Hartshorne Exercise II.7.4.
(a) (a) Use (7.6) to show that if $X$ is a scheme of finite type over a noetherian ring $A$, and if $X$ admits an ample invertible sheaf, then $X$ is separated.
(b) Let $X$ be the affine line over a field $k$ with the origin doubled (4.0.1). Calculate $\op{Pic}X$, determine which invertible sheaves are generated by global sections, and then show directly (without using (a)) that there is no ample invertible sheaf on $X$.
(a) Let $X$ be a scheme of finite type over a noetherian ring $A$ that admits an ample invertible sheaf $\mathcal L$. Let $\varphi :X\ra \PP{\C}{N}$ be the closed immersion corresponding to some power of $\mathcal L$. Then $\varphi$ is separated, and since compositions of separated morphisms are separated, $X$ is separated over $k$.
(b) Let $U=\A{x}{1}$ and $V=\A{y}{1}$ be the two affine lines, glued over ${x\ne 0}\cong {y\ne 0}$ by $x\mapsto y$. Let $\mathcal L\in\op{Pic}X$. The Picard group of the affine line is trivial, so $\mathcal L|_U\cong \OO_U$ and $\mathcal L|_V\cong \OO_V$. Therefore, $\mathcal L$ is determined by its transition function $g\in k[x,x^{-1}]^\times$. Using the identification $\op{Pic}X\cong \op{CaCl}X$ (because $X$ is integral), there is a unique $n\in \Z$ such that for open $W\subset X$,
$$\mathcal L(W) \cong \{(f(x),g(y))\in \OO(W\cap U)\times \OO(W\cap V)|f=x^ng\in \OO(W\cap U\cap V)\}$$
and isomorphisms of these line bundles are determined modulo $k^\times$. Denote these line bundles as $\mathcal L_n$, and one can easily verify that $\mathcal L_n\otimes\mathcal L_m\cong \mathcal L_{n+m}$ and that $\mathcal L_0\cong \OO_X$. It follows that $\op{Pic}X\cong \Z$. To see that there are no ample line bundles, it suffices to show that any morphism given by an $\mathcal L_n$ is not a closed immersion. The problem occurs where we might expect: at the doubled origin. Hartshorne proposition II.7.3 gives equivalent conditions for being a closed immersion when $X$ is projective over an algebraically closed field.
However, the projective hypothesis is only used to prove that the conditions are sufficient for a closed immersion, and it is clearly still necessary that the sections that give a closed immersion must separate tangent vectors on Zariski tangent spaces at closed points that can be listed as tuples in $k$. Moreover, it suffices to show that that $\mathcal L_n$ does not give an immersion for $n>>0$ and for $n<<0$. Hence, I claim that for $n>1$ and $n<-1$, any list of global sections of $\mathcal L_n$ fails to separate tangent vectors at one of the origins. The global sections of $\mathcal L_n$ are pairs $s=(f(x),g(y))\in k[x]\times k[y]$ such that $g(x)=x^nf(x)\in k[x,x^{-1}]$ for some $n\in \Z$. Let one origin be $P=(x)$ and the other $Q=(y)$. Then $s(P)=f(x)\in k[x]_{(x)}$. If $f\in (x)$ then $g\notin (y)$ $\iff$ $n=-\op{degl}f$, where $\op{degl}f$ is the lowest degree term of $f$. A general section $s$ has $s(Q)\in \mathfrak m_Q$ $\iff$ $n+\op{degl}f\ge 1$, and $s$ spans $\mathfrak m_Q/\mathfrak m_Q^2$ $\iff$ equality holds. When $n>1$, equality never holds. When $n<-1$ the co-discussion gives the same for tangent vectors at $P$.
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