## 109.63 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

109.63.0.1
$$\label{examples-equation-commutes-up-to-homotopy} \vcenter { \xymatrix{ A^\bullet \ar[d]_{1 + \epsilon } \ar[r] & B^\bullet \ar[r] \ar[d]_1 & C^\bullet \ar[d]_1 \\ A^\bullet \ar[r] & B^\bullet \ar[r] & C^\bullet } }$$

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,

$\text{Tr}_{A^\bullet }(1 + \epsilon ) + \text{Tr}_{C^\bullet }(1) \not= \text{Tr}_{B^\bullet }(1).$

Lemma 109.63.1. 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 $\text{Tr}_ K(a) + \text{Tr}_ M(c) \not= \text{Tr}_ L(b)$.

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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