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
\[ \xymatrix{ 0 \ar[r] & \mathop{\mathrm{Ker}}(b^\bullet ) \ar[r] \ar[d]_{c^\bullet } & K^\bullet \ar[r]_{b^\bullet } \ar[d]_{k^\bullet } & L^\bullet \ar[r] \ar[d]^{\text{id}} & 0 \\ 0 \ar[r] & \mathop{\mathrm{Ker}}(a^\bullet ) \ar[r] & K^\bullet \ar[r]^{a^\bullet } & L^\bullet \ar[r] & 0 } \]
of complexes, where $c^\bullet $ is the induced isomorphism of kernels. Using Lemma 15.118.3 we see that
\[ \xymatrix{ \det (\mathop{\mathrm{Ker}}(b^ i)) \otimes \det (L^ i) \ar[r] \ar[d]_{\det (c^ i) \otimes 1} & \det (K^ i) \ar[d]^{\det (k^ i)} \\ \det (\mathop{\mathrm{Ker}}(a^ i)) \otimes \det (L^ i) \ar[r] & \det (K^ i) } \]
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)
\[ \det (k^0) = \det (k^{-1}) \]
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 calculation1 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
\[ \alpha , \beta : R^{\oplus n} \longrightarrow R^{\oplus m} \]
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. $\square$
[1] $a^{-1}(k^{-1}(u')) = a^{-1}(u') + a^{-1}(\tilde h(d(u'))) = a^{-1}(u') + h(d(u')) = b^{-1}(u') = a^{-1}(u)$
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