MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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

What are the (cont.) endomorphisms resp. automorphisms of the adeles (for a given global field)

1) as a topological abelian group and

2) as a topological ring?

3) What are the endomorphisms and the automorphisms group of the ideles?

4) What is known for the adelic points of an algebraic group?

share|cite|improve this question
Random comment: All ring automorphisms of $\mathbf{Q}_p$ and $\mathbf{R}$ ara automatically continuous. It might follow that all ring automorphisms of the adeles are continuous? – Kevin Buzzard May 13 '11 at 18:29
what is funny though that eg. $Q_2$ and $Q_3$ are abstractly isomorphic without considering topolgy or with considering topology without considering group struture;) – Marc Palm May 14 '11 at 0:01
All ring endomorphisms of R and local fields are continuous, from which one can show that any ring endomorphism of the adele ring of a totally real number field is continuous. There are discontinuous endomorphisms of the adele ring of a non-totally real number field, since you can use a wacky discontinuous automorphism of C on one complex coordinate and the identity in the other coordinates. – KConrad May 14 '11 at 4:12
up vote 2 down vote accepted
  1. There are no nontrivial continuous homomorphisms between factors of different residue characteristic, so any endomorphism/automorphism decomposes into a collection of endomorphisms of factors, with a global condition that $(1,1,\cdots)$ has bounded denominators. Each factor over a rational prime $p$ (including infinity) is a product of copies of $\mathbb{Q}_p$, so continuous endomorphisms are elements of $M_{n_p}(\mathbb{Q}_p)$, where $n_p$ is the sum of inertia degrees over primes over $p$. Therefore, the endomorphism ring is the restricted product of $M_{n_p}(\mathbb{Q}_p)$, i.e., all but finitely many factors lie in $M_{n_p}(\mathbb{Z}_p)$ (when $p$ is finite). The automorphism group is given by replacing $M_{n_p}$ with $GL_{n_p}$.

  2. An automorphism is a product of local automorphisms for a given residue charactristic, which are made out of permutations of $p$-adic fields in the local factors and Galois automorphisms of the fields. Endomorphisms also include projections that kill individual fields in the factors.

  3. You can decompose the ideles as $\prod_{r_1} \mathbb{R}^\times \oplus \prod_{r_2} \mathbb{C}^\times \oplus \bigoplus_\mathfrak{p} \mathbb{Z} \oplus \prod_\mathfrak{p} \prod_{n,k=1}^\infty \mathbb{Z}/p^k\mathbb{Z} \oplus \prod_\mathfrak{p} \mathbb{Z}_p^{f_\mathfrak{p}}$, using the decomposition of units of a $p$-adic field into valuations and units of integers. It shouldn't be too hard to work it out on your own from here.

share|cite|improve this answer
Are you sure that you aren't just describing linear endomorphism? What about $z \mapsto \overline{z}$? at a complex place? – Marc Palm May 13 '11 at 16:59
Sorry, I was only looking at the adeles of $\mathbb{Q}$. I'll revise. – S. Carnahan May 13 '11 at 17:06
Okay, I see your point. What is a refernece for $\mathbb{Q}$ or is so straight forward? If so, could you provide at least an argument for the additive group $\mathbb{Q}_p$ then? I will be really grateful. – Marc Palm May 13 '11 at 17:15
A continuous group endomorphism of $\mathbb{Q}_p$ is uniquely determined by the image of 1. – S. Carnahan May 13 '11 at 17:31
@pm: The compact open topology on the ideles (via their action on the adeles by multiplication) does agree with the "right" (read: not the subspace) topology on the ideles. – Daniel Litt May 13 '11 at 17:47

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.