0
$\begingroup$

Let $X$ be an integral scheme over $\mathrm{Spec}(\mathbb{Z})$, we denote $X(\mathbb{C})$ as the set of $\mathbb{C}$-valued points in $X$. Then $Y\rightarrow \mathrm{Spec}(\mathbb{Z})$ is flat if $Y(\mathbb{C})\not=\emptyset$.

Can anyone show me how to prove this?

$\endgroup$
2
$\begingroup$

Since $\mathbb{Z}$ is a pid, flatness is same as torsion free. Since $Y$ is an integral scheme, torsion free just means the map is dominant. Dominance is equivalent to to $Y(\mathbb{C})\neq\emptyset$.

This is just to answer the query by the OP in the comments. $f:Y\to\mathbb{Z}$ is dominant means the generic point is in the image. That is, there exists a morphism $\mathrm{Spec}\, K\to Y$ which when composed with $f$ factors through the generic point $\mathrm{Spec}\,\mathbb{Q}\subset\mathrm{Spec}\,\mathbb{Z}$. Base changing to $\mathrm{Spec}\,\mathbb{C}\to \mathrm{Spec}\,\mathbb{Q}$, we get a morphism $\mathrm{Spec}\,(K\otimes_{\mathbb{Q}}\mathbb{C})\to Y\times_{\mathbb{Z}}\mathrm{Spec}\, \mathbb{C}$. Hope rest is clear.

$\endgroup$
  • $\begingroup$ always so prompt and excellent answer $\endgroup$ – Intoks Liobein Nov 17 '18 at 15:11
  • $\begingroup$ Can you explain a little bit for your very last step? $\endgroup$ – Intoks Liobein Nov 21 '18 at 0:50

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.