Hartshorne exercise III.6.2 (b) - $\mathfrak{Qco}(X)$ need not have enough projectives

Question

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.

Answer 1

Notate $X:=\mathbb{P}^1$

Claim: If $\mathcal{P}\to \mathcal{O}_X\to 0$ where $\mathcal{P}$ quasi-coherent, then there exists some $n$ such that there is a surjection on global sections $H^0(\mathcal{P}(n))\to H^0(\mathcal{O}_X(n))\to 0$

Suppose there is a surjection $\mathcal{P}\stackrel{\phi}{\to} \mathcal{O}_X\to 0$. By Exercise 5.15 in Chapter II we can write $\mathcal{P}$ as a strictly ascending union of its coherent subsheaves, say $\cup \mathcal{P}_i$.

Thus $\mathcal{O}_X=\cup \operatorname{im}\phi_i$ (where $\phi_i$ the appropriate sheaf image). Thus some coherent sheaf surjects onto $\mathcal{O}_X$.

Relabel and consider the exact sequence of coherent sheaves $0\to \mathcal{K}\to \mathcal{P}\to \mathcal{O}_X\to 0$. By Serre's theorem (III.5.2) we can find a sufficiently large integer so that $H^1(X, \mathcal{K}(n))=0$.

Twisting the whole short exact sequence by $\mathcal{O}_X(n)$ (i.e. tensoring with this locally free sheaf) is exact and hence keeps it short exact. Taking the long exact sequence associated to this short exact sequence gives us that $$\cdots\to H^0(X, \mathcal{P}(n))\to H^0(X, \mathcal{O}_X(n))\to H^1(X, \mathcal{K}(n))=0$$ Thus on global sections we have a surjection $H^0(\mathcal{P}(n))\to H^0(\mathcal{O}_X(n))$.

Now use the projective assumption (and the hint). Since we have a surjection $\mathcal{O}_X(-n-1)\to k(x)\to 0$, and by assumption we have a map $\mathcal{P}\to \mathcal{O}_X\to k(x)$ there exists a map $\mathcal{P}\to \mathcal{O}_X(-n-1)$ making the appropriate diagram commute.

Now twist everything by $n$ and take global sections. You get a surjection onto a non-zero group factoring through $0$ (since $H^0(X, \mathcal{O}_X(-1))=0$) a contradiction. Thus no such surjection $\mathcal{P}\to \mathcal{O}_X$ can exist.

Answer 2

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.