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.

  • 1
    $\begingroup$ 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? $\endgroup$ – Jason Starr Aug 11 '15 at 12:01
  • 1
    $\begingroup$ ... My last sentence should have been: What are the inverse images of these open affines with respect to your (putative) section? $\endgroup$ – 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.