Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $X=\mathbf{A}^1_{\overline{\mathbf{Q}}}-\{0\} = \mathbf{G}_{m,\overline{\mathbf{Q}}}$ be the multiplicative over the field of algebraic numbers. Each finite etale cover $Y\to X$ (with $Y$ connected) is isomorphic to the finite etale morphism $X\to X$ given by $z\mapsto z^n$ for some $n\geq 1$.

The universal covering space $\widetilde{X}$ of $X$ is the projective limit over all finite etale covers of $X$. It's not a scheme. (If $\widetilde{X}$ were a scheme, the "morphism" $\widetilde{X}\to X$ would be an etale morphism with non-finite fibres. That's not possible.)

Is it endowed with a morphism $\widetilde{X}\to X$? In which category should I consider this "morphism"?

Can we describe $\widetilde{X}$ a bit more explicitly using the above description of all finite etale covers of $X$?

share|cite|improve this question
Since you have taken a projective limit, the map is probably in the pro-category which is just a formal construction where essentially the objects are projective systems and morphisms are maps that make all diagrams commute. – Matt May 8 '12 at 20:07
Alternatively, one could pass to an appropriate completion of such a category: let $\mathcal{C}$ be the category of schemes of finite presentation over $X$, and consider the category of presheaves $\widehat{\mathcal{C}} = [\mathcal{C}^\textrm{op}, \textbf{Set}]$; the Yoneda lemma implies that the Yoneda embedding $\mathcal{C} \hookrightarrow \widehat{\mathcal{C}}$ is full and faithful, so you have $X$ sitting in this larger category $\widehat{\mathcal{C}}$; but $\widehat{\mathcal{C}}$ contains all small limits, so you can form the presheaf which would "represent" $\widetilde{X}$ too. – Zhen Lin May 8 '12 at 21:48
up vote 6 down vote accepted

One can form the projective limit $\tilde{X}$ in the category of schemes without any problem; it is Spec $\overline{\mathbb Q}(\{z^{1/n}\}_{n \geq 1})$. It is not finite type over $X$, and so in particular is not etale, but so what? There is no theorem saying that a projective limit of finite etale maps is etale.

share|cite|improve this answer

Your Answer


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

Not the answer you're looking for? Browse other questions tagged or ask your own question.