# Inclusion of multiplicative group of one local field into the idele class group is a closed embedding, but inclusion of more than one isn't?

Let $k$ be a global field, $J$ its idele group, and $C = J/k^\times$ the idele class group. For any place $v$ of $k$, we have the familiar closed embedding $k_v^\times \hookrightarrow J$. More generally, we can let $S$ be a finite set of places of $k$, consider $P = \prod_{v \in S} k_v^\times$ with the product topology, and still obtain a closed embedding $P \hookrightarrow J$. This much is clear from the definition of the topology on $J$.

I've seen it asserted various places that the continuous injection $k_v^\times \rightarrow C$ is still a closed embedding, and always treated this statement as 'obvious' without much thought. Recently, while reading the Artin-Tate notes (Ch. X, section 2, p.76), I learned the corresponding continuous injection $P \rightarrow C$ is not a topological embedding when $|S| > 1$, and that the image is not even closed! I found this surprising and upon reflection realized that I do not know how to prove either statement. Worse yet, I can't find any discussion of the matter in the standard references (e.g. Cassels-Fröhlich, Neukirch, Weil's Basic Number Theory, ...)

Can someone please explain to me why we get a closed embedding when $|S| = 1$, and what goes wrong in the more general case? (bonus points for explicit examples).

EDIT

Thanks to GH from MO for addressing the "embedding" part of this question in the special case that $S$ contains the archimedean places. I expect that this is the basic explanation for what's going on here. However, the question is not fully answered:

1. I'm still very interested in why the map $P \rightarrow C$ is closed when $|S| = 1$ but not otherwise. It seems to me that a slightly different argument is needed here - I'm not sure what it should look like.

2. It would be good to remove the restriction about the archimedean places - after all, one is frequently interested in the case where $S$ consists of a single non-archimedean place, or in the case where there is more than one archimedean place. Here's my attempt at extending the argument to this case (copied from my comment on the answer):

My guess is the answer to the second question goes something like this. Let $|S| = s_\infty + s_f$ with $s_\infty$ the number of archimedean places in $S$. We can force the log of the absolute value of $t$ to be close to $0$ at the $r-s_\infty$ archimedean places outside of $S$. The logs of absolute values of $S$-units span a lattice of dimension $r + s_f - 1 = (r - s_\infty) + |S| - 1$, and we win if we force the image of $t$ to be $0$ in this lattice. So we have "enough dimensions to work with" outside of $S$ iff $|S| = 1$. But this isn't quite rigorous and I'm confused how to make it so (The statement I'm looking for is something like "the projection of a $d$-dimensional lattice in $\mathbb{R}^n$ to an $m$-dimensional subspace has discrete image iff $m \geq d$ - neither direction is obvious to me.)

• It is certainly not true that the image of $P\to C$ is closed when $|S|=1$. For example, in the case of $k=\mathbb{Q}(i)$, the kernel of the canonical homomorphism $J\to\mathrm{Gal}(k_\text{ab}/k)$ is the closure of $k^\times k_\infty^\times$, which implicitly means that $k^\times k_\infty^\times$ is not closed in $J$. Sep 11, 2018 at 23:56
• I think in this case, the connected component of the identity in $C$ is exactly $k^\times k_\infty^\times$, e.g. by the explicit description in Ch. IX of the Artin-Tate notes. Sep 12, 2018 at 1:10
• What is $J$? All ideles or just norm 1? $\mathbb A^\times_1/k^\times$ is compact, so no field embeds. The natural map doesn't land there, either, since the local field doesn't have norm 1. Any single field embeds in $\mathbb A^\times/k^\times$, but the quotient is compact, so a second field cannot embed. Sep 12, 2018 at 3:14
• Oh great, that's very direct. Yeah, $J$ is all ideles. Why does any single field embed in $\mathbb{A}^\times/k^\times$? Sep 12, 2018 at 3:50
• @pierredefermat a single $k_v^\times$ embeds because $k^\times$ is in $\mathbb A_k^\times$ diagonally, so $k^\times$ intersects any one component $k_v^\times$ of $\mathbb A_k^\times$ only in the identity. Apr 28, 2019 at 15:13

Let $J^1$ be the ideles of idelic norm one; the global points $k^*$ is contained in $J^1$. In the number field case, we have a surjection $J/k^* \rightarrow \mathbb R _{>0}$ with kernel $J^1/k^*$ (the latter is compact).

Now let $v$ be a place and $O_v^*$ the unit group of $k_v^*$ (if $v$ is complex, $O_v^*=S^1$ ; if $v$ is real, $O_v^*=\{\pm 1\}$; if $v$ is non-archimedean, $O_v^*$ is the group of units). Then under the above quotient map $J/k^*$ the inclusion of $k_v^*$ in $J$ induces a map $k_v^*\rightarrow \mathbb R_{>0}$ whose kernel is the compact group $O_v^* \subset J^1/k^*$. The map $O_v^* \rightarrow J^1/k^*$ is a continuous injection of compact Hausdorff spaces and is hence a homeomorphism onto the image.Since $k_v^*$ is the product of $O_v^*$ with a closed subgroup of $R_{>0}$ (in the archimedean case, it is all of $\mathbb R_{>0}$ and in the non-archimedean case it is a discrete subgroup of $\mathbb R_{>0}$ generated by the uniformizing parameter of $k_v^*$) it follows that we need only check that the map $k_v^* \rightarrow \mathbb R_{>0}$ has closed image. The image is always a closed subgroup and hence if $Card S=1$, $k_v^* \rightarrow J/k^*$ is a closed embedding.

In the non-archimedean case, the image is of the form $\mid \pi \mid ^{\mathbb Z}$ with $\pi$ a uniformizing parameter. If $S$ has two non-archimedean places (which lie above different primes of $\mathbb Q$), the image is of the form $\mid \pi _v \mid ^{\mathbb Z} \mid \pi _w\mid ^{\mathbb Z} \subset \mathbb R_{>0}$ and is not a closed subgroup in $\mathbb R_{>0}$ and hence the embedding of $k_v^*\times k_w^*$ is not closed.

• I guess $K$ is $k$. Sep 12, 2018 at 11:01
• @Chris Wuthrich Thank you. I have tried to rectify it. Sep 12, 2018 at 11:38
• You can generalize the theorem that an injective map from a compact to a Hausdorff is a homeomorphism to its image to: an injective, proper map from a locally compact space to a Hausdorff space is a homeomorphism to its image. (I'm not sure that this is actually useful in a proof without circularity, but I think it's a good statement to keep in mind.) Sep 12, 2018 at 23:03
• Ah, wonderful! If I'm not mistaken, this argument also shows that when all of the non-archimedean places in $S$ lie above the same rational prime, the image is closed (though likely still not an embedding, e.g. by the argument in GH from MO's answer). Sep 13, 2018 at 1:43

For simplicity, let me assume that $S$ contains the archimedean places. Then the brief answer is that the group of $S$-units is finite when $|S|=1$, but infinite when $|S|>1$. Let me give more detail.

A fundamental system of neighborhoods of the identity in $J$ is provided by the product sets $U=\prod_v U_v$ in $J$, where $U_v$ is an open neighborhood of the identity in $k_v^\times$, and $U_v=r_v^\times$ for all but finitely many $v$'s. Without loss of generality, for any nonarchimedean place, either $U_v=r_v^\times$ or $U_v=1+p_v^{n_v}$ for some $n_v\geq 1$. The corresponding neighborhood in $C$ is $Uk^\times/k^\times$, which intersects the embedded copy of $P$ in $(Pk^\times\cap Uk^\times)/k^\times=(P\cap Uk^\times)k^\times/k^\times$. So in the induced topology of the embedded $P$, a fundamental system of neighborhoods of the identity is provided by the intersections $P\cap Uk^\times$. This intersection consists of those elements $p\in P$, which can be decomposed as $p=ut$ with $u\in U$ and $t\in k^\times$. For any $v\not\in S$, we have that $1=p_v=u_vt_v$, so $t_v=u_v^{-1}\in r_v^\times$. That is, $t$ is an $S$-unit in $k^\times$. Moreover, $t$ is congruent to $1$ modulo a large ideal (determined by the $n_v$'s), but these are all the constraints on $t$. If $|S|=1$, then the group of $S$-units is finite, hence we can force that $t=1$ with an appropriate congruence, which shows that the induced topology on $P$ is the usual one. If $|S|>1$, then the group of $S$-units has rank $|S|-1>0$, which property survives any congruence, so the induced topology will be cruder than the usual one. Indeed, in this case, any $\prod_{v\in S}U_v$ gets translated by infinitely many $t$'s, and only the union of these translates will be open in the induced topology of the embedded $P$, not the translates themselves.

• Great - this certainly helps explain what's going on. There's some things still unclear to me though. First, what about closed-ness of the image? Second, the case where $|S| = 1$ but does not contain all the archimedean places seems typical, but I don't exactly see how to extend the argument. Third, it's not clear to me why, when $|S| > 1$, there are infinitely many $S$-units congruent to any given modulus disjoint from $S$. Sep 11, 2018 at 22:24
• My guess is the answer to the second question goes something like this. Let $|S| = s_\infty + s_f$ with $s_\infty$ the number of archimedean places in $S$. We can force the log of the absolute value of $t$ to be close to $0$ at the $r - s_\infty$ archimedean places outside of $S$. The logs of absolute values of $S$-units span a lattice of dimension $r + s_f - 1 = (r - s_\infty) + |S| - 1$, and we win if we force the image of $t$ to be $0$ in this lattice. So we have "enough dimensions to work with" outside of $S$ iff $|S| = 1$. But this isn't quite rigorous and I'm confused how to make it so. Sep 11, 2018 at 22:30
• @pierredefermat: I have not addressed closedness of the image (I think the image is never closed, not even when $|S|=1$), or the case of general $S$. The answer to your last question is easy: any congruence condition defines a finite index subgroup of $S$-units, hence the rank of the subgroup is still $|S|-1$. In more precise terms, reducing $S$-units modulo any ideal $\mathfrak{m}$ is a homomorphism with finite image, so the kernel has rank $|S|-1$. Note that in my post I talked about $S$-units congruent to $1$ modulo $\prod_v\mathfrak{p}_v^{n_v}$. Sep 11, 2018 at 22:31
• Fantastic, thanks! If you'd rather not address the other two questions in your answer, I'll edit the question to emphasize that they're the part I'm still looking for help with. (Especially the closedness of the image - this is the part that really surprised me initially). Is that ok with you? Sep 11, 2018 at 22:38
• @pierredefermat: Do as you wish, I am glad if I could help. I am quite busy, so I don't plan to think more on this, but perhaps others can add their ideas. BTW it is usually not recommended to ask more than one question in one post, e.g. because it causes confusion when each question is answered by a different person (you can only accept one answer and make it green). Sep 11, 2018 at 22:45