# $\mathbb{C}$-valued points and flatness

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.