Lemma 15.122.1. Let $R$ be a ring. Let $a^\bullet : K^\bullet \to L^\bullet$ be a map of complexes of $R$-modules satisfying (1), (2), (3) above. If $L^\bullet$ has rank $0$, then $\det (a^\bullet )$ maps the canonical element $\delta (K^\bullet )$ to $\delta (L^\bullet )$.

Proof. Write $M^ i = \mathop{\mathrm{Ker}}(a^ i)$. Thus we have a map of short exact sequences

$\xymatrix{ 0 \ar[r] & M^{-1} \ar[r] \ar[d]_{d_ M} & K^{-1} \ar[r] \ar[d]_{d_ K} & L^{-1} \ar[r] \ar[d]_{d_ L} & 0 \\ 0 \ar[r] & M^0 \ar[r] & K^0 \ar[r] & L^0 \ar[r] & 0 }$

By Lemma 15.118.3 we know that $\det (d_ K)$ corresponds to $\det (d_ M) \otimes \det (d_ L)$ as maps. Unwinding the definitions this gives the required equality. $\square$

There are also:

• 2 comment(s) on Section 15.122: Determinants of complexes

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).