A scheme, by definition, is a locally ringed space in which every point has an open neighborhood $U$ such that the topology space $U$, together with the restricted sheaf, is an affine scheme(isomorphic to the spectrum of some ring). And as we know, a variety can by covered by affine open subsets so that we can deal with problems locally. Thus, I wonder whether a scheme can be covered by affine schemes and using the local property of spectrum to solve problems?
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1$\begingroup$ Yes. A scheme can be covered by affine schemes by definition. Problems of a local nature can often be reduced to the spectrum of a ring, and thus to commutative algebra. But not all problems are of a local nature. This is what I know, from my limited knowledge of this area. $\endgroup$– MalkounNov 26, 2018 at 11:44
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$\begingroup$ Thanks! That explains a lot. $\endgroup$– Yuyi ZhangNov 26, 2018 at 12:00
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$\begingroup$ A scheme, from the functor of points POV, is canonically a colimit over all affine schemes mapping into it. This essentially means that a scheme is covered by affines. $\endgroup$– Dat Minh HaDec 5, 2019 at 7:21
2 Answers
Yes:
A scheme is a locally ringed space $X$ admitting a covering by open sets $U_i$, such that each $U_i$ (as a locally ringed space) is an affine scheme. In particular, $X$ comes with a sheaf $O_X$, which assigns to every open subset $U$ a commutative ring $O_X(U)$ called the ring of regular functions on $U$. One can think of a scheme as being covered by "coordinate charts" which are affine schemes. The definition means exactly that schemes are obtained by gluing together affine schemes using the Zariski topology.
Yes, it is already in the definition you give. If $(X,\mathcal{O}_X)$ is a scheme, then by definition every point $x\in X$ has an open neighbourhood $U_x$ such that $(U_x,\mathcal{O}_X\vert_{U_x})$ is an affine scheme. Then $\{U_x\}_{x\in X}$ is an affine open cover of $X$.