Lemma 15.118.2. Let $R$ be a ring. Let

$0 \to M' \to M \to M'' \to 0$

be a short exact sequence of finite projective $R$-modules. Then there is a canonical isomorphism

$\gamma : \det (M') \otimes \det (M'') \longrightarrow \det (M)$

First proof. First proof. Decompose $R$ into a product of rings $R_{ij}$ such that $M' = \prod M'_{ij}$ and $M'' = \prod M''_{ij}$ where $M'_{ij}$ has rank $i$ and $M''_{ij}$ has rank $j$. Of course then $M = \prod M_{ij}$ and $M_{ij}$ has rank $i + j$. This reduces us to the case where $M'$ and $M''$ have constant rank say $i$ and $j$. In this case we have to construct a canonical map

$\wedge ^ i(M') \otimes \wedge ^ j(M'') \longrightarrow \wedge ^{i + j}(M)$

To do this choose $m'_1, \ldots , m'_ i$ in $M'$ and $m''_1, \ldots , m''_ j$ in $M''$. Denote $m_1, \ldots , m_ i \in M$ the images of $m'_1, \ldots , m'_ i$ and denote $m_{i + 1}, \ldots , m_{i + j} \in M$ elements mapping to $m''_1, \ldots , m''_ j$ in $M''$. Our rule will be that

$m'_1 \wedge \ldots \wedge m'_ i \otimes m''_1 \wedge \ldots \wedge m''_ j \longmapsto m_1 \wedge \ldots \wedge m_{i + j}$

We omit the detailed proof that this is well defined and an isomorphism. $\square$

Second proof. We will use the description of $\det (M)$, $\det (M')$, and $\det (M'')$ given in Remark 15.118.1. Consider the $R$-algebra maps $\wedge ^*_ R(M') \to \wedge ^*_ R(M)$ and $\wedge ^*_ R(M) \to \wedge ^*_ R(M'')$. The first is injective and the second is surjective. Take an element $x' \in \det (M') \subset \wedge ^*_ R(M')$ and an element $x'' \in \det (M'') \subset \wedge ^*_ R(M'')$. Choose an element $y'' \in \wedge ^*(M)$ mapping to $x''$ and set

$\gamma (x' \otimes x'') = x' \wedge y'' \in \det (M) \subset \wedge ^*_ R(M)$

The reader verifies easily by looking at localizations at primes that this well defined and an isomorphism. Moreover, this construction gives the same map as the construction given in the first proof. $\square$

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