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.)



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 $J^1$ 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. 


