Is the derived category of a full subcategory full?

Question

Certainly not, but I cannot find a good counterexample. I tried to do something like $\mathbf Z\text{-Mod}\subseteq \mathbf Z[x]\text{-Mod}$, but without success. Does anyone have a short counterexample?

Answer 1

Thanks to the comments by @MarianoSuárez-Álvarez, I can formulate the following answers:

Consider a field $k$ and the ring $k[x]$. Of course, $\hom_{D(k)}(k, k[i])=\begin{cases}k & \text{if i=0,}\\0& \text{o/w}\end{cases}$ in the derived category $D(k)$. Now in the category $k[x]\text{-Mod}$, there is a nontrivial extension $0\to k\xrightarrow{x} k[x]/(x^2) \to k\to 0$, where we consider $k$ a $k[x]$-module by $k=k[x]/(x^2)$. This extension yields a map

$$\begin{matrix}k\\\simeq\:\downarrow\phantom{\simeq} \\k[x]/(x^2) &\rightarrow& k\\&&\downarrow\\&&k\end{matrix}$$

which is a nontrivial map $k\to k[1]$ in $D(k[x])$.