Example 48.3.3. If $f : X \to Y$ is a separated finite type morphism of Noetherian schemes, then the right adjoint of $Rf_* : D_\mathit{QCoh}(\mathcal{O}_ X) \to D_\mathit{QCoh}(\mathcal{O}_ Y)$ does not map $D_{\textit{Coh}}(\mathcal{O}_ Y)$ into $D_{\textit{Coh}}(\mathcal{O}_ X)$. Namely, let $k$ be a field and consider the morphism $f : \mathbf{A}^1_ k \to \mathop{\mathrm{Spec}}(k)$. By Example 48.3.2 this corresponds to the question of whether $R\mathop{\mathrm{Hom}}\nolimits (B, -)$ maps $D_{\textit{Coh}}(A)$ into $D_{\textit{Coh}}(B)$ where $A = k$ and $B = k[x]$. This is not true because
\[ R\mathop{\mathrm{Hom}}\nolimits (k[x], k) = \left(\prod \nolimits _{n \geq 0} k\right)[0] \]
which is not a finite $k[x]$-module. Hence $a(\mathcal{O}_ Y)$ does not have coherent cohomology sheaves.
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