# Connected components of schemes over $\mathbb{F}_1$

I'm reading Deitmar's paper on Schemes over $\mathbb{F}_1$. 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:

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 $\mathcal{O}_{X,x} = \mathcal{O}_{X,x}^*$, 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?

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.

I think this is a simple topological statement using the following facts:

1. Every affine set has a unique generic point.

2. Every open set contains an affine open subset.

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.

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.

We therefore get a bijection between the set of connected components of $X$ and the set of generic points of affine open subschemes.

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 $U$.

We conclude that $Hom(Spec(F_1),X)$ stands in bijection to $\pi_0(X)$.

• What do you mean by a generic point? If $k$ is a field, how many generic points does the affine scheme ${\rm Spec}(k+k)$ have? – M P Feb 4 '12 at 16:55
• A generic point of a topological space is a point that is contained in every non-empty open subset. – user1688 Feb 4 '12 at 17:29
• @Xogn: Thank you! One also has to show that the sets $Y_{\eta}$ you describe partition $X$, but this is easy. And the $\eta$ correspond to the points I mention in my question, this gives an alternative finish for the proof. @MP: A point $x \in X$ is called generic if $\{x\}$ is dense in $X$. For a scheme $\mathrm{Spec}(A)$ over $\mathbb{Z}$, this point exists and is unique iff $\mathrm{rad}(A)$ is a prime ideal. For a scheme $\mathrm{Spec}(A)$ over $\mathbb{F}_1$ à la Deitmar, this always exists and is given by the prime ideal $\emptyset \subseteq A$. – Martin Brandenburg Feb 4 '12 at 17:41