Example 48.4.2. There is a finite morphism $f : X \to Y$ of Noetherian schemes such that (48.4.1.1) is not an isomorphism when evaluated on some $K \in D_{\textit{Coh}}(\mathcal{O}_ Y)$. Namely, let $X = \mathop{\mathrm{Spec}}(B) \to Y = \mathop{\mathrm{Spec}}(A)$ with $A = k[x, \epsilon ]$ where $k$ is a field and $\epsilon ^2 = 0$ and $B = k[x] = A/(\epsilon )$. For $n \in \mathbf{N}$ set $M_ n = A/(\epsilon , x^ n)$. Observe that

because $B$ has the free periodic resolution $\ldots \to A \to A \to A$ with maps given by multiplication by $\epsilon $. Consider the object $K = \bigoplus M_ n[n] = \prod M_ n[n]$ of $D_{\textit{Coh}}(A)$ (equality in $D(A)$ by Derived Categories, Lemmas 13.33.5 and 13.34.2). Then we see that $a(K)$ corresponds to $R\mathop{\mathrm{Hom}}\nolimits (B, K)$ by Example 48.3.2 and

by the above. But this module has elements which are not annihilated by any power of $x$, whereas the complex $K$ does have every element of its cohomology annihilated by a power of $x$. In other words, for the map (48.4.1.1) with $V = D(x)$ and $U = D(x)$ and the complex $K$ cannot be an isomorphism because $(j')^*(a(K))$ is nonzero and $a'(j^*K)$ is zero.

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