TITLE: Hartshorne exercise III.6.2 (b) - $\mathfrak{Qco}(X)$ need not have enough projectives
QUESTION [14 upvotes]: Let $X=\mathbb P^1_k$, with $k$ an infinite field. Show that there does not exist a projective object $\mathcal P$ either in $\mathfrak{Qco}(X)$ or $\mathfrak{Coh}(X)$ together with a surjection $\mathcal P\to \mathcal O_X\to 0$.
The author suggests to consider surjections of the form $\mathcal L\to \mathcal L\otimes k(x)\to 0$, where $x$ is a closed point, and $\mathcal L$ is an invertible sheaf on $X$.
I don't know how to choose $\mathcal L$, and I don't see how to use (quasi-)coherence here. Any suggestion? Thank you in advance.

REPLY [9 votes]: I'll show  the non-existence of  projectives in the category of coherent sheaves on $X$.    
Let $\mathcal P$ be a non-zero projective coherent sheaf and $\mathcal P \stackrel {p}\to \mathcal O_X$ a surjective morphism.
Let $\mathcal L$ be a line bundle that we will determine later and $\mathcal L \stackrel {ev}{\rightarrow} \underline {\mathcal L(x)} \to 0$ the evaluation at $x$, with values in the sky-scraper sheaf associated to the fiber of $\mathcal L$ at $x$.
We choose a surjective morphism $\mathcal O \stackrel {c}{\rightarrow}\underline {\mathcal L(x)}$ and obtain by projectivity of $\mathcal P$ a commutative diagram
$$\begin{matrix}
\mathcal P&\stackrel {p}\rightarrow&\mathcal O_X\\
\downarrow&&\downarrow{c}\\
  \mathcal L&\stackrel {ev}{\rightarrow}&\underline {\mathcal L(x)} 
\end{matrix}
$$ 
Now take  $n$ big enough for the induced morphism on global sections $\Gamma(X,\mathcal P(n))\to \Gamma(X,\mathcal O_X(n))$ to be surjective.
We then have a commutative diagram 
$$\begin{matrix}
\mathcal P(n)&\rightarrow&\mathcal O_X(n)\\
\downarrow&&\downarrow\\
\mathcal L(n)& \rightarrow &k 
\end{matrix}
$$ 
We decide now to take $\mathcal L=\mathcal O_X(-n-1)$, we take global sections in the above diagram and we obtain $$\begin{matrix}
\Gamma(X,\mathcal P(n))&\rightarrow&\Gamma(X,\mathcal O_X(n))\\
\downarrow&&\downarrow\\
\Gamma(X,\mathcal L(n))=\Gamma(X,\mathcal O(-1))=0& \rightarrow &k 
\end{matrix}
$$
This is a contradiction because the lower composition $ \Gamma(X,\mathcal P(n))\rightarrow  \Gamma(X,\mathcal L(n))=0 \rightarrow k    $ is zero whereas  the upper compsition $ \Gamma(X,\mathcal P(n))\rightarrow  \Gamma(X,\mathcal O_X(n))  \rightarrow k    $ is not zero.