt}}(I)$ is the ideal of $\widetilde {R}$ generated by all elements of the form $\det \left (w_1,\ldots ,w_{m+n}\right )\in \widetilde {R}$ for $w_1,\ldots ,w_{m+n}\in K\subseteq \widetilde {R}^{m+n}$ .

Now, given any $u_1,\ldots ,u_m\in K_1$ and $v_1,\ldots ,v_n\in K_2$ it’s easy to see that $\displaystyle \binom {u_i\otimes 1}{0}, \binom {0}{1\otimes v_j}\in K$ for all i and j, and so $\operatorname {\mathrm {Fitt}}(I_{})$ contains the element

$$\begin{align*}\det\begin{pmatrix} u_1\otimes 1&\cdots & u_m\otimes 1&0&\cdots&0\\ 0&\cdots&0&1\otimes v_1&\otimes&1\otimes v_n \end{pmatrix} =\det(u_1,\ldots,u_m)\otimes\det(v_1,\ldots,v_m). \end{align*}$$

It follows that $\operatorname {\mathrm {Fitt}}(I_{1}){\widehat {\otimes }}_{{\mathcal O}}\operatorname {\mathrm {Fitt}}(I_{2})\subseteq \operatorname {\mathrm {Fitt}}(I_{})$ and so $\lambda _1\left (\operatorname {\mathrm {Fitt}}(I_{1})\right )\lambda _2\left (\operatorname {\mathrm {Fitt}}(I_{2})\right )\subseteq \lambda \left (\operatorname {\mathrm {Fitt}}(I_{})\right )$ .

For the reverse inclusion, we will use the following simple lemma:

Proof. As $w\in K$ , we have $(A\otimes \operatorname {\mathrm {Id}})(w_1)-(\operatorname {\mathrm {Id}}\otimes B)(w_2) = C(w) = 0$ so let $r = (A\otimes \operatorname {\mathrm {Id}})(w_1) = (\operatorname {\mathrm {Id}}\otimes B)(w_2) \in \widetilde {R}$ . By the definitions of A and B, we have $r = (A\otimes \operatorname {\mathrm {Id}})(w_1) \in I_{1}{\widehat {\otimes }}_{{\mathcal O}}\widetilde {R}_{2}$ and $r = (\operatorname {\mathrm {Id}}\otimes B)(w_2) \in \widetilde {R}_{1}{\widehat {\otimes }}_{{\mathcal O}}I_{2}$ and so

$$\begin{align*}r \in \left(I_{1}{\widehat{\otimes}}_{{\mathcal O}}\widetilde{R}_{2}\right)\cap \left(\widetilde{R}_{1}{\widehat{\otimes}}_{{\mathcal O}}I_{2}\right) = I_{1}{\widehat{\otimes}}_{{\mathcal O}}I_{2}\end{align*}$$

Now, as $\lambda _1(I_{1}) = \lambda _2(I_{2})=0$ by assumption, we get that $(\lambda _1\otimes \operatorname {\mathrm {Id}})(r) = (\operatorname {\mathrm {Id}}\otimes \lambda _2)(r) = 0$ . Now, let $u = (\operatorname {\mathrm {Id}}\otimes \lambda _2)(w_1)\in \widetilde {R}_{1}^m$ and $v = (\lambda _1\otimes \operatorname {\mathrm {Id}})(w_2)\in \widetilde {R}_{2}^n$ so that

$$ \begin{align*} \lambda_1(u) &= \lambda_1((\operatorname{\mathrm{Id}}\otimes\lambda_2)(w_1)) = (\lambda_1\otimes\lambda_2)(w_1) = \lambda(w_1)\\ \lambda_2(v) &= \lambda_2((\lambda_1\otimes\operatorname{\mathrm{Id}})(w_2)) = (\lambda_1\otimes\lambda_2)(w_2) = \lambda(w_2) \end{align*} $$

and

$$ \begin{align*} A(u) &= (A\otimes\operatorname{\mathrm{Id}})(\operatorname{\mathrm{Id}}\otimes\lambda_2)(w_1) = (\operatorname{\mathrm{Id}}\otimes\lambda_2)(A\otimes\operatorname{\mathrm{Id}})(w_1) = (\operatorname{\mathrm{Id}}\otimes\lambda_2)(r) = 0\\ B(v) &= (\operatorname{\mathrm{Id}}\otimes B)(\lambda_1\otimes\operatorname{\mathrm{Id}})(w_2) = (\lambda_1\otimes\operatorname{\mathrm{Id}})(\operatorname{\mathrm{Id}}\otimes B)(w_2) = (\lambda_1\otimes\operatorname{\mathrm{Id}})(r) = 0. \end{align*} $$

So now $w_1\in \ker A = K_1$ and $w_2 \in \ker B = K_2$ , as desired.

So now take any $w_1,\ldots ,w_{m+n}\in K$ . The lemma allows us to write $\displaystyle \lambda (w_i) = \binom {\lambda _1(u_i)}{\lambda _2(v_1)}$ for $u_i\in K_1$ and $v_i\in K_2$ , which gives

$$\begin{align*}\lambda(\det(w_1,\ldots,w_{m+n})) = \det \begin{pmatrix} \lambda_1(u_1)&\cdots&\lambda_1(u_{m+n})\\ \lambda_2(v_1)&\cdots&\lambda_2(v_{m+n}) \end{pmatrix}. \end{align*}$$

But now by standard properties of determinants, the determinant of this $(m+n)\times (m+n)$ matrix may be written as an alternating sum in the form

$$\begin{align*}\sum_{X,Y}(\pm1) \det\big((\lambda_1(u_i))_{i\in X}\big)\det\big((\lambda_2(v_j))_{j\in Y}\big) = \sum_{X,Y}(\pm1) \lambda_1\left(\det\big((u_i)_{i\in X}\big)\right)\lambda_2\left(\det\big((v_j)_{j\in Y}\big)\right) \end{align*}$$

(where the sum is taken over partitions $X\sqcup Y = \{1,\ldots ,m+n\}$ with $|X|=m$ and $|Y|=n$ ). As this sum is in $\lambda _1\left (\operatorname {\mathrm {Fitt}}(I_{1})\right )\lambda _2\left (\operatorname {\mathrm {Fitt}}(I_{2})\right )$ , it follows that $\lambda \left (\operatorname {\mathrm {Fitt}}(I_{})\right )\subseteq \lambda _1\left (\operatorname {\mathrm {Fitt}}(I_{1})\right ) \lambda _2\left (\operatorname {\mathrm {Fitt}}(I_{2})\right )$ , giving the desired equality $\lambda \left (\operatorname {\mathrm {Fitt}}(I_{})\right )=\lambda _1\left (\operatorname {\mathrm {Fitt}}(I_{1})\right )\lambda _2\left (\operatorname {\mathrm {Fitt}}(I_{2})\right )$ , and completing the proof.

Proof. The proof follows from the aruguments given in the proof of [Lemma 3.2, Reference Barnet-Lamb, Geraghty, Harris and TaylorBLGHT11] which contains a proof of Lemma 4.1.