The Stacks project

42.68.12 Periodic complexes and determinants

Let $R$ be a local ring with residue field $\kappa $. Let $(M, \varphi , \psi )$ be a $(2, 1)$-periodic complex over $R$. Assume that $M$ has finite length and that $(M, \varphi , \psi )$ is exact. We are going to use the determinant construction to define an invariant of this situation. See Subsection 42.68.1. Let us abbreviate $K_\varphi = \mathop{\mathrm{Ker}}(\varphi )$, $I_\varphi = \mathop{\mathrm{Im}}(\varphi )$, $K_\psi = \mathop{\mathrm{Ker}}(\psi )$, and $I_\psi = \mathop{\mathrm{Im}}(\psi )$. The short exact sequences

\[ 0 \to K_\varphi \to M \to I_\varphi \to 0, \quad 0 \to K_\psi \to M \to I_\psi \to 0 \]

give isomorphisms

\[ \gamma _\varphi : \det \nolimits _\kappa (K_\varphi ) \otimes \det \nolimits _\kappa (I_\varphi ) \longrightarrow \det \nolimits _\kappa (M), \quad \gamma _\psi : \det \nolimits _\kappa (K_\psi ) \otimes \det \nolimits _\kappa (I_\psi ) \longrightarrow \det \nolimits _\kappa (M), \]

see Lemma 42.68.6. On the other hand the exactness of the complex gives equalities $K_\varphi = I_\psi $, and $K_\psi = I_\varphi $ and hence an isomorphism

\[ \sigma : \det \nolimits _\kappa (K_\varphi ) \otimes \det \nolimits _\kappa (I_\varphi ) \longrightarrow \det \nolimits _\kappa (K_\psi ) \otimes \det \nolimits _\kappa (I_\psi ) \]

by switching the factors. Using this notation we can define our invariant.

Definition 42.68.13. Let $R$ be a local ring with residue field $\kappa $. Let $(M, \varphi , \psi )$ be a $(2, 1)$-periodic complex over $R$. Assume that $M$ has finite length and that $(M, \varphi , \psi )$ is exact. The determinant of $(M, \varphi , \psi )$ is the element

\[ \det \nolimits _\kappa (M, \varphi , \psi ) \in \kappa ^* \]

such that the composition

\[ \det \nolimits _\kappa (M) \xrightarrow {\gamma _\psi \circ \sigma \circ \gamma _\varphi ^{-1}} \det \nolimits _\kappa (M) \]

is multiplication by $(-1)^{\text{length}_ R(I_\varphi )\text{length}_ R(I_\psi )} \det \nolimits _\kappa (M, \varphi , \psi )$.

Remark 42.68.14. Here is a more down to earth description of the determinant introduced above. Let $R$ be a local ring with residue field $\kappa $. Let $(M, \varphi , \psi )$ be a $(2, 1)$-periodic complex over $R$. Assume that $M$ has finite length and that $(M, \varphi , \psi )$ is exact. Let us abbreviate $I_\varphi = \mathop{\mathrm{Im}}(\varphi )$, $I_\psi = \mathop{\mathrm{Im}}(\psi )$ as above. Assume that $\text{length}_ R(I_\varphi ) = a$ and $\text{length}_ R(I_\psi ) = b$, so that $a + b = \text{length}_ R(M)$ by exactness. Choose admissible sequences $x_1, \ldots , x_ a \in I_\varphi $ and $y_1, \ldots , y_ b \in I_\psi $ such that the symbol $[x_1, \ldots , x_ a]$ generates $\det _\kappa (I_\varphi )$ and the symbol $[x_1, \ldots , x_ b]$ generates $\det _\kappa (I_\psi )$. Choose $\tilde x_ i \in M$ such that $\varphi (\tilde x_ i) = x_ i$. Choose $\tilde y_ j \in M$ such that $\psi (\tilde y_ j) = y_ j$. Then $\det _\kappa (M, \varphi , \psi )$ is characterized by the equality

\[ [x_1, \ldots , x_ a, \tilde y_1, \ldots , \tilde y_ b] = (-1)^{ab} \det \nolimits _\kappa (M, \varphi , \psi ) [y_1, \ldots , y_ b, \tilde x_1, \ldots , \tilde x_ a] \]

in $\det _\kappa (M)$. This also explains the sign.

Lemma 42.68.15. Let $R$ be a local ring with residue field $\kappa $. Let $(M, \varphi , \psi )$ be a $(2, 1)$-periodic complex over $R$. Assume that $M$ has finite length and that $(M, \varphi , \psi )$ is exact. Then

\[ \det \nolimits _\kappa (M, \varphi , \psi ) \det \nolimits _\kappa (M, \psi , \varphi ) = 1. \]

Proof. Omitted. $\square$

Lemma 42.68.16. Let $R$ be a local ring with residue field $\kappa $. Let $(M, \varphi , \varphi )$ be a $(2, 1)$-periodic complex over $R$. Assume that $M$ has finite length and that $(M, \varphi , \varphi )$ is exact. Then $\text{length}_ R(M) = 2 \text{length}_ R(\mathop{\mathrm{Im}}(\varphi ))$ and

\[ \det \nolimits _\kappa (M, \varphi , \varphi ) = (-1)^{\text{length}_ R(\mathop{\mathrm{Im}}(\varphi ))} = (-1)^{\frac{1}{2}\text{length}_ R(M)} \]

Proof. Follows directly from the sign rule in the definitions. $\square$

Lemma 42.68.17. Let $R$ be a local ring with residue field $\kappa $. Let $M$ be a finite length $R$-module.

  1. if $\varphi : M \to M$ is an isomorphism then $\det _\kappa (M, \varphi , 0) = \det _\kappa (\varphi )$.

  2. if $\psi : M \to M$ is an isomorphism then $\det _\kappa (M, 0, \psi ) = \det _\kappa (\psi )^{-1}$.

Proof. Let us prove (1). Set $\psi = 0$. Then we may, with notation as above Definition 42.68.13, identify $K_\varphi = I_\psi = 0$, $I_\varphi = K_\psi = M$. With these identifications, the map

\[ \gamma _\varphi : \kappa \otimes \det \nolimits _\kappa (M) = \det \nolimits _\kappa (K_\varphi ) \otimes \det \nolimits _\kappa (I_\varphi ) \longrightarrow \det \nolimits _\kappa (M) \]

is identified with $\det _\kappa (\varphi ^{-1})$. On the other hand the map $\gamma _\psi $ is identified with the identity map. Hence $\gamma _\psi \circ \sigma \circ \gamma _\varphi ^{-1}$ is equal to $\det _\kappa (\varphi )$ in this case. Whence the result. We omit the proof of (2). $\square$

Lemma 42.68.18. Let $R$ be a local ring with residue field $\kappa $. Suppose that we have a short exact sequence of $(2, 1)$-periodic complexes

\[ 0 \to (M_1, \varphi _1, \psi _1) \to (M_2, \varphi _2, \psi _2) \to (M_3, \varphi _3, \psi _3) \to 0 \]

with all $M_ i$ of finite length, and each $(M_1, \varphi _1, \psi _1)$ exact. Then

\[ \det \nolimits _\kappa (M_2, \varphi _2, \psi _2) = \det \nolimits _\kappa (M_1, \varphi _1, \psi _1) \det \nolimits _\kappa (M_3, \varphi _3, \psi _3). \]

in $\kappa ^*$.

Proof. Let us abbreviate $I_{\varphi , i} = \mathop{\mathrm{Im}}(\varphi _ i)$, $K_{\varphi , i} = \mathop{\mathrm{Ker}}(\varphi _ i)$, $I_{\psi , i} = \mathop{\mathrm{Im}}(\psi _ i)$, and $K_{\psi , i} = \mathop{\mathrm{Ker}}(\psi _ i)$. Observe that we have a commutative square

\[ \xymatrix{ & 0 \ar[d] & 0 \ar[d] & 0 \ar[d] & \\ 0 \ar[r] & K_{\varphi , 1} \ar[r] \ar[d] & K_{\varphi , 2} \ar[r] \ar[d] & K_{\varphi , 3} \ar[r] \ar[d] & 0 \\ 0 \ar[r] & M_1 \ar[r] \ar[d] & M_2 \ar[r] \ar[d] & M_3 \ar[r] \ar[d] & 0 \\ 0 \ar[r] & I_{\varphi , 1} \ar[r] \ar[d] & I_{\varphi , 2} \ar[r] \ar[d] & I_{\varphi , 3} \ar[r] \ar[d] & 0 \\ & 0 & 0 & 0 & } \]

of finite length $R$-modules with exact rows and columns. The top row is exact since it can be identified with the sequence $I_{\psi , 1} \to I_{\psi , 2} \to I_{\psi , 3} \to 0$ of images, and similarly for the bottom row. There is a similar diagram involving the modules $I_{\psi , i}$ and $K_{\psi , i}$. By definition $\det _\kappa (M_2, \varphi _2, \psi _2)$ corresponds, up to a sign, to the composition of the left vertical maps in the following diagram

\[ \xymatrix{ \det _\kappa (M_1) \otimes \det _\kappa (M_3) \ar[r]^\gamma \ar[d]^{\gamma ^{-1} \otimes \gamma ^{-1}} & \det _\kappa (M_2) \ar[d]^{\gamma ^{-1}} \\ \det \nolimits _\kappa (K_{\varphi , 1}) \otimes \det \nolimits _\kappa (I_{\varphi , 1}) \otimes \det \nolimits _\kappa (K_{\varphi , 3}) \otimes \det \nolimits _\kappa (I_{\varphi , 3}) \ar[d]^{\sigma \otimes \sigma } \ar[r]^-{\gamma \otimes \gamma } & \det \nolimits _\kappa (K_{\varphi , 2}) \otimes \det \nolimits _\kappa (I_{\varphi , 2}) \ar[d]^\sigma \\ \det \nolimits _\kappa (K_{\psi , 1}) \otimes \det \nolimits _\kappa (I_{\psi , 1}) \otimes \det \nolimits _\kappa (K_{\psi , 3}) \otimes \det \nolimits _\kappa (I_{\psi , 3}) \ar[d]^{\gamma \otimes \gamma } \ar[r]^-{\gamma \otimes \gamma } & \det \nolimits _\kappa (K_{\psi , 2}) \otimes \det \nolimits _\kappa (I_{\psi , 2}) \ar[d]^\gamma \\ \det _\kappa (M_1) \otimes \det _\kappa (M_3) \ar[r]^\gamma & \det _\kappa (M_2) } \]

The top and bottom squares are commutative up to sign by applying Lemma 42.68.6 (2). The middle square is trivially commutative (we are just switching factors). Hence we see that $\det \nolimits _\kappa (M_2, \varphi _2, \psi _2) = \epsilon \det \nolimits _\kappa (M_1, \varphi _1, \psi _1) \det \nolimits _\kappa (M_3, \varphi _3, \psi _3) $ for some sign $\epsilon $. And the sign can be worked out, namely the outer rectangle in the diagram above commutes up to

\begin{eqnarray*} \epsilon & = & (-1)^{\text{length}(I_{\varphi , 1})\text{length}(K_{\varphi , 3}) + \text{length}(I_{\psi , 1})\text{length}(K_{\psi , 3})} \\ & = & (-1)^{\text{length}(I_{\varphi , 1})\text{length}(I_{\psi , 3}) + \text{length}(I_{\psi , 1})\text{length}(I_{\varphi , 3})} \end{eqnarray*}

(proof omitted). It follows easily from this that the signs work out as well. $\square$

Example 42.68.19. Let $k$ be a field. Consider the ring $R = k[T]/(T^2)$ of dual numbers over $k$. Denote $t$ the class of $T$ in $R$. Let $M = R$ and $\varphi = ut$, $\psi = vt$ with $u, v \in k^*$. In this case $\det _ k(M)$ has generator $e = [t, 1]$. We identify $I_\varphi = K_\varphi = I_\psi = K_\psi = (t)$. Then $\gamma _\varphi (t \otimes t) = u^{-1}[t, 1]$ (since $u^{-1} \in M$ is a lift of $t \in I_\varphi $) and $\gamma _\psi (t \otimes t) = v^{-1}[t, 1]$ (same reason). Hence we see that $\det _ k(M, \varphi , \psi ) = -u/v \in k^*$.

Example 42.68.20. Let $R = \mathbf{Z}_ p$ and let $M = \mathbf{Z}_ p/(p^ l)$. Let $\varphi = p^ b u$ and $\varphi = p^ a v$ with $a, b \geq 0$, $a + b = l$ and $u, v \in \mathbf{Z}_ p^*$. Then a computation as in Example 42.68.19 shows that

\begin{eqnarray*} \det \nolimits _{\mathbf{F}_ p}(\mathbf{Z}_ p/(p^ l), p^ bu, p^ av) & = & (-1)^{ab}u^ a/v^ b \bmod p \\ & = & (-1)^{\text{ord}_ p(\alpha )\text{ord}_ p(\beta )} \frac{\alpha ^{\text{ord}_ p(\beta )}}{\beta ^{\text{ord}_ p(\alpha )}} \bmod p \end{eqnarray*}

with $\alpha = p^ bu, \beta = p^ av \in \mathbf{Z}_ p$. See Lemma 42.68.37 for a more general case (and a proof).

Example 42.68.21. Let $R = k$ be a field. Let $M = k^{\oplus a} \oplus k^{\oplus b}$ be $l = a + b$ dimensional. Let $\varphi $ and $\psi $ be the following diagonal matrices

\[ \varphi = \text{diag}(u_1, \ldots , u_ a, 0, \ldots , 0), \quad \psi = \text{diag}(0, \ldots , 0, v_1, \ldots , v_ b) \]

with $u_ i, v_ j \in k^*$. In this case we have

\[ \det \nolimits _ k(M, \varphi , \psi ) = \frac{u_1 \ldots u_ a}{v_1 \ldots v_ b}. \]

This can be seen by a direct computation or by computing in case $l = 1$ and using the additivity of Lemma 42.68.18.

Example 42.68.22. Let $R = k$ be a field. Let $M = k^{\oplus a} \oplus k^{\oplus a}$ be $l = 2a$ dimensional. Let $\varphi $ and $\psi $ be the following block matrices

\[ \varphi = \left( \begin{matrix} 0 & U \\ 0 & 0 \end{matrix} \right), \quad \psi = \left( \begin{matrix} 0 & V \\ 0 & 0 \end{matrix} \right), \]

with $U, V \in \text{Mat}(a \times a, k)$ invertible. In this case we have

\[ \det \nolimits _ k(M, \varphi , \psi ) = (-1)^ a\frac{\det (U)}{\det (V)}. \]

This can be seen by a direct computation. The case $a = 1$ is similar to the computation in Example 42.68.19.

Example 42.68.23. Let $R = k$ be a field. Let $M = k^{\oplus 4}$. Let

\[ \varphi = \left( \begin{matrix} 0 & 0 & 0 & 0 \\ u_1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & u_2 & 0 \end{matrix} \right) \quad \varphi = \left( \begin{matrix} 0 & 0 & 0 & 0 \\ 0 & 0 & v_2 & 0 \\ 0 & 0 & 0 & 0 \\ v_1 & 0 & 0 & 0 \end{matrix} \right) \quad \]

with $u_1, u_2, v_1, v_2 \in k^*$. Then we have

\[ \det \nolimits _ k(M, \varphi , \psi ) = -\frac{u_1u_2}{v_1v_2}. \]

Next we come to the analogue of the fact that the determinant of a composition of linear endomorphisms is the product of the determinants. To avoid very long formulae we write $I_\varphi = \mathop{\mathrm{Im}}(\varphi )$, and $K_\varphi = \mathop{\mathrm{Ker}}(\varphi )$ for any $R$-module map $\varphi : M \to M$. We also denote $\varphi \psi = \varphi \circ \psi $ for a pair of morphisms $\varphi , \psi : M \to M$.

Lemma 42.68.24. Let $R$ be a local ring with residue field $\kappa $. Let $M$ be a finite length $R$-module. Let $\alpha , \beta , \gamma $ be endomorphisms of $M$. Assume that

  1. $I_\alpha = K_{\beta \gamma }$, and similarly for any permutation of $\alpha , \beta , \gamma $,

  2. $K_\alpha = I_{\beta \gamma }$, and similarly for any permutation of $\alpha , \beta , \gamma $.

Then

  1. The triple $(M, \alpha , \beta \gamma )$ is an exact $(2, 1)$-periodic complex.

  2. The triple $(I_\gamma , \alpha , \beta )$ is an exact $(2, 1)$-periodic complex.

  3. The triple $(M/K_\beta , \alpha , \gamma )$ is an exact $(2, 1)$-periodic complex.

  4. We have

    \[ \det \nolimits _\kappa (M, \alpha , \beta \gamma ) = \det \nolimits _\kappa (I_\gamma , \alpha , \beta ) \det \nolimits _\kappa (M/K_\beta , \alpha , \gamma ). \]

Proof. It is clear that the assumptions imply part (1) of the lemma.

To see part (1) note that the assumptions imply that $I_{\gamma \alpha } = I_{\alpha \gamma }$, and similarly for kernels and any other pair of morphisms. Moreover, we see that $I_{\gamma \beta } =I_{\beta \gamma } = K_\alpha \subset I_\gamma $ and similarly for any other pair. In particular we get a short exact sequence

\[ 0 \to I_{\beta \gamma } \to I_\gamma \xrightarrow {\alpha } I_{\alpha \gamma } \to 0 \]

and similarly we get a short exact sequence

\[ 0 \to I_{\alpha \gamma } \to I_\gamma \xrightarrow {\beta } I_{\beta \gamma } \to 0. \]

This proves $(I_\gamma , \alpha , \beta )$ is an exact $(2, 1)$-periodic complex. Hence part (2) of the lemma holds.

To see that $\alpha $, $\gamma $ give well defined endomorphisms of $M/K_\beta $ we have to check that $\alpha (K_\beta ) \subset K_\beta $ and $\gamma (K_\beta ) \subset K_\beta $. This is true because $\alpha (K_\beta ) = \alpha (I_{\gamma \alpha }) = I_{\alpha \gamma \alpha } \subset I_{\alpha \gamma } = K_\beta $, and similarly in the other case. The kernel of the map $\alpha : M/K_\beta \to M/K_\beta $ is $K_{\beta \alpha }/K_\beta = I_\gamma /K_\beta $. Similarly, the kernel of $\gamma : M/K_\beta \to M/K_\beta $ is equal to $I_\alpha /K_\beta $. Hence we conclude that (3) holds.

We introduce $r = \text{length}_ R(K_\alpha )$, $s = \text{length}_ R(K_\beta )$ and $t = \text{length}_ R(K_\gamma )$. By the exact sequences above and our hypotheses we have $\text{length}_ R(I_\alpha ) = s + t$, $\text{length}_ R(I_\beta ) = r + t$, $\text{length}_ R(I_\gamma ) = r + s$, and $\text{length}(M) = r + s + t$. Choose

  1. an admissible sequence $x_1, \ldots , x_ r \in K_\alpha $ generating $K_\alpha $

  2. an admissible sequence $y_1, \ldots , y_ s \in K_\beta $ generating $K_\beta $,

  3. an admissible sequence $z_1, \ldots , z_ t \in K_\gamma $ generating $K_\gamma $,

  4. elements $\tilde x_ i \in M$ such that $\beta \gamma \tilde x_ i = x_ i$,

  5. elements $\tilde y_ i \in M$ such that $\alpha \gamma \tilde y_ i = y_ i$,

  6. elements $\tilde z_ i \in M$ such that $\beta \alpha \tilde z_ i = z_ i$.

With these choices the sequence $y_1, \ldots , y_ s, \alpha \tilde z_1, \ldots , \alpha \tilde z_ t$ is an admissible sequence in $I_\alpha $ generating it. Hence, by Remark 42.68.14 the determinant $D = \det _\kappa (M, \alpha , \beta \gamma )$ is the unique element of $\kappa ^*$ such that

\begin{align*} [y_1, \ldots , y_ s, \alpha \tilde z_1, \ldots , \alpha \tilde z_ s, \tilde x_1, \ldots , \tilde x_ r] \\ = (-1)^{r(s + t)} D [x_1, \ldots , x_ r, \gamma \tilde y_1, \ldots , \gamma \tilde y_ s, \tilde z_1, \ldots , \tilde z_ t] \end{align*}

By the same remark, we see that $D_1 = \det _\kappa (M/K_\beta , \alpha , \gamma )$ is characterized by

\[ [y_1, \ldots , y_ s, \alpha \tilde z_1, \ldots , \alpha \tilde z_ t, \tilde x_1, \ldots , \tilde x_ r] = (-1)^{rt} D_1 [y_1, \ldots , y_ s, \gamma \tilde x_1, \ldots , \gamma \tilde x_ r, \tilde z_1, \ldots , \tilde z_ t] \]

By the same remark, we see that $D_2 = \det _\kappa (I_\gamma , \alpha , \beta )$ is characterized by

\[ [y_1, \ldots , y_ s, \gamma \tilde x_1, \ldots , \gamma \tilde x_ r, \tilde z_1, \ldots , \tilde z_ t] = (-1)^{rs} D_2 [x_1, \ldots , x_ r, \gamma \tilde y_1, \ldots , \gamma \tilde y_ s, \tilde z_1, \ldots , \tilde z_ t] \]

Combining the formulas above we see that $D = D_1 D_2$ as desired. $\square$

Lemma 42.68.25. Let $R$ be a local ring with residue field $\kappa $. Let $\alpha : (M, \varphi , \psi ) \to (M', \varphi ', \psi ')$ be a morphism of $(2, 1)$-periodic complexes over $R$. Assume

  1. $M$, $M'$ have finite length,

  2. $(M, \varphi , \psi )$, $(M', \varphi ', \psi ')$ are exact,

  3. the maps $\varphi $, $\psi $ induce the zero map on $K = \mathop{\mathrm{Ker}}(\alpha )$, and

  4. the maps $\varphi $, $\psi $ induce the zero map on $Q = \mathop{\mathrm{Coker}}(\alpha )$.

Denote $N = \alpha (M) \subset M'$. We obtain two short exact sequences of $(2, 1)$-periodic complexes

\[ \begin{matrix} 0 \to (N, \varphi ', \psi ') \to (M', \varphi ', \psi ') \to (Q, 0, 0) \to 0 \\ 0 \to (K, 0, 0) \to (M, \varphi , \psi ) \to (N, \varphi ', \psi ') \to 0 \end{matrix} \]

which induce two isomorphisms $\alpha _ i : Q \to K$, $i = 0, 1$. Then

\[ \det \nolimits _\kappa (M, \varphi , \psi ) = \det \nolimits _\kappa (\alpha _0^{-1} \circ \alpha _1) \det \nolimits _\kappa (M', \varphi ', \psi ') \]

In particular, if $\alpha _0 = \alpha _1$, then $\det \nolimits _\kappa (M, \varphi , \psi ) = \det \nolimits _\kappa (M', \varphi ', \psi ')$.

Proof. There are (at least) two ways to prove this lemma. One is to produce an enormous commutative diagram using the properties of the determinants. The other is to use the characterization of the determinants in terms of admissible sequences of elements. It is the second approach that we will use.

First let us explain precisely what the maps $\alpha _ i$ are. Namely, $\alpha _0$ is the composition

\[ \alpha _0 : Q = H^0(Q, 0, 0) \to H^1(N, \varphi ', \psi ') \to H^2(K, 0, 0) = K \]

and $\alpha _1$ is the composition

\[ \alpha _1 : Q = H^1(Q, 0, 0) \to H^2(N, \varphi ', \psi ') \to H^3(K, 0, 0) = K \]

coming from the boundary maps of the short exact sequences of complexes displayed in the lemma. The fact that the complexes $(M, \varphi , \psi )$, $(M', \varphi ', \psi ')$ are exact implies these maps are isomorphisms.

We will use the notation $I_\varphi = \mathop{\mathrm{Im}}(\varphi )$, $K_\varphi = \mathop{\mathrm{Ker}}(\varphi )$ and similarly for the other maps. Exactness for $M$ and $M'$ means that $K_\varphi = I_\psi $ and three similar equalities. We introduce $k = \text{length}_ R(K)$, $a = \text{length}_ R(I_\varphi )$, $b = \text{length}_ R(I_\psi )$. Then we see that $\text{length}_ R(M) = a + b$, and $\text{length}_ R(N) = a + b - k$, $\text{length}_ R(Q) = k$ and $\text{length}_ R(M') = a + b$. The exact sequences below will show that also $\text{length}_ R(I_{\varphi '}) = a$ and $\text{length}_ R(I_{\psi '}) = b$.

The assumption that $K \subset K_\varphi = I_\psi $ means that $\varphi $ factors through $N$ to give an exact sequence

\[ 0 \to \alpha (I_\psi ) \to N \xrightarrow {\varphi \alpha ^{-1}} I_\psi \to 0. \]

Here $\varphi \alpha ^{-1}(x') = y$ means $x' = \alpha (x)$ and $y = \varphi (x)$. Similarly, we have

\[ 0 \to \alpha (I_\varphi ) \to N \xrightarrow {\psi \alpha ^{-1}} I_\varphi \to 0. \]

The assumption that $\psi '$ induces the zero map on $Q$ means that $I_{\psi '} = K_{\varphi '} \subset N$. This means the quotient $\varphi '(N) \subset I_{\varphi '}$ is identified with $Q$. Note that $\varphi '(N) = \alpha (I_\varphi )$. Hence we conclude there is an isomorphism

\[ \varphi ' : Q \to I_{\varphi '}/\alpha (I_\varphi ) \]

simply described by $\varphi '(x' \bmod N) = \varphi '(x') \bmod \alpha (I_\varphi )$. In exactly the same way we get

\[ \psi ' : Q \to I_{\psi '}/\alpha (I_\psi ) \]

Finally, note that $\alpha _0$ is the composition

\[ \xymatrix{ Q \ar[r]^-{\varphi '} & I_{\varphi '}/\alpha (I_\varphi ) \ar[rrr]^-{\psi \alpha ^{-1}|_{I_{\varphi '}/\alpha (I_\varphi )}} & & & K } \]

and similarly $\alpha _1 = \varphi \alpha ^{-1}|_{I_{\psi '}/\alpha (I_\psi )} \circ \psi '$.

To shorten the formulas below we are going to write $\alpha x$ instead of $\alpha (x)$ in the following. No confusion should result since all maps are indicated by Greek letters and elements by Roman letters. We are going to choose

  1. an admissible sequence $z_1, \ldots , z_ k \in K$ generating $K$,

  2. elements $z'_ i \in M$ such that $\varphi z'_ i = z_ i$,

  3. elements $z''_ i \in M$ such that $\psi z''_ i = z_ i$,

  4. elements $x_{k + 1}, \ldots , x_ a \in I_\varphi $ such that $z_1, \ldots , z_ k, x_{k + 1}, \ldots , x_ a$ is an admissible sequence generating $I_\varphi $,

  5. elements $\tilde x_ i \in M$ such that $\varphi \tilde x_ i = x_ i$,

  6. elements $y_{k + 1}, \ldots , y_ b \in I_\psi $ such that $z_1, \ldots , z_ k, y_{k + 1}, \ldots , y_ b$ is an admissible sequence generating $I_\psi $,

  7. elements $\tilde y_ i \in M$ such that $\psi \tilde y_ i = y_ i$, and

  8. elements $w_1, \ldots , w_ k \in M'$ such that $w_1 \bmod N, \ldots , w_ k \bmod N$ are an admissible sequence in $Q$ generating $Q$.

By Remark 42.68.14 the element $D = \det _\kappa (M, \varphi , \psi ) \in \kappa ^*$ is characterized by

\begin{eqnarray*} & & [z_1, \ldots , z_ k, x_{k + 1}, \ldots , x_ a, z''_1, \ldots , z''_ k, \tilde y_{k + 1}, \ldots , \tilde y_ b] \\ & = & (-1)^{ab} D [z_1, \ldots , z_ k, y_{k + 1}, \ldots , y_ b, z'_1, \ldots , z'_ k, \tilde x_{k + 1}, \ldots , \tilde x_ a] \end{eqnarray*}

Note that by the discussion above $\alpha x_{k + 1}, \ldots , \alpha x_ a, \varphi w_1, \ldots , \varphi w_ k$ is an admissible sequence generating $I_{\varphi '}$ and $\alpha y_{k + 1}, \ldots , \alpha y_ b, \psi w_1, \ldots , \psi w_ k$ is an admissible sequence generating $I_{\psi '}$. Hence by Remark 42.68.14 the element $D' = \det _\kappa (M', \varphi ', \psi ') \in \kappa ^*$ is characterized by

\begin{eqnarray*} & & [\alpha x_{k + 1}, \ldots , \alpha x_ a, \varphi ' w_1, \ldots , \varphi ' w_ k, \alpha \tilde y_{k + 1}, \ldots , \alpha \tilde y_ b, w_1, \ldots , w_ k] \\ & = & (-1)^{ab} D' [\alpha y_{k + 1}, \ldots , \alpha y_ b, \psi ' w_1, \ldots , \psi ' w_ k, \alpha \tilde x_{k + 1}, \ldots , \alpha \tilde x_ a, w_1, \ldots , w_ k] \end{eqnarray*}

Note how in the first, resp. second displayed formula the first, resp. last $k$ entries of the symbols on both sides are the same. Hence these formulas are really equivalent to the equalities

\begin{eqnarray*} & & [\alpha x_{k + 1}, \ldots , \alpha x_ a, \alpha z''_1, \ldots , \alpha z''_ k, \alpha \tilde y_{k + 1}, \ldots , \alpha \tilde y_ b] \\ & = & (-1)^{ab} D [\alpha y_{k + 1}, \ldots , \alpha y_ b, \alpha z'_1, \ldots , \alpha z'_ k, \alpha \tilde x_{k + 1}, \ldots , \alpha \tilde x_ a] \end{eqnarray*}

and

\begin{eqnarray*} & & [\alpha x_{k + 1}, \ldots , \alpha x_ a, \varphi ' w_1, \ldots , \varphi ' w_ k, \alpha \tilde y_{k + 1}, \ldots , \alpha \tilde y_ b] \\ & = & (-1)^{ab} D' [\alpha y_{k + 1}, \ldots , \alpha y_ b, \psi ' w_1, \ldots , \psi ' w_ k, \alpha \tilde x_{k + 1}, \ldots , \alpha \tilde x_ a] \end{eqnarray*}

in $\det _\kappa (N)$. Note that $\varphi ' w_1, \ldots , \varphi ' w_ k$ and $\alpha z''_1, \ldots , z''_ k$ are admissible sequences generating the module $I_{\varphi '}/\alpha (I_\varphi )$. Write

\[ [\varphi ' w_1, \ldots , \varphi ' w_ k] = \lambda _0 [\alpha z''_1, \ldots , \alpha z''_ k] \]

in $\det _\kappa (I_{\varphi '}/\alpha (I_\varphi ))$ for some $\lambda _0 \in \kappa ^*$. Similarly, write

\[ [\psi ' w_1, \ldots , \psi ' w_ k] = \lambda _1 [\alpha z'_1, \ldots , \alpha z'_ k] \]

in $\det _\kappa (I_{\psi '}/\alpha (I_\psi ))$ for some $\lambda _1 \in \kappa ^*$. On the one hand it is clear that

\[ \alpha _ i([w_1, \ldots , w_ k]) = \lambda _ i[z_1, \ldots , z_ k] \]

for $i = 0, 1$ by our description of $\alpha _ i$ above, which means that

\[ \det \nolimits _\kappa (\alpha _0^{-1} \circ \alpha _1) = \lambda _1/\lambda _0 \]

and on the other hand it is clear that

\begin{eqnarray*} & & \lambda _0 [\alpha x_{k + 1}, \ldots , \alpha x_ a, \alpha z''_1, \ldots , \alpha z''_ k, \alpha \tilde y_{k + 1}, \ldots , \alpha \tilde y_ b] \\ & = & [\alpha x_{k + 1}, \ldots , \alpha x_ a, \varphi ' w_1, \ldots , \varphi ' w_ k, \alpha \tilde y_{k + 1}, \ldots , \alpha \tilde y_ b] \end{eqnarray*}

and

\begin{eqnarray*} & & \lambda _1[\alpha y_{k + 1}, \ldots , \alpha y_ b, \alpha z'_1, \ldots , \alpha z'_ k, \alpha \tilde x_{k + 1}, \ldots , \alpha \tilde x_ a] \\ & = & [\alpha y_{k + 1}, \ldots , \alpha y_ b, \psi ' w_1, \ldots , \psi ' w_ k, \alpha \tilde x_{k + 1}, \ldots , \alpha \tilde x_ a] \end{eqnarray*}

which imply $\lambda _0 D = \lambda _1 D'$. The lemma follows. $\square$

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