6
$\begingroup$

I'd like to check a definition:

If $X$ is a scheme, what does it mean to say that $X$ is "defined over $\textrm{Spec }\mathbb{Z}$"? Is this a precise statement? Certainly this statement requires that $X$ is finite type over $\textrm{Spec }\mathbb{Z}$.

If $X$ is a projective or affine variety over $\textrm{Spec }\mathbb{C}$ (with choice of embedding) we can ask if the coefficients of the equations defining it are integers, and maybe call such a scheme to defined over $\textrm{Spec }\mathbb{Z}$. Other than not being "coordinate free", it is also easy to get schemes that are better said to be "defined over $\textrm{Spec }\mathbb{Z}[1/N]$" (in particular, there are lots of examples when a construction is defined over the latter scheme, and a goal is to make it defined over $\textrm{Spec }\mathbb{Z}$).

$\textbf{Question:}$ If $X$ is a scheme, does the phrase "defined over $\textrm{Spec }\mathbb{Z}$" mean that the structure map to $\textrm{Spec }\mathbb{Z}$ is fppf? (or is the base change to say $\textrm{Spec }\mathbb{Q}$ of a scheme with such a structure map).

$\endgroup$
4
  • 1
    $\begingroup$ This is certainly not a precise statement to say that a scheme is defined over $\mathbf{Z}$, since for every scheme $X$ there is a unique morphism from $X$ to $\mathbf{Z}$. Could you point us to specific places where this terminology is used? $\endgroup$ Nov 29 '12 at 8:29
  • $\begingroup$ I mean from $X$ to $\mathrm{Spec} \mathbf{Z}$. $\endgroup$ Nov 29 '12 at 8:30
  • 11
    $\begingroup$ Dear LMN, There is no precise definition in general; this is an informal phrase whose meaning you have to interpret on a case-by-case basis. (To see some of the complications in doing so, you could look at the literature on canonical models of Shimura varieties, and see the difficulties that arise there, in the non-proper case, in giving a good definition of canonical model over an integer ring.) Regards, $\endgroup$
    – Emerton
    Nov 29 '12 at 11:44
  • $\begingroup$ Emerton, Thanks a lot for clarifying this! $\endgroup$
    – LMN
    Nov 29 '12 at 16:23
2
$\begingroup$

A scheme $X$ is defined over $\mathbb Z$ if there is a scheme $X_0$ over $\mathbb Z$ (usually of finite type) such that $X$ is obtained from $X_0$ by some base change $S \to Spec(\mathbb Z)$. For example, if $X$ is a scheme over a field $K$ then $X$ should be isomorphic to $X_0\times_{\mathbb Z}K$.

$\endgroup$
3
  • 3
    $\begingroup$ Sasha, your definition doesn't distinguish between things defined over $\textrm{Spec } \mathbb{Z}$ and those over $\textrm{Spec } \mathbb{Z}[1/N]$. I've certainly seen uses of this phrase that distinguish between these two (as above). $\endgroup$
    – LMN
    Nov 29 '12 at 5:56
  • $\begingroup$ The fact that the requirement "finite type" doesn't distinguish between these is the reason for my question. Requiring fppf for the structure map does distinguish, and my question is if this is the definition in arithmetic geometry. $\endgroup$
    – LMN
    Nov 29 '12 at 6:07
  • $\begingroup$ There's a bad ambiguity between "defined over $\mathbf{Z}$" and "definable over $\mathbf{Z}$", which is not fixed by this answer, since "there is" is highly ambiguous. If it means "there exists", it should be "definable over $\mathbf{Z}$", often said "defined", and if it means "there is a given/ endowed with a given", it really means "defined over $\mathbf{Z}$". $\endgroup$
    – YCor
    Jun 13 at 10:31

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.