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Let $X$ be a one-dimensional Noetherian scheme over an algebraically closed field $k$. Suppose $X$ is reduced and let $X=\bigcup X_i$ be the composition of $X$ into irreducible components. Then, is the following homomorphism surjective?

$\mathrm{Pic} X\to \bigoplus \mathrm{Pic} X_i$.

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    $\begingroup$ Yes. Use the description of $Pic(X)$ as the cohomology group $H^1(X,O_X^*)$. You may want to see the thread mathoverflow.net/questions/57127/…. $\endgroup$
    – J.C. Ottem
    Commented Mar 4, 2011 at 22:22
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    $\begingroup$ This is problem III.5.8 in Hartshorne by the way. $\endgroup$
    – J.C. Ottem
    Commented Mar 4, 2011 at 22:25

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Yes! This can be shown using the isomorphism $H^1(X,\mathcal{O}_X) \cong \mathrm{Pic} X$. First, look at the short exact sequence:

$ 0 \to \mathcal{O}_X^* \to \bigoplus \mathcal{O}_{X_i}^* \to \mathcal{C} \to 0$

From the long exact sequence of cohomological groups associated to the short exact sequence, it suffices to show that $H^1(X,\mathcal{C})\cong 0$. However, this is clear since from the short exact sequence above, we can see that the support of $\mathcal{C}$ is a finite number of points (points that belong to more than just 1 irreducible component) and hence, of dimension 0. Now, use Grothendieck's vanishing theorem and we are done.

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  • $\begingroup$ Yes, although the Grothendieck vanishing theorem is perhaps a bit of a sledgehammer here.. $\endgroup$
    – J.C. Ottem
    Commented Mar 4, 2011 at 22:48
  • $\begingroup$ True. But using this, we get a strong result: it seems to work for Noetherian locally ringed space of dimension 1 in general. $\endgroup$
    – Brian
    Commented Mar 4, 2011 at 22:52

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