Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

In Milne - Lectures on étale cohomology, example 6.10 i came across the following.

We fix a variety $X$ and work in the category $Var/X$ of varieties over $X$ (so with fixed morphisms to $X$!) and furthermore we assume these morphisms to $X$ to be étale. Assume for simplicity that our base field $k = \mathbb{C}$ and all varieties are affine.

Define $\mu_n = \operatorname{Spec}\big[ k[t]/(t^n - 1) ]$ as the reduced group variety given by the $n$-th roots in the affine line.

Milne now claims that for all varieties $U$ over $X$, the set $$ \operatorname{Hom}_{Var/X}(U, \; \mu_n) $$ is exactly the group of $n$-th roots in $\mathcal{O}(U)$, i.e. there is a bijective relation between these two groups.

Using anti-equivalence of rings and affine varieties, this is easily shown to be true in the categories of varieties. However, we are working in the category of varieties over $X$! It seems to me that Milne's claim is therefore not true: take for example $U = X = \operatorname{Spec}(k[x,y])$ the affine plane with the morphism $U \rightarrow X$ being the identity. Then a morphism from $U$ to $\mu_n$ in the category of varieties over $X$ is equivalent to a commutative diagram

$$ \begin{matrix} k[x,y] & & \leftarrow && k[t]/(t^n - 1)\\ & \nwarrow& & \nearrow\\ & &k[x,y] \end{matrix} $$ where the most left arrow is the identity. There is obviously no such diagram, hence in this case $\operatorname{Hom}_{Var/X}(U, \mu_n)$ is empty. This is in contradiction with Milne's claim that this should be the collection of $n$-th roots of unity in $k[x,y]$, of which there are $n$.

So there are a few possibilities, of which i guess these are not all:

  • I am stupidly missing something: perhaps the assumption that $\mu_n \rightarrow X$ is étale changes things?

  • Milne meant morphisms in varieties. However, this does not really make sense with the rest of the theory;

  • Milne meant to say "If there is a morphism $U \rightarrow \mu_n$ in varieties over $X$, it corresponds to a choice of $n$-th root in $\mathcal{O}(U)$, but not all such roots give rise to such a morphism".

I actually hope it is the first one haha. Can anyone help me out here?

Thanks!

Edit: I just realized that it is in fact probably the first one. So an alternative question: does any morphism $k[x,y] \rightarrow k[t]/(t^n - 1)$ which is defined by choosing two $n$-th roots of unity in $k \subset k[t]/(t^n-1)$ and sending $x$ and $y$ to them, define an étale morphism? I will check this now so might repost about this later. (this is probably something semi-general about closed immersions)

share|improve this question
2  
What happens if you replace $\mu$ by the pullpack of $\mu$ to $X$? (You have to do something like this, notice, because otherwise $\mu_n$ is not a variety over $X$, so the $\hom$ does not even make sense...) –  Mariano Suárez-Alvarez Feb 7 '13 at 15:58
    
@MarianoSuárez-Alvarez, thanks. I was being stupid, of course there are no etale morphisms from the zero dimensional variety $\mu_n$ to any variety of positive dimension, so the whole thing did not make sense. I think your suggestion is exactly what i should do, i will check it out now. –  Joachim Feb 7 '13 at 16:33

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.