The Stacks project

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

and $\alpha _1$ is the composition

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

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

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

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

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

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

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

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

and

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

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

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

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

and on the other hand it is clear that

and

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