Lemma 15.102.4. Let $R$ be a ring. There is a map

which maps $[M]$ to the class of the invertible module $\wedge ^ n(M)$ if $M$ is a finite locally free module of rank $n$.

Lemma 15.102.4. Let $R$ be a ring. There is a map

\[ \det : K_0(R) \longrightarrow \mathop{\mathrm{Pic}}\nolimits (R) \]

which maps $[M]$ to the class of the invertible module $\wedge ^ n(M)$ if $M$ is a finite locally free module of rank $n$.

**Proof.**
Let $M$ be a finite projective $R$-module. There exists a product decomposition $R = R_0 \times \ldots \times R_ t$ such that in the corresponding decomposition $M = M_0 \times \ldots \times M_ t$ of $M$ we have that $M_ i$ is finite locally free of rank $i$ over $R_ i$. This follows from Algebra, Lemma 10.77.2 (to see that the rank is locally constant) and Algebra, Lemmas 10.20.3 and 10.23.3 (to decompose $R$ into a product). In this situation we define

\[ \det (M) = \wedge ^0_{R_0}(M_0) \times \ldots \times \wedge ^ t_{R_ t}(M_ t) \]

as an $R$-module. This is a finite locally free module of rank $1$ as each term is finite locally free of rank $1$. To finish the proof we have to show that

\[ \det (M' \oplus M'') \cong \det (M') \otimes \det (M'') \]

whenever $M'$ and $M'"$ are finite projective $R$-modules. 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 $j$. This reduces us to the case where $M'$ and $M''$ have constant rank say $i$ and $j$. In this case we have to prove that

\[ \wedge ^{i + j}(M' \oplus M'') \cong \wedge ^ i(M') \otimes \wedge ^ j(M'') \]

the proof of which we omit. $\square$

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