# Endomorphism ring of the adeles and ideles?

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?

• 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

## 1 Answer

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.

• 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