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Let $\Lambda=\mathbf{Z}_p[[T]]$. The group $M[p]$ is dual to $M^\vee/pM^\vee$. So if $M[p]$ is finite, then $M^\vee/pM^\vee$ is finite, and the form of Nakayama's lemma applicable to profinite modules …

Yes, there is a way to get the finiteness of class number as well as the S-unit theorem, avoiding using Minkowski's theorem explicitly. However, it doesn't give you the Minkowski bound. The idea is to …

What you know as the Chinese remainder theorem for the abelian group $\mathbb{Z}/n\mathbb{Z}$ (which you probably don't want to call a ``simple group" unless $n$ is prime, as this term has a technical …

You can definitely talk about integration on $\overline{\mathbb{Q}}$. The question is "with respect to what measure?" The reason integration theory works so well over a completion of $\mathbb{Q}$ is t …

If $G$ is any profinite group, and $a\in\widehat{\mathbf{Z}}$, then for any sequence $(a_n)$ of integers converging in $\widehat{\mathbf{Z}}$ to $a$, the sequence $(g^{a_n})$ converges in $g$ to an el …

Let $G$ be a connected reductive group over a number field $F$, $G_\infty=\prod_{v\mid\infty} G(F_v)$, $\mathbf{A}$ the adèles of $F$, $\mathbf{A}_f$ the finite adèles of $F$. Fix a maximal compact su …

If $K$ is an imaginary quadratic field, then the $\mathbf{Z}_p$-rank of $K$ is $2$, meaning that the Galois group of the compositum of all the $\mathbf{Z}_p$-extensions of $K$ in an algebraic closure …

Any complex-valued function on $\mathbb{N}$ can be extended to an entire function, so the answer is "yes." This follows from Theorem 15.13 of Rudin's Real and Complex Analysis, which states that for a …

If $p$ is a rational prime, then choosing a prime $v$ of $\overline{\mathbf{Q}}$ lying over $p$ amounts to choosing an embedding $i:\overline{\mathbf{Q}}\hookrightarrow\overline{\mathbf{Q}}_p$. This g …