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.
Let \(X\) be the affine line over a field \(k\) with the origin doubled (4.0.1). Calculate \(\Pic X\), determine which invertible sheaves are generated by global sections, and then show directly, without using (a), that there is no ample invertible sheaf on \(X\).
Solution
For part (b), write the doubled-origin line as \[ X=U\cup V,\qquad U=\Spec k[t],\quad V=\Spec k[t], \quad U\cap V=\Spec k[t,t^{-1}], \] with the two copies identified by the identity on the intersection. Let \(0_U\) and \(0_V\) denote the two origins.
1. Part (a) follows from Theorem II.7.6.
Let \(\mathcal L\) be ample on the finite-type \(A\)-scheme. By the ample/very-ample criterion [@Har10a, Theorem II.7.6], some positive tensor power of \(\mathcal L\) is the pullback of \(\OO(1)\) under an immersion \(X\hookrightarrow\PP_A^N\). An immersion is separated, projective space is separated over \(\Spec A\), and composites of separated morphisms are separated [@Har10a, Chapter II, §4]. Thus \(X\to\Spec A\) is separated. The argument only uses an immersion; it does not require \(X\) to be proper or the immersion to be closed.
2. The Picard group of the doubled-origin line is \(\boxed{\Pic(X)\cong\ZZ}\).A representative \(\mathcal L_a\) of \(a\in\ZZ\) is given by frames \(e_U,e_V\) with transition \(e_V=t^a e_U\).
Every invertible sheaf is trivial on \(U\) and \(V\): both have coordinate ring the PID \(k[t]\), so their rank-one projective modules are free. It is therefore determined by a gluing unit in \(k[t,t^{-1}]^*\). Every such unit is \(ct^a\) for a unique \(a\in\ZZ\) and \(c\in k^*\), since the largest and smallest exponents in a Laurent polynomial add under multiplication. Changing either frame multiplies the gluing unit by a unit of \(k[t]\), which is a nonzero constant. We may remove \(c\), and such frame changes never change \(a\).
Conversely, every transition \(t^a\) glues two trivial invertible sheaves to an invertible sheaf on \(X\). Tensor products multiply transitions, so \[ \mathcal L_a\otimes\mathcal L_b\cong\mathcal L_{a+b}, \qquad\mathcal L_0\cong\OO_X. \] This proves the claimed group isomorphism and exhausts all isomorphism classes.
3. An invertible sheaf on this \(X\) is generated by global sections exactly when it is trivial.
A section of \(\mathcal L_a\) is a pair \((p,q)\in k[t]^2\) satisfying \[ p(t)e_U=q(t)e_V, \qquad p(t)=t^a q(t)\text{ in }k[t,t^{-1}]. \] For \(a>0\), every such \(p\) is divisible by \(t\), so all global sections have zero image in the fibre at \(0_U\). They cannot generate the stalk there: every linear combination remains in its maximal-ideal multiple. For \(a<0\), the equivalent identity \(q=t^{-a}p\) shows that all global sections vanish at \(0_V\), giving the same obstruction there. For \(a=0\), the pair \((1,1)\) is a global frame and generates everywhere. Thus \[ \boxed{\mathcal L_a\text{ is globally generated}\iff a=0.} \]
4. No invertible sheaf on the doubled-origin line is ample.
3.This contradiction proves the assertion directly, without using step 1 or the nonseparatedness of \(X\).
5. Q.E.D.
Step 1 proves (a). Steps <1>2–<1>4 compute the Picard group, determine its globally generated classes, and exclude every ample class, proving all of (b).