Given a scheme $X$, its structure sheaf on the elements of the cover by affines is pretty easy to define, say $\mathcal{O}_X(Spec(A))=A$. But how is it then defined on the union of two of these sets? In general, it is clear how to define open subfunctors of affines, and the definition of an open subfunctor of $X$ is completely understandable, but what does the "union" of two open subfunctors look like?



The structure sheaf is actually not so simple. On base open affines $U_f$, we define it to be $\mathcal{O}_X(U_f) = A_f$ (localization), but as you say this doesn't tell us what $\mathcal{O}_X(U)$ should be if $U$ is open but not affine.

If you look at Hartshorne, for example, you'll see that the better way to proceed is to first define the stalks of $\mathcal{O}_X$, and then take $\mathcal{O}_X(U)$ to be the ring of local sections (maps from $U$ to the stalks over points in $U$ satisfying some nice conditions). It is not too hard to check that on an open affine $U_f$, this is equivalent to $\mathcal{O}_X(U_f) = A_f$. This is very similar to sheafification of a presheaf.

Another way to think of this, if it helps, is that since any open set $U$ can be covered by affine open subsets $\{U_{f_i}\}_{i \in I}$, so a section of $\mathcal{O}_X(U)$ can be thought of as a collection of sections $s_i \in A_{f_i}$, satisfying the condition that $s_i$ and $s_j$ agree on the intersection of $U_{f_i}$ and $U_{f_j}$ (and we can cover $U_{f_i} \cap U_{f_j}$ by open affines if necessary, to make sense of this condition). The point is that since the base open affines $U_f$ form a base of the topology on $X$, any sheaf on $X$ is uniquely determined by its values on these base open affines.

There was a good MO thread on this, which can be found here.

  • $\begingroup$ Thanks very much for this! $\endgroup$ – Jonathan Beardsley Mar 16 '12 at 23:26

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