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Let $K$ be an algebraic number field. Let $\mathbb{I}$ be the group of ideles and let $\mathbb{I}_f$ be the group of finite ideles. We embed $K^\times$ diagonally in both.

It is know that $K^\times$ is a discrete subgroup of $\mathbb{I}$. Hence it is also closed in $\mathbb{I}$. However, if the ring of integers of $K$ has an infinite unit group, then $K^\times$ is not discrete in $\mathbb{I}_f$.

Question: Is $K^\times$ at least closed in $\mathbb{I}_f$?

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    Does $\mathbb{I}_f$ inherit its topology from $\mathbb{I}$? If so, isn't the assertion immediate?2015-07-18

0 Answers 0