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?


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.

  • $\begingroup$ 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

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