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Let $D$ be the Weil divisor $D = -2[(x)] + [(x-1)] + [(x-2)]$ on $\Bbb{A}^1_k = \operatorname{Spec} k[x]$. I want to show that $\mathcal{O}_X(D) \cong \mathcal{O}_X$. To do this, it is enough to specify isomorphisms $$\mathcal{O}_X(D)(D(f_i)) \to k[x]_{f_i}$$ that agree on $D(f_i) \cap D(f_j) = D(f_if_j)$. Now to do this I calculate the following: \begin{eqnarray*} \mathcal{O}_X(D)(D(x^2)) &=& k[x]\cdot \frac{1}{(x-1)(x-2)} \\ \mathcal{O}_X(D)(D(x-2)) &=& k[x] \cdot \frac{x^2}{(x-1)} \\ \mathcal{O}_X(D)(D(x-1)) &=& k[x]\cdot \frac{x^2}{(x-2)} \\ \mathcal{O}_X(D)(D(f(x))) &=& k[x]\cdot \frac{x^2}{(x-1)(x-2)f(x)}\end{eqnarray*}

where $f$ is not one of the earlier three polynomials. Now I want to say that the map out of $\mathcal{O}_X(D)D(x^2)$ is multiplication by $(x-1)(x-2)/x^2$ and the map out of $\mathcal{O}_X(D)D(x-2)$ is mutiplication by $(x-1)/(x^2(x-2))$. However on the intersection which is $D((x-2)x^2)$, the maps don't seem to agree. What's the problem here?

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    $O_X(D)=fO_X$ where $f=x^2/((x-1)(x-2))$. – Cantlog Oct 16 '13 at 14:17
  • Precisely, dear @Cantlog: +1. – Georges Elencwajg Oct 16 '13 at 14:47
  • @Cantlog It was an exercise in Ravi's notes to prove $\mathcal{O}_X(D) \cong \mathcal{O}_X$ and perhaps my confusion came from not interpreting the question correctly. I tried to set up an isomorphism directly to $\mathcal{O}_X$ and not $f\mathcal{O}_X$. –  Oct 16 '13 at 22:19
  • Dear @user38268: I think that in this concrete example, it is more interesting to find a basis than just knowning the sheaf is free of rank 1. – Cantlog Oct 17 '13 at 07:03

1 Answers1

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If $\phi$ is the rational function $\phi=\frac {(x-1)(x-2)}{x^2}$, the sheaf $\mathcal O_X(D)$ is exactly equal to $\frac {1}{\phi}\mathcal O_X= \frac {x^2}{(x-1)(x-2)}\mathcal O_X$ .
Notice that we have an equality , not just an isomorphism.
This means more precisely that on open subsets $U\subset \mathbb A^1_k$ we have the equality $$\Gamma(U,\mathcal O_X(D) )= (\frac {1}{\phi|U})\cdot\Gamma(U,\mathcal O_X) $$

So the natural isomorphism $\mathcal O_X(D) \stackrel {\cong}{\to} \mathcal O_X$ is the isomorphism of sheaves $$\phi \cdot : \mathcal O_X(D) \stackrel {\cong}{\to} \mathcal O_X $$ given on open subsets $U\subset \mathbb A^1_k$ by $$\Gamma(U,\mathcal O_X(D) )\to \Gamma(U,\mathcal O_X): s\mapsto \phi|U \cdot s = \frac {(x-1)(x-2)}{x^2} \cdot s$$

Your equalities however seem not to be correct: for example $\mathcal{O}_X(D)(D(f)) = k[x]\cdot \frac{1}{f}$ is not true and must be replaced by $\mathcal{O}_X(D)(D(f)) = \frac {x^2}{(x-1)(x-2)}\cdot k[x,\frac {1}{f(x)}] $

Conclusion
This is the kind of questions where it is essential to carefully distinguish between equality and isomorphism.

Georges Elencwajg
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