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
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2 Answers
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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.
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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 $X$.
