Explicit defining equations for the Leopoldt locus

2012-08-11T01:12:06Z

Let $F$ be a number field, which we assume for simplicity to be Galois and totally real. Set $\mathcal{O}_p=\mathcal{O}_F\otimes_{\mathbf{Z}}\mathbf{Z}_p$ . The norm map on $\mathcal{O}_F$ extends uniquely to a multiplicative function $N: \mathcal{O}_p \to \mathbf{Z}_p$, and we set $\mathcal{O}_p^{1}=x\in \mathcal{O}_p \, \mathrm{with}\, Nx=1$ . The famous Leopoldt conjecture asserts that the image of $\mathcal{O}_F^\times$ in $\mathcal{O}_p$, a priori contained in $\mathcal{O}_p^1$, is actually topologically dense in $\mathcal{O}_p^1$. Let us define the Leopoldt locus $\mathcal{L}$ as the topological closure of $\mathcal{O}_F^\times$ in $\mathcal{O}_p^1$.

Now, the Leopoldt conjecture is equivalent to other basic statements, for example the vanishing of $H^2_{\mathrm{cts}}(G_{F'},\mathbf{Q}_p)$ where $F'/F$ is the maximal extension unramified outside the primes dividing $p$ and $\infty$. In fact, the rank of this Galois cohomology group is equal to the codimension of $\mathcal{L}$ inside $\mathcal{O}_{p}^1$ , so it makes sense to ask: can we start with a basis of $H^2(G_{F'}, \mathbf{Q}_p)$ , i.e. a bunch of 2-cocycles, and use them to write down analytic functions on $\mathcal{O}_p^1$ which cut out $\mathcal{L}$ as their common zero locus?

I'm assuming the answer is "yes, and this is a baby version of some ideas of Minhyong Kim", but I don't know well enough to say.

Explicit defining equations for the Leopoldt locus - MathOverflow

Explicit defining equations for the Leopoldt locus