Base change of a $\mathbb{Z}$-scheme to $\mathrm{Spec}(k)$, $k$ a field Consider a $\mathbb{Z}$-scheme $\chi$. I hope I am correct that this is just a scheme $\chi \mapsto \mathrm{Spec}(\mathbb{Z})$. Now let $k$ be any field. As we have a morphism $\mathrm{Spec}(k) \mapsto \mathrm{Spec}(\mathbb{Z})$, we can form the fiber product $\chi \otimes_{\mathrm{Spec}(\mathbb{Z})}\mathrm{Spec}(k) =: \chi_k$. Is this correct ?
Two elementary questions then:


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*Consider the $\mathbb{Z}$-scheme $\chi = \mathrm{Spec}(\mathbb{C}[x,y])$; how does $\chi_{\mathbb{F}_q}$ look like, with $\mathbb{F}_q$ any finite field ?

*Consider any $\mathbb{Z}$-scheme, and any finite field $\mathbb{F}_q$. Is it then true that $\chi(\mathbb{F}_q) = \mathrm{hom}(\mathrm{Spec}(\mathbb{F}_q),\chi) = \mathrm{hom}(\mathrm{Spec}(\mathbb{F}_q),\chi_{\mathbb{F}_q})$ ?
 A: Adding to Mohan's answer, 


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*$\chi_{\mathbb{F}_q}$ is empty, since the ring $\mathbb{C}[x,y]\otimes_\mathbb{Z} \mathbb{F}_q$ is the zero ring. Indeed, if $p = \text{char}(\mathbb{F}_q)$, $1 = 1\otimes 1 = pp^{-1}\otimes 1 = p^{-1}\otimes p = p^{-1}\otimes 0 = 0$. Of course you can replace $\mathbb{C}[x,y]$ with any ring in which $p$ is invertible.

*By the universal property of the fiber product (and the fact that $\text{Spec}(\mathbb{Z})$ is the terminal object in the category of schemes), $$\text{hom}(\text{Spec}(\mathbb{F}_q),\chi_{\mathbb{F}_q}) \cong \text{hom}(\text{Spec}(\mathbb{F}_q),\chi)\times \text{hom}(\text{Spec}(\mathbb{F}_q),\text{Spec}(\mathbb{F}_q)).$$ When $q$ is prime, this is naturally isomorphic to $\text{hom}(\text{Spec}(\mathbb{F}_q),\chi)$, but here you're using the fact that $\mathbb{F}_q$ has no nontrivial endomorphisms. If you instead use a field $k$ which has nontrivial endomorphisms (like $\mathbb{F}_{p^n}$ for $n>1$ or $\overline{\mathbb{F}_p}$, or $\mathbb{C}$), then a $k$-point of the base change $\chi_k$ is a $k$-point of $\chi$ together with a "twist" of the structure sheaf coming from a map $k\to k$. This is fixed by thinking of $\chi_k$ as a $k$-scheme, and looking at morphisms of $k$-schemes (which preserve the structure map $\chi_k\to \text{Spec}(k)$). So the correct general statement is $$\text{hom}_\mathbb{Z}(\text{Spec}(k),\chi) \cong \text{hom}_k(\text{Spec}(k),\chi_k)$$
A: *

*$\chi_{\mathbb{F}_q}$ is empty.

*This is correct.
