About two isomorphic schemes My question is related to an answer I read on MO:
https://mathoverflow.net/questions/157973/classical-algebraic-varieties-vs-k-schemes-vs-schemes
In the accepted answer, the user Julian Rosen claims that if $\sigma\in\operatorname{Aut}\mathbb C$ exchanges $e$ and $\pi$, then the two schemes associated to the varieties $V_1$ and $V_2$ are isomorphic. I don't undestand why this is true.
Thanks in advance.
Edit: As I say in some comments, the problem is that when one sends $V_1$ and $V_2$ in the category of schemes with the functor $t(\cdot)$ (see Hartshorne Proposition 2.6), I don't understand how the element $\sigma$ works.
 A: Well, suppose we had such an automorphism of $\mathbb{C}$. This extends to an automorphism of $\mathbb{P}^1$, and since it comes from a field automorphism, it must fix each of $\{ 0, 1, \infty \}$. By hypothesis it exchanges $e$ and $\pi$, so this induces an isomorphism of the subschemes $\mathbb{P}^1 \setminus \{ 0, 1, e, \infty \}$ and $\mathbb{P}^1 \setminus \{ 0, 1, \pi, \infty \}$. 
The point is that, given any ring homomorphism $\sigma : A \to B$, we get an induced graded ring homomorphism $A [x_0, x_1] \to B [x_0, x_1]$ and hence a scheme morphism $\sigma^* : \mathbb{P}^1_B \to \mathbb{P}^1_A$. (Note the direction!) For each pair $(a_0, a_1)$ of elements of $A$, we get a commutative diagram of the following form,
$$\begin{array}{ccc}
A [x_0, x_1] & \rightarrow & B [x_0, x_1] \\
\downarrow & & \downarrow \\
A [t] & \rightarrow & B [t]
\end{array}$$
where the horizontal arrows are induced by $\sigma : A \to B$, $A [x_0, x_1] \to A [t]$ is defined by $f (x_0, x_1) \mapsto f (a_0 t, a_1 t)$, and $B [x_0, x_1] \to B [t]$ is defined by $g (x_0, x_1) \mapsto g (\sigma (a_0) t, \sigma (a_1) t)$. Applying $\operatorname{Proj}$, we get a commutative diagram of the form below:
$$\begin{array}{ccc}
\operatorname{Spec} B & \rightarrow & \operatorname{Spec} A \\
\downarrow & & \downarrow \\
\mathbb{P}^1_B & \rightarrow & \mathbb{P}^1_A
\end{array}$$
Thus, if $A = B = k$ is an algebraically closed field and $\sigma$ is an automorphism, then the induced automorphism of $\mathbb{P}^1_k$ is defined on closed points by $(b_0 : b_1) \mapsto (\sigma^{-1} (b_0) : \sigma^{-1} (b_1))$.
Perhaps you are objecting to the claim that such an automorphism of $\mathbb{C}$ exists. This is a more non-obvious fact but it doesn't really matter: we could equally well replace $e$ and $\pi$ with some pair of transcendental numbers for which there is an automorphism exchanging them. And there are lots of automorphisms of that form: see here.
