# invertible sheaves of the projective line

I'm trying to do the following exercise: "Consider the projective line $$X=\mathbb{P}^1_R$$ over a ring $$R$$. Describe Serre's twisted sheaves $$\mathcal{O}_X(n)$$, $$n\in\mathbb{Z}$$, via Cech cocycles and show that there cannot exist any further invertible sheaves on $$X$$, up to isomorphism."

For the first part I guess it's ok: $$\mathcal{O}_X(n)$$ correspond to the element of $$\tilde{H}^1(\mathcal{U},\mathcal{O}_X(n))$$ (with the covering $$\mathcal{U}=(D_+(T_0),D_+(T_1))$$) given by $$\lambda_{i,j}=\left(\frac{T_j}{T_i}\right)^n$$.

For the second part I can't get it: I take $$\mathcal{L}$$ corresponding to $$(\overline{\eta_{\alpha,\beta}})\in\tilde{H}^1(\mathcal{V},\mathcal{O}_X(n))$$. We can suppose that the covering $$\mathcal{V}$$ is finer than $$\mathcal{U}$$ and is given by principal open sets $$D_+(g_\alpha)$$ with all $$g_\alpha$$ having the same degree (because $$D+(g_\alpha)=D_+(g_\alpha^k)$$) and $$g_\alpha=T_{i(\alpha)}h_\alpha$$ (because $$\mathcal{V}$$ is obtained by intersection with $$\mathcal{U}$$ and $$D_+(T_i)\cap D_+(h_\alpha)=D_+(T_i h_\alpha)$$).

We have $$\eta_{\alpha,\beta}\in\mathcal{O}_X(D_+(g_\alpha)\cap D_+(g_\beta))^*$$ ie $$\eta_{\alpha,\beta}\in R[T_0,T_1]_{(g_\alpha g_\beta)}^*$$ ie $$\eta_{\alpha,\beta}=\frac{h_{\alpha,\beta}}{(g_\alpha g_\beta)^q}$$.

If I understand well we want $$(\eta_{\alpha,\beta})\sim\left(\left( \frac{T_{i(\beta)}}{T_{i(\alpha)}} \right)^n \right)$$ for some $$n\in\mathbb{Z}$$ that is we want the existence of $$f_\alpha\in\mathcal{O}_X(D_+ g_\alpha)^*=R[T_0,T_1]_{(g_\alpha)}^*$$ with $$\eta_{\alpha,\beta}\left( \frac{T_{i(\alpha)}}{T_{i(\beta)}} \right)^n=f_\beta f_\alpha^{-1}$$

I don't see how obtain that, the princpal problem is $$h_{\alpha,\beta}$$. I would like to have $$\eta_{\alpha,\beta}=\left(\frac{g_\beta}{g_\alpha}\right)^n$$ but I'm not sure that all the inversible of $$R[T_0,T_1]_{(g_\alpha g_\beta)}^*$$ are of this form.

This exercise is false as stated - if $$R$$ is not a sufficiently nice ring, then there is no hope of this at all. We'll outline how this exercise is probably supposed to be solved at the level of the other part of the question, and then discuss the general situation at the end of the post.

Proof using Cech cocycles: Assume $$R$$ is sufficiently nice (we'll say what we mean by this later - for now, know that "nice" includes being a domain). Write $$\Bbb P^1_R=\operatorname{Proj} R[x_0,x_1]$$ and $$\mathcal{F}$$ for our invertible sheaf. Then we claim that $$\mathcal{F}$$ is trivial when restricted to the standard affine opens $$D(x_0)$$ and $$D(x_1)$$. This is because every invertible sheaf on any affine space over $$R$$ for a sufficiently nice ring $$R$$ is trivial (that is, isomorphic to the structure sheaf). So the data of the gluing cocycle on $$D(x_0x_1)$$ is the only thing that specifies this sheaf.

It remains to identify what this cocycle can possibly be. We need it to be an invertible element in $$R[\frac{x_1}{x_0},\frac{x_0}{x_1}]=R[x,x^{-1}]$$ (the functions on $$D(x_0x_1)$$). We can write every element in this ring as a fraction $$\frac{p(x)}{x^n}$$ for some polynomial $$p(x)$$ and some nonnegative integer $$n$$. It is then clear that if this expression is invertible, it's inverse is $$\frac{x^n}{p(x)}$$. Write $$p(x)=x^mq(x)$$ where $$q(x)$$ has nonvanishing constant term. $$p(x)$$ is invertible inside $$R[x,x^{-1}]$$ iff $$q(x)$$ is, so now we analyze this case.

The inverse of $$q(x)$$ cannot have any terms of positive degree in $$x$$: if it did, multiplication by the highest degree part of $$q(x)$$ would give a surviving term of positive degree in the product $$q(x)\cdot(q(x))^{-1}$$, which is a contradiction. Similarly, the inverse cannot have any terms of negative degree in $$x$$ either: multiplication by the constant term would give a surviving term of negative degree in $$x$$ in the product, again a contradiction. (These two statements are where we use $$R$$ a domain.) So the inverse of $$q(x)$$ is just a constant in $$R$$, and by uniqueness of inverses, we have that $$q(x)$$ is just a unit $$r\in R$$ and the cocycle is of the form $$rx^d$$ for some integer $$d$$. We can now get rid of the $$r$$ by multiplying sections on one open by $$r$$, which is an isomorphism as $$r$$ is a unit.

So our cocycle is of the form $$\left(\frac{x_0}{x_1}\right)^d$$ for some integer $$d$$ and as two invertible sheaves trivial on the same covering with the same cocycles are isomorphic, our invertible sheaf $$\mathcal{F}$$ is some $$\mathcal{O}(d)$$ and we're done.

Promised discussion of the issues with this exercise: Every invertible sheaf on $$\Bbb P^n_S = \Bbb P^n_\Bbb Z \times_\Bbb Z S$$ is of the form $$p_1^*(\mathcal{O}_{\Bbb P^n_\Bbb Z}(n)) \otimes p_2^*(L)$$ for $$L$$ an invertible sheaf on $$S$$ - see here, with a reference to Mumford's Geometric Invariant Theory and some other helpful discussion). So if $$S$$ has more interesting invertible sheaves (for example, take $$S=\operatorname{Spec} R$$ with $$R$$ an affine model of a curve of genus $$>0$$ or your favorite Dedekind domain) then there is no hope for what you want to be true.

Clearly we should want "$$R$$ nice" to mean $$Pic(R)=0$$, and then we also use "niceness" in the statement that every invertible sheaf on $$\Bbb A^n_R$$ should be trivial. This gives us some hints about what collection of properties we really want when we say "$$R$$ nice". First, $$R$$ should be a domain, as otherwise we could do silly things that are different on each irreducible component. Next, if $$R$$ is normal, then $$Pic(R)=Pic(R\times \Bbb A^1)$$ (see here, though the argument is a little involved), so normal plus domain plus vanishing Picard group will work.

One example of a class of rings satisfying these conditions where we can prove the niceness properties a little more easily is noetherian UFDs - this just requires some material from Hartshorne chapter II section 6. Briefly, the property of being a UFD is equivalent to normal + vanishing class group, we have an isomorphism $$Cl(X)\cong Cl(X\times \Bbb A^1)$$ for noetherian integral normal separated $$X$$, and for any noetherian integral separated locally factorial scheme, the class group and the Picard group coincide, so our argument works.

• Dear KReiser, you write "... its inverse is $\frac{x^n}{p(x)}$, but this is impossible unless $p(x)=x^m$ for some nonnegative integer $m$." This is not true: for example $1+\epsilon X\in \mathbb C[\epsilon][X]$ is invertible over the ring of dual complex numbers ($\epsilon^2=0$). Of course this does not happen in the classes of rings you mention, but it is better to be precise in the general statement. Nevertheless +1 for this excellent answer. Mar 7, 2020 at 10:19
• @GeorgesElencwajg thank you for the correction - I believe I have now fixed the issue. Mar 7, 2020 at 19:12
• Dear KReiser: yes, everything is all right now. Mar 7, 2020 at 20:47