We know that there is an equivalence of categories between the two following categories:

$1)$ Classical varieties over $k$, where $k$ is an algebraically closed field. (Informally I mean locally ringed spaces formed by patching affine irriducible algebraic sets over $k$. This is the definition of algebraic variety present for example in Perrin's book.)

$2)$ Schemes over $k$ which are integral and of finite type over $k$

Now I need an example, if it does exist, of two non isomorphic (as classical varieties) algebraic sets that are non isomorphic as $k$-schemes (this is obvious by the previous functor) but isomorphic in the category of schemes.

To be more specific I need two non isomorphic algebraic sets such that when I use the above functor $1)\longrightarrow 2)$ and then I forget the structure morphism on Spec $k$, I finally obtain two isomorphic schemes.

Thanks in advance

  • 1
    $\begingroup$ I've edited. $k$ It is an algebraically closed field. I'm sorry. $\endgroup$
    – Dubious
    Feb 18, 2014 at 21:05
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    $\begingroup$ To obtain what you want, you just have to take an algebraic variety an let an automorphism of $k$ act on the coefficients. This way you get plenty of examples. $\endgroup$ Feb 18, 2014 at 22:35

1 Answer 1


Let's say $k=\mathbb{C}$ (although something like this should work over any algebraically closed field). Let $V_1=\mathbb{P}^1-\{0,1,\infty,\pi\}$ and $V_2 = \mathbb{P}^1-\{0,1,\infty,e\}$. One can see that $V_1$ and $V_2$ are not isomorphic as varieties/schemes over $\mathbb{C}$: such an isomorphism would extend to a map $\mathbb{P}^1\to\mathbb{P}^1$ taking $\{0,1,\infty,\pi\}$ to $\{0,1,\infty,e\}$, which cannot exist because the cross ratio of any permutation of $\{0,1,\infty,\pi\}$ does not equal the cross ratio of $\{0,1,\infty,e\}$. However, there exists an abstract field automorphism $\sigma$ of $\mathbb{C}$ taking $\pi$ to $e$, so there will be an isomorphism of schemes $V_1\to V_2$ which is $\sigma$-semilinear.

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    $\begingroup$ Nice, probably most simple, example! $\endgroup$ Feb 18, 2014 at 21:39

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