# In how many ways can one extend the zero section of the affine line with a double origin

Let $$X$$ be the affine line with a double origin over $$\mathrm{Spec}\,\mathbb Z$$. Let $$X_\eta$$ be its generic fibre, the affine line with a double origin over $$\mathrm{Spec}\,\mathbb Q$$.

Let $$0$$ be one of the origins of $$X_\eta$$.

Are there infinitely many ways of extending $$0$$ to a section of $$X$$?

My feeling is that besides taking the closure of $$0$$ in $$X$$, we can also alter our choice of origin at some primes. If that feeling is correct, then we would get infinitely many sections of $$X\to \mathrm{Spec}\,\mathbb Z$$ inducing the same section generically.

• You should think about this on your own. Do you know a finite atlas of open affines for $X$? For each such open, what is the relative closure of $\{0\}$ in that open? – Jason Starr Aug 11 '15 at 12:01
• ... My last sentence should have been: What are the inverse images of these open affines with respect to your (putative) section? – Jason Starr Aug 11 '15 at 13:51

The answer is "no", two generically equal sections are equal.

First prove that the image of a morphism $$f:\mathrm{Spec}\,\mathbb{Z}\rightarrow X$$ can not intersect both origins. Since $$\mathrm{Spec}\,\mathbb{Z}$$ is the terminal object in the category of schemes, any morphism $$\mathrm{Spec}\,\mathbb{Z}\rightarrow X$$ is a section (in particular, a locally closed immersion).

Note that $$\mathrm{Spec}\,\mathbb{Z}$$ is irreducible, the image of an irreducible space is irreducible and the closure of an irreducible subset is irreducible–so the closure $$Y$$ of $$f(\mathrm{Spec}\,\mathbb{Z})$$ is irreducible. It can not have Krull dimension 0 (because $$\mathrm{Spec}\,\mathbb{Z}$$ is one-dimensional) and it can not have dimension 2 (a closed irreducible subset of top dimension has to be the whole space; because both $$\mathrm{Spec}\,\mathbb{Z}$$ and $$\mathrm{Spec}\,\mathbb{Z}[x]$$ are reduced, this would mean that the former is an open subscheme of the latter–but this is not true because function fields are not isomorphic). Therefore, $$Y$$ is a closed irreducible subscheme of dimension 1. It also has codimension 1 in $$X$$ because $$X$$ is a catenary space.

The intersection of $$Y$$ with a standard affine open $$U_1\subset X$$ is a non-empty closed subscheme of $$U_1$$. It is irreducible beucase it can also be considered as a non-empty open subscheme of $$Y$$. It has dimension 1 because codimension is preserved under taking intersection with opens and $$U_1$$ is catenary. This means that it is cut out by a prime ideal of height 1; such a prime ideal is generated either by a prime number $$p$$ or a non-zero irreducible polynomial $$f(x)$$. It can not be the first because $$f(\mathrm{Spec}\,\mathbb{Z})\approx \mathrm{Spec}\,\mathbb{Z}$$ is an open subscheme of $$Y$$ so its function field has to be $$\mathbb{Q}$$; it can not be the second because $$Y$$ then could not intersect both $$x=0$$ and $$x\neq 0$$ (and it has to intersect the latter since otherwise $$f(\mathrm{Spec}\,\mathbb{Z})$$ would be reducible). Contradiction.

So the image of any morphism $$\mathrm{Spec}\,\mathbb{Z}\rightarrow X$$ has to lie in one of the two standard affine opens. It is clear that generically equal morphisms $$\mathrm{Spec}\,\mathbb{Z}\rightarrow \mathrm{Spec}\,\mathbb{Z}[x]$$ are equal.

EDIT: since there are many down-votes and no comments, I would guess that the last step seems unjustified (if this edit seems to be useless, feel free to down-vote but please leave a comment).

Recall that a morphism $$\mathrm{Spec}\,\mathbb{Z}\rightarrow \mathrm{Spec}\,\mathbb{Z}[x]$$ is equivalent to the data of a morphism of commutative unital rings $$\mathbb{Z}[x]\rightarrow \mathbb{Z}$$. If two morphisms $$f_1$$, $$f_2:\mathbb{Z}[x]\rightarrow \mathbb{Z}$$ define generically equal sections, then $$f_1(a)\otimes b-f_2(a)\otimes b=0\in \mathbb{Q}$$ for all $$a\in \mathbb{Z}[x], b \in \mathbb{Q}$$. By linearity of tensor product, this implies $$(f_1(a)-f_2(a))\otimes b=0$$ for all $$a\in \mathbb{Z}[x], b \in \mathbb{Q}$$. But there is no non-zero $$c\in \mathbb{Z}$$ such that $$c\otimes b=0$$ for some non-zero $$b \in \mathbb{Q}$$. Therefore, $$f_1(a)-f_2(a)=0$$, i.e. two morphisms are equal.