TITLE: Koszul (exterior/symmetric) duality for a 1-dim vector space
QUESTION [7 upvotes]: The simplest example of Koszul duality (see introduction of Beilinson, Ginzburg, and Soergel - Koszul Duality Patterns in Representation Theory) is as follows.
Let $V = \mathbb{C}x$ be a $1$ dimensional vector space. Then the exterior algebra is $A=\mathbb{C}[x]/(x^2)$, and the symmetric algebra is $A^! = \mathbb{C}[x]$. Koszul duality states that there exists an isomorphism (where "mod" means the category of finitely generated modules):
$F: D^b(A\mbox{-mod}) \rightarrow D^b(A^!\mbox{-mod})$
The heart of the category on the left is $A\mbox{-mod}$; what is the image of this on the right? It should be possible to work this out by working through the above paper, but there are many things in that paper which I don't understand (and this toy example should be easy).

REPLY [4 votes]: There's actually a general answer to this question, given by Mazorchuk, Ovsienko and Stroppel - Quadratic duals, Koszul dual functors, and applications (though I assume some version of it was known earlier).  The category of graded representations of $A^!$ is the same as the (abelian!) category of linear projective complexes over $A$, and vice versa.  So, in Dag Oskar Madsen's answer, I would write not $\mathbb{C}[n]\langle n\rangle$, but rather its projective resolution   $A^![n+1]\langle n+1\rangle\overset{x}\to A^![n]\langle n\rangle$.  Similarly, the $A^!$ modules $\mathbb{C}[x]/(x^{n+1})$ are sent by Koszul duality to the complexes
$$A[n]\langle n\rangle \overset{x} \to A[n-1]\langle n-1\rangle \overset{x} \to\dotsb \overset{x} \to A.$$