Lemma 110.64.1. There exists a ring R, a distinguished triangle (K, L, M, \alpha , \beta , \gamma ) in the homotopy category K(R), and an endomorphism (a, b, c) of this distinguished triangle, such that K, L, M are perfect complexes and \text{Tr}_ K(a) + \text{Tr}_ M(c) \not= \text{Tr}_ L(b).
110.64 Example of non-additivity of traces
Let k be a field and let R = k[\epsilon ] be the ring of dual numbers over k. In other words, R = k[x]/(x^2) and \epsilon is the congruence class of x in R. Consider the short exact sequence of complexes
\xymatrix{ 0 \ar[d] \ar[r] & R \ar[d]^\epsilon \ar[r]_1 & R \ar[d] \\ R \ar[r]^1 & R \ar[r] & 0 }
Here the columns are the complexes, the first row is placed in degree 0, and the second row in degree 1. Denote the first complex (i.e., the left column) by A^\bullet , the second by B^\bullet and the third C^\bullet . We claim that the diagram
110.64.0.1
\begin{equation} \label{examples-equation-commutes-up-to-homotopy} \vcenter { \xymatrix{ A^\bullet \ar[d]_{1 + \epsilon } \ar[r] & B^\bullet \ar[r] \ar[d]_1 & C^\bullet \ar[d]_1 \\ A^\bullet \ar[r] & B^\bullet \ar[r] & C^\bullet } } \end{equation}
commutes in K(R), i.e., is a diagram of complexes commuting up to homotopy. Namely, the square on the right commutes and the one on the left is off by the homotopy 1 : A^1 \to B^0. On the other hand,
\text{Tr}_{A^\bullet }(1 + \epsilon ) + \text{Tr}_{C^\bullet }(1) \not= \text{Tr}_{B^\bullet }(1).
Proof. Consider the example above. The map \gamma : C^\bullet \to A^\bullet [1] is given by multiplication by \epsilon in degree 0, see Derived Categories, Definition 13.10.1. Hence it is also true that
\xymatrix{ C^\bullet \ar[d] \ar[r]_\gamma & A^\bullet [1] \ar[d] \\ C^\bullet \ar[r]^\gamma & A^\bullet [1] }
commutes in K(R) as \epsilon (1 + \epsilon ) = \epsilon . Thus we indeed have a morphism of distinguished triangles. \square
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