The Stacks project

Lemma 15.104.4. 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. $\square$


Comments (0)


Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0AFX. Beware of the difference between the letter 'O' and the digit '0'.