Lemma 15.122.2. 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. Let $h : K^0 \to L^{-1}$ be a map such that $b^0 = a^0 + d \circ h$ and $b^{-1} = a^{-1} + h \circ d$ are surjective. Then $\det (a^\bullet ) = \det (b^\bullet )$ as maps $\det (K^\bullet ) \to \det (L^\bullet )$.

**Proof.**
Suppose there exists a map $\tilde h : K^0 \to K^{-1}$ such that $h = a^{-1} \circ \tilde h$ and such that $k^0 = \text{id} + d \circ \tilde h : K^0 \to K^0$ and $k^1 = \text{id} + \tilde h \circ d : K^{-1} \to K^{-1}$ are isomorphisms. Then we obtain a commutative diagram

of complexes, where $c^\bullet $ is the induced isomorphism of kernels. Using Lemma 15.118.3 we see that

commutes. Since $\det (c^\bullet )$ maps the canonical trivialization of $\det (\mathop{\mathrm{Ker}}(a^\bullet ))$ to the canonical trivializatio of $\mathop{\mathrm{Ker}}(b^\bullet )$ (Lemma 15.122.1) we see that we conclude if (and only if)

as elements of $R$ which follows from Lemma 15.118.6.

Suppose there exists a direct summand $U \subset K^{-1}$ such that both $a^{-1}|_ U : U \to L^{-1}$ and $b^{-1}|_ U : U \to L^{-1}$ are isomorphisms. Define $\tilde h$ as the composition of $h$ with the inverse of $a^{-1}|_ U$. We claim that $\tilde h$ is a map as in the first paragraph of the proof. Namely, we have $h = a^{-1} \circ \tilde h$ by construction. To show that $k^{-1} : K^{-1} \to K^{-1}$ is an isomorphism it suffices to show that it is surjective (Algebra, Lemma 10.16.4). Let $u \in U$. We may choose $u' \in U$ such that $b^{-1}(u') = a^{-1}(u)$. Then $u = k^{-1}(u')$. Namely, both $u$ and $k^{-1}(u')$ are in $U$ and $a^{-1}(u) = a^{-1}(k^{-1}(u'))$ by a calculation^{1} Since $a^{-1}|_ U$ is an isomorphism we get the equality. Thus $U \subset \mathop{\mathrm{Im}}(k^{-1})$. On the other hand, if $x \in \mathop{\mathrm{Ker}}(a^{-1})$ then $x = k^{-1}(x) \bmod U$. Since $K^{-1} = \mathop{\mathrm{Ker}}(a^{-1}) + U$ we conclude $k^{-1}$ is surjective. Finally, we show that $k^0 : K^0 \to K^0$ is surjective. First, since $a^0 \circ k^0 = b^0$ we see that $a^0 \circ k^0$ is surjective. If $x \in \mathop{\mathrm{Ker}}(a^0)$, then $x = d(y)$ for some $y \in \mathop{\mathrm{Ker}}(a^{-1})$. We may write $y = k^{-1}(z)$ for some $z \in K^{-1}$ by the above. Then $x = k^0(d(z))$ and we conclude.

Final step of the proof. It suffices to find $U$ as in the preceding paragraph, but this may not always be possible. However, in order to show equality of two maps of $R$-modules, it suffices to do so after localization at primes of $R$. Hence we may assume $R$ is local. Then we get the following problem: suppose

are two surjective $R$-linear maps. Find a direct summand $U \subset R^{\oplus n}$ such that both $\alpha |_ U$ and $\beta |_ U$ are isomorphisms. If $R$ is a field, this is possible by linear algebra. In general, one takes a solution over the residue field and lifts this to a solution over the local ring $R$. Some details omitted. $\square$

## 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.

## Comments (0)

There are also: