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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.

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Ginevra Carbone
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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.

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Zhen Lin
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  • I have proved that such automorphism exists, but the problem is the following: $V_1$ and $V_2$ are subsets of $\mathbb P_1$, but not two schemes. First one should pass to the schemes associated to $V_1$ and $V_2$ and then discuss the isomorphism. Practically I have problem to understand what is the isomorphism induced by $\sigma$. – Ginevra Carbone Mar 01 '14 at 18:26
  • They are open subsets and thus inherit the structure of an open subscheme. Morphisms of schemes can obviously be restricted to open subschemes. – Zhen Lin Mar 01 '14 at 19:14
  • We are considering $\mathbb P^1(\mathbb C)=\frac{\mathbb C^2\setminus{0}}{\sim}$ not ${\mathbb P^1}_{\mathbb C}=\operatorname{Proj}(\mathbb C[T_0,T_1])$. The first object is in bijection with the closed point of the second. – Ginevra Carbone Mar 01 '14 at 19:17
  • No, I do mean $\mathbb{P}^1_\mathbb{C}$. $\mathbb{P}^1 (\mathbb{C})$ is not a scheme. – Zhen Lin Mar 01 '14 at 19:18
  • Topologically $t(V_1)$ is the set of al closed irreducible subspaces of $V_1$... – Ginevra Carbone Mar 01 '14 at 21:35
  • No, just do things directly with the schemes! – Zhen Lin Mar 01 '14 at 21:40
  • I think you might be missing the point that Zhen doesn't highlight, which is that the graded ring maps $A[x_0,x_1]\to B[x_0,x_1]$ are in bijection with the scheme morphisms $\mathbb{P}^1_B\to\mathbb{P}^1_A$. – Kevin Carlson Mar 01 '14 at 23:07