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I am trying to understand Leopoldt's conjecture as formulated in section 5.5 of Washington's "Introduction to Cyclotomic Fields".

For an algebraic number field $K$ let $E$ denote the global units, for each prime $\mathfrak p$ over $p$ let $U_\mathfrak p$ be the local units of the completion $K_\mathfrak p$, and let $U_{1,\mathfrak p}$ be the principal units, i.e. the units $\varepsilon \equiv 1$ modulo $\mathfrak p$. Let $$U = \prod_{\mathfrak p|p} U_\mathfrak p \quad \text{and} \quad U_1 = \prod_{\mathfrak p|p} U_{1,\mathfrak p}.$$ The units $E$ are embedded diagonally in $U$ and the units $\varepsilon$ that embed to $U_1$ are denoted by $E_1$. Then $E_1$ is a $\mathbb Z$-module of rank $r_1+r_2-1$ (where $r_1$ is the number of real embeddings and $2r_2$ is the number of complex embeddings). Leopoldt's conjecture says that $\overline{E_1}$ (the closure of $E_1$ in the topology of $U_1$) is a $\mathbb Z_p$-module of rank $r_1+r_2-1$

Now in the proof of Washington's theorem 5.31 (and the example before that) he uses the fact that a $\mathbb Z$-basis of $E_1$ modulo torsion generates $\overline{E_1}$ modulo torsion as a $\mathbb Z_p$-module. This seems plausible to me, however I haven't been able to prove it.

So my question is: Why does a $\mathbb Z$-basis of $E_1$ (modulo torsion) generate $\overline{E_1}$ over $\mathbb Z_p$ (modulo torsion)?

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I eventually figured it out by myself. Let $\varepsilon_1,\ldots,\varepsilon_r$ be a $\mathbb Z$-basis of $E_1$ modulo torsion, let $\langle \varepsilon_1,\ldots,\varepsilon_r \rangle$ denote the generated subgroup of $U_1$ and $\langle \varepsilon_1,\ldots,\varepsilon_r \rangle_{\mathbb Z_p}$ the generated $\mathbb Z_p$-module of $U_1$. We want to show $$\overline{\langle \varepsilon_1,\ldots,\varepsilon_r \rangle} = \langle \varepsilon_1,\ldots,\varepsilon_r \rangle_{\mathbb Z_p}.$$ The inclusions $$\langle \varepsilon_1,\ldots,\varepsilon_r \rangle \subseteq \langle \varepsilon_1,\ldots,\varepsilon_r \rangle_{\mathbb Z_p},\\ \overline{\langle \varepsilon_1,\ldots,\varepsilon_r \rangle} \supseteq \langle \varepsilon_1,\ldots,\varepsilon_r \rangle_{\mathbb Z_p}$$ are clear so it suffices to show that $\langle \varepsilon_1,\ldots,\varepsilon_r \rangle_{\mathbb Z_p}$ is closed. But $\langle \varepsilon_1,\ldots,\varepsilon_r \rangle_{\mathbb Z_p}$ is the image of the $\mathbb Z_p$-linear map \begin{align*} \mathbb Z_p^r &\to U_1,\; \\ (a_1,\ldots,a_r) &\mapsto \varepsilon_1^{a_1}\cdots \varepsilon_r^{a_r}. \end{align*} This map is continuous since for fixed $\varepsilon \in U_1$ the $\mathbb Z_p$-operation $\mathbb Z_p \to U_1, a \mapsto \varepsilon^a$ is continuous. Since $\mathbb Z_p^r$ is compact the image $\langle \varepsilon_1,\ldots,\varepsilon_r \rangle_{\mathbb Z_p}$ is also compact, hence closed in $U_1$.

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