We know from Beilinson that there's an equivalence of derived categories
$D^b Rep(Q) \simeq D^b Coh(\mathbb{P}^1)$
where the lefthandside is the derived category of bounded complexes of representations of the Kronecker quiver
$* => *$
and the righthandside is the derived category of bounded complexes of coherent sheaves on projective space.
My question is:
Is there a proof that
$Rep(Q) \not \simeq Coh(\mathbb{P}^1)$
as abelian categories?
Yes: As a module category, $\text{Rep}(Q)$ has enough projectives, while $\text{Coh}({\mathbb P}^1)$ does not have any nonzero projective objects at all: By Serre duality we have $\text{Ext}^1_{\text{Coh}({\mathbb P}^1)}({\mathscr F},{\mathscr G})\cong\text{Hom}_{\text{Coh}({\mathbb P}^1)}({\mathscr G}(2),{\mathscr F})^{\ast},$ so if ${\mathscr F}$ is projective, then $\text{Hom}(-,{\mathscr F})\equiv 0$, hence ${\mathscr F}=0$ by considering $\text{id}_{\mathscr F}\in\text{Hom}({\mathscr F},{\mathscr F})$.
Hanno's reason is probably the canonical one, but an arguably more elementary reason is that $\text{Rep}(Q)$ has only two simple objects up to isomorphism, but $\text{Coh}({\mathbb P}^1)$ has infinitely many simple objects, the skyscraper sheaves at points of ${\mathbb P}^1$.
Or alternatively, $\text{Rep}(Q)$ is an artinian category, but $\text{Coh}({\mathbb P}^1)$ is not.
