A pair $(X,O_X)$ is a ringed space if $X$ is a topological space and $O_X$ is a sheaf of rings. If every stalk $O_{X,x}$ is a local ring, then we say that $(X,O_X)$ is a locally ringed space.

In the case of $X$ being an abstract algebraic variety, not necessarily irreducible, and $O_X$ its sheaf of regular functions:

1) Is it possible to have a ringed space which is not a locally ringed one? And if $X$ is irreducible? My guess is NO for the first question and YES for the second.

Correct me if I am wrong. My reasoning goes as follows, since stalks are local we can work on the affine case. All irreducible algebraic varieties are quotient of $K[x_1,\ldots,x_n]$ by a prime ideal and prime ideals always have a unique maximal ideal, corresponding to a point, so the quotient on the point is a field. This seems to work for $K=\mathbb{C}$. In the irreducible case I am not sure of what happens, but for the complex case you can have two lines crossing and I get each of them may work independently giving two local ring ideals? Pure speculative...

2) Is this true for non-algebraically closed fields or fields with positive characteristic?

3) Could you please provide with any example of non-locally ringed spaces? In varieties or schemes, if it is possible.

4) If $X$ is a variety/scheme, is there any example of morphism of ringed spaces which are locally ringed but which is not a morphism of locally ringed spaces? (i.e. for $f\colon X\rightarrow Y$ there is a sheaf morphism $f^\sharp\colon O_Y \rightarrow f_* O_X$ which is not a morphism of local rings on the stalks).

These questions are a bit vague but hopefully you understand what I mean.

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    $\begingroup$ Schemes have open coverings by affine schemes, which are locally ringed spaces, so schemes are locally ringed spaces. There are trivial examples of ringed spaces which are not locally ringed. Consider the one point space $X=\{x_0\}$. Given any ring $R$ there is a sheaf of rings on $X$ with global sections $R$. If $R$ is not local, these ringed space is not a locally ringed space. Of course, such examples are nothing like the ones that arise geometrically. $\endgroup$ Sep 10, 2010 at 14:49
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    $\begingroup$ Jesus, perhaps I am being dumb here, but would you explain the paragraph after 1) a little: what do you mean by "prime ideals always have unique maximal ideal" and "two local rings ideals"? Thanks. $\endgroup$ Sep 10, 2010 at 15:41
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    $\begingroup$ I wrote some notes giving motivation for the notion of "locality" for morphisms of ringed spaces. $\endgroup$ Sep 10, 2010 at 19:19
  • $\begingroup$ A counterexample to what you talk about in 4) is to be found in Example 6.17 here. $\endgroup$ Jul 5, 2023 at 9:55

1 Answer 1

  1. No, all schemes are locally ringed spaces (cf. Robin Chapman's comment above). In fact, schemes are often defined by using locally ringed spaces.
  2. It depends on what you mean by "this". If you have a pair of lines crossing, the local ring at the crossing has zero divisors, but it's still local.
  3. Take a point, and set its ring of functions to be the ring of rational integers (see Robin Chapman's comment again). It's not a scheme. There are more interesting examples in, e.g., work by Kato and Nakayama on log analytic spaces.
  4. Yes. Hartshorne produces one as example 2.3.2 in his Algebraic Geometry: The spectrum of a discrete valuation ring $R$ is a locally ringed space made of an open dense point (whose stalk is the fraction field of $R$) and a closed point (whose stalk is $R$). There is a ringed space map from the spectrum of $R$ to itself that takes both points to the closed point, where the restriction to the closed stalk is identity, and the restriction to the open dense point is defined by the inclusion of $R$ into its fraction field.

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