Are chain complexes over a field always injective? Question: Let $\mathbb{F}$ be an algebraically closed field of characteristic zero and let $\mathrm{Ch}_{\mathbb{F}}$ be the category whose objects are chain complexes (of $\mathbb{F}$-modules) and whose morphisms are chain maps. Is it true that every object in $\mathrm{Ch}_{\mathbb{F}}$ is injective? If so, does it remain true if $\mathbb{F}$ is not algebraically closed? What if $\mathbb{F}$ has positive characteristic?
On one hand this seems to good to be true, but on the other hand things are often exceedingly nice when dealing with vector spaces.
Edit: removed a flawed proof. Thanks darij for spotting the error.
 A: More abstractly, in an abelian category, saying either that every object is injective or that every object is projective is equivalent to saying that every short exact sequence splits (semisimplicity). 
Categories of chain complexes essentially never have this property, e.g. because the chain complex $c \xrightarrow{{id}_c} c$ is always a nontrivial extension of its degree-$0$ and degree-$1$ parts provided that $c$ is not itself zero. 
A: No.
Let $X$ be the chain complex with $\mathbb{k}$ in degree $n$ (and all differentials zero) and let $Y$ be the chain complex with $\mathbb{k}$ in degrees $n$ and $n+1$ with the identity as differential $Y_{n+1}\to Y_n$ (I choose for my convention that the differential lowers degree).
Then $X$ is not injective because there is no chain map $Y\to X$ which factors the identity on $X$ through the evident inclusion of $X$ into $Y$.
To be injective over a field it is necessary and sufficient to have vanishing homology.
A similar argument shows that not every chain complex is projective, and that to be projective over a field it is necessary and sufficient to have vanishing homology.
