Connected components of schemes over $\mathbb{F}_1$ - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T17:56:21Z http://mathoverflow.net/feeds/question/87460 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/87460/connected-components-of-schemes-over-mathbbf-1 Connected components of schemes over $\mathbb{F}_1$ Martin Brandenburg 2012-02-03T17:51:01Z 2012-02-04T16:39:08Z <p>I'm reading Deitmar's paper on <a href="http://arxiv.org/abs/math/0404185v7" rel="nofollow">Schemes over $\mathbb{F}_1$</a>. Proposition 2.4. states that for a scheme $X$ over $\mathbb{F}_1$ there is a bijection between $X(\mathbb{F}_1)$ and the set of connected components of $X$. I don't understand the proof, which is quite sketchy. Here is what I think:</p> <p>Elements of $X(\mathbb{F}_1)$ correspond to morphisms $\mathrm{Spec}(\mathbb{F}_1) \to X$, where $\mathrm{Spec}(F_1)$ is the point together with the trivial monoid sheaf $1$. These morphisms correspond to a point $x \in X$ together with a local homomorphism $\mathcal{O}_{X,x} \to \{1\}$. But this is unique and exists iff <code>$\mathcal{O}_{X,x} = \mathcal{O}_{X,x}^*$</code>, i.e. iff the stalk is actually a group. Now to such a point we should associate to irreducible closed subset $\overline{\{x\}} \subseteq X$. But why should it be a connected component, and why does every one arise this way?</p> <p>I can show that every irreducible scheme over $\mathbb{F}_1$ has exactly one generic point. So perhaps Proposition 2.4 should talk about irreducible components? I'm a bit confused. Also Deitmar's proof suggests implicitly that every $X$ is the disjoint union of its connected components, i.e. that they are open - but why should this be true? For ordinary schemes this is true at least in the noetherian case.</p> http://mathoverflow.net/questions/87460/connected-components-of-schemes-over-mathbbf-1/87535#87535 Answer by anton for Connected components of schemes over $\mathbb{F}_1$ anton 2012-02-04T16:39:08Z 2012-02-04T16:39:08Z <p>I think this is a simple topological statement using the following facts:</p> <ol> <li><p>Every affine set has a unique generic point.</p></li> <li><p>Every open set contains an affine open subset.</p></li> </ol> <p>Let $X$ be a scheme over $F_1$. If $U$ and $V$ are affine open subsets with $U\cap V\ne \emptyset$, then their generic points coincide, as both generic points also form a generic point of any affine subset of $V\cap U$, which is unique.</p> <p>Let now $U$ be an affine open subset of $X$ with generic point $\eta$. Let $Y$ be the union of all open affine subsets $V\subset X$ which contain $\eta$. Then every open subset of $Y$ contains $\eta$, which implies that $Y$ is connected. Further, $Y$ is open. We show that it also is closed. Let $Z$ be its complement and let $z\in Z$ a point. Let $W$ be an affine neighborhood of $z$. If $W$ has non-empty intersection with $Y$, then the generic point of $W$ coincides with $\eta$, therefore $W$ is a subset of $Y$ by the definition of $Y$. Hence $W\cap Y=\emptyset$, so $W\subset Z$, and $Z$ contains an open neighborhood of each of its points, so $Z$ is open.</p> <p>We therefore get a bijection between the set of connected components of $X$ and the set of generic points of affine open subschemes.</p> <p>Next for any affine open subscheme $U$ there is a unique morphism from $Spec(F_1)$ to $U$ mapping the unique point of $Spec(F_1)$ to the generic point of $U$.</p> <p>We conclude that $Hom(Spec(F_1),X)$ stands in bijection to $\pi_0(X)$.</p>