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Let $A_\Bbb{Q}$ be the adele group of $\Bbb{Q}$. Let $C_n=\{x \in A_\Bbb{Q}: x_\infty=0 \text{ and }x_p \in p^{\operatorname{ord}_p(n)}\Bbb{Z}_p \text{ for prime }p\}$. I want to show that $\Bbb{R}/n\Bbb{Z}$ is isomorphic to $A_\Bbb{Q}/(\Bbb{Q}+C_n)$.

We can construct a map $q$ from $\Bbb{R}/n\Bbb{Z}$ to $A_\Bbb{Q}/(\Bbb{Q}+C_n)$ sending $x$ to the class of the adele with $x$ as the Archimedean component and zero for the non archimedian component. I showed that this map is well defined and injective. But I cannot show that this map is surjective. Please help...

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    $\begingroup$ It's too late for me to type up everything, but isn't the idea that if you take an adele $x=(x_\infty,x_2,x_3,x_5,\ldots)$ then we get a finite set $S$ of primes such that $p\in S$, iff $x_p\notin\Bbb{Z}_p$ or $p\mid n$. Then (apply Chineser Remainder Theorem or a suitable approximation theorem) we can find a $q\in\Bbb{Q}$ such that all the prime divisor of the denominator of $q$ are in $S$ and that $x_p-q\in p^{\operatorname{ord}_p(n)}\Bbb{Z}_p$ for all primes $p$. Hence the non-archimedean part of $x-(q,q,q,\ldots)$ is in $C_n$, and thus $x$ is in the coset of $(x_\infty-q,0,0,0,\ldots)$. $\endgroup$ – Jyrki Lahtonen Oct 27 '14 at 20:47
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So the overarching question is showing that the quotient $A_{\mathbb{Q}}$ can be identified with the pro-universal covering of $\mathbb{R}/\mathbb{Z}$ $$A_{\mathbb{Q}}/\mathbb{Q} = \varprojlim_n\, \mathbb{R}/n\mathbb{Z},$$the projective limit being taken over the set of natural integers ordered by the divisibility order.

There is an exact sequence $$0 \to \mathbb{R} \times \hat{\mathbb{Z}} \to A_\mathbb{Q} \to \bigoplus_p \mathbb{Q}_p/\mathbb{Z}_p \to 0$$where $\bigoplus_p \mathbb{Q}_p/\mathbb{Z}_p $ is the subgroup of $\prod_p \mathbb{Q}_p/\mathbb{Z}_p$ consisting of sequences $(x_p)$ whose members $x_p \in \mathbb{Q}_p/\mathbb{Z}_p$ vanish for almost all $p$. Consider now the homomorphism between the two exact sequences:

$$\require{AMScd} \begin{CD} 0 @>>> 0 @>>> \mathbb{Q} @>>> \mathbb{Q} @>>> 0\\ @VVV @VVV @VVV @VVV @VVV\\ 0 @>>> \hat{\mathbb{Z}} \times \mathbb{R} @>>> A_\mathbb{Q} @>>> \bigoplus_p \mathbb{Q}_p/\mathbb{Z}_p @>>> 0 \end{CD} $$

Because the middle vertical arrow is injective and the right vertical arrow is surjective with kernel $\mathbb{Z}$, the snake lemma induces an exact sequence$$0 \to \mathbb{Z} \to \hat{\mathbb{Z}} \times \mathbb{R} \to A_{\mathbb{Q}}/\mathbb{Q} \to 0,$$ $\mathbb{Z}$ being diagonally embedded in $\hat{\mathbb{Z}} \times \mathbb{R}$. In other words, there is a canonical isomorphism$$A_\mathbb{Q}/\mathbb{Q} \to (\hat{\mathbb{Z}} \times \mathbb{R})/\mathbb{Z}.$$Dividing both sides by $\hat{\mathbb{Z}}$, we get an isomorphism $$A_{\mathbb{Q}}/(\mathbb{Q} + \hat{\mathbb{Z}}) \to \mathbb{R}/\mathbb{Z}.$$By the same argument, for every $n \in \mathbb{N}$ we can identify the covering $$A_\mathbb{Q}/(\mathbb{Q} + n\hat{\mathbb{Z}}) = A_\mathbb{Q}/(\mathbb{Q} + C_n)$$ of $A_\mathbb{Q}/(\mathbb{Q} + \hat{\mathbb{Z}})$ with the covering $\mathbb{R}/n\mathbb{Z}$ of $\mathbb{R}/\mathbb{Z}$. Hence $A_{\mathbb{Q}}/\mathbb{Q}$ is the pro-universal covering of $\mathbb{R}/\mathbb{Z}$.

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