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Let $X$ be a noetherian separated scheme over the spectrum of a field $K$. By definition, I can find an open covering of $X$ by finitely many affine schemes $Spec(R_i)$ where each $R_i$ is a $K$-algebra. A scheme is a bunch of rings glued together. Is a scheme also a bunch of local rings glued together? More precisely:

Can every such $X$ be covered by finitely many spectra of local rings (which are also $K$-algebras) such that the intersection of any number of them is either empty or again the spectrum of a local ring?

My first try was to cover $X$ by finitely many $D(f)$ where each $f$ is an element of some $R_i$ and $D(f)\cong Spec((R_i)_f)$. However, the prime ideals of $(R_i)_f$ correspond one-to-one to those of $R_i$ not containing a power of $f$. In particular, $(R_i)_f$ does not have to be a local ring.

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  • $\begingroup$ Considering curves: how can one glue points without considering a neighborhood? $\endgroup$ – Hagen Knaf Feb 27 '13 at 10:25
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    $\begingroup$ The answer is yes iff $X$ has finitely many closed points. $\endgroup$ – user18119 Feb 27 '13 at 14:16
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The answer to your question is no: take $X = Spec( \mathbb{C}[t])$, the affine line over $\mathbb C$. The set-theoretic image of any map $$Spec(R) \to X$$ for a local ring $R$, contains a single maximal ideal of $X$, so you definitely can't cover $X$ by finitely many of these.

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    $\begingroup$ Rather than "a single maximal ideal", you should say "at most one maximal ideal". (Take $R = \mathbb{C}(t)$.) $\endgroup$ – Zhen Lin Feb 27 '13 at 9:43

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