Zariski covering

We also have a good many coverings which are not Zariski coverings— consider the case R = k. But the analogy is close enough that we say a functor F satisfying (a) and (b) is a sheaf in the faithfully flat or fpqc topology. 

Notice that if X = Spec (R), T = M, then for all prime ideals P C R, [P] is an associated point of T if and only if P is an associated prime ideal of M. (For the definition and theory of these, see Bourbaki, Ch. 4, 1 or Zariski-Samuel, vol. 1, pp. 252-3, where these are called the prime ideals associated to the 0 submodule of Mn.) The only problem in proving this is to check that, when X is affine, to find the associated points of T it suffices to look at the supports of global sections s G r(X, T) we leave this point to the reader to check. One of the main facts in the theory of noetherian decompositions is that a finite. R-module has only a finite number of associated prime ideals. Since any noetherian scheme can be covered by a finite set of open affines, this implies... 

Example 1.10. - - Let us give an example which shows that Proposition 1.9 is false for Zariski topology. Let xq, xi be two closed points of over a field k. I et S be the spectrum of the semilocal ring of xo,xi. Any Zariski open covering for S has a refinement which consists of exaedy two open subsets and therefore F) = 0 for... 

To prove the general case we proceed as follows. Fisrt of all since Z — X is a closed embedding of smooth schemes there exists a finite Zariski open covering X = UU, such that for any i the embedding ZPlU, —> U, satisfies the condition of Lemma 2.28. Note also that if this condition holds for Z —> X it also holds for Z fl LT —> U where U is any open subset of X. In particular, it holds for all intersections of the form... 

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