I'm doing some Galois cohomology stuff (specifically, trying to calculate $H^1(\mathbb{Q}_3,E[\varphi])$, where $\varphi:E\to E'$ is an isogeny of elliptic curves), and it involves calculating $\mathbb{Q}_3(\sqrt{-6})^{\times}/(\mathbb{Q}_3(\sqrt{-6})^{\times})^3$. Here's what I've done so far.

Let $K=\mathbb{Q}(\sqrt{-6})$. As $-6\not\equiv 1$ (mod 4), we have that $\mathcal{O}_K=\mathbb{Z}[\sqrt{-6}]$. Let $v$ be the finite place of $K$ corresponding to the (non-principal) prime ideal $(3,\sqrt{-6})$. It's fairly easy to see that $K_v=\mathbb{Q}_3(\sqrt{-6})$, and that the residue field is $k_v\cong\mathcal{O}_{K}/(3,\sqrt{-6})\cong\mathbb{F}_3$. Now, by Hensel's lemma, $\sqrt{-2}\in\mathbb{Q}_3$, so it follows that $\sqrt{3}\in\mathbb{Q}_2(\sqrt{-6})$. In $\mathcal{O}_K$, $(3)$ decomposes as $(3,\sqrt{-6})^2$, so $v(3)=2$, and hence $v(\sqrt{3})=1$. So we can legitimately choose $\sqrt{3}$ as a uniformizer of $\mathcal{O}_{K_v}$. This means that every element of $\mathcal{O}_{K_v}$ has a unique representation $$\sum_{n=0}^{\infty}a_n\sqrt{3}^n, \text{ where } a_n\in\{-1,0,1\}.$$ The elements of $\mathcal{O}_{K_v}^{\times}$ are the ones where $a_0=\pm1$. Now, using Hensel's lemma I went ahead and showed that $$(\mathcal{O}_{K_v}^{\times})^3=\{\pm1+\sum_{n=3}^{\infty}a_n\sqrt{3}^n~|~a_n\in\{-1,0,1\}\}.$$ But how do I find distinct representatives for $\mathcal{O}_{K_v}^{\times}/(\mathcal{O}_{K_v}^{\times})^3$? Does modding out by the group above mean that I can just look up to sign and ignore everything past $\sqrt{3}^3$, so that a set of representatives would be $\{1,1\pm\sqrt{3},1\pm3,1\pm\sqrt{3}\pm3\}$, which has size 9 (the 2 $\pm$'s are independent in the last expression)? Perhaps my working is not useful, because I've written things additively but the groups are multiplicative. Also, I could just as well have chosen $\sqrt{-6}$ as my uniformizer. Can I replace $\sqrt{3}$ with $\sqrt{-6}$ everywhere and still get a set of representatives for $\mathcal{O}_{K_v}^{\times}/(\mathcal{O}_{K_v}^{\times})^3$? I'm very confused!

Of course, once $\mathcal{O}_{K_v}^{\times}/(\mathcal{O}_{K_v}^{\times})^3$ is determined, finding $K_v^{\times}/(K_v^{\times})^3$ is easy.

  • $\begingroup$ $(1+x)^3 = 1+3x+3x^2+x^3$ so for any $m \ge 3$, $(1+\sqrt{3} O)^3$ contains $(1+\sqrt{3}^{m-2})^3 = 1+\sqrt{3}^m+O(\sqrt{3}^{m+1})$ thus $(1+\sqrt{3} O)^3 = 1+ 3\sqrt{3} O$ and $(1+ \sqrt{3} O)/(1+\sqrt{3} O)^3$ has order $9$, it is a product of two cyclic groups of order $3$, generated by $1+\sqrt{3}$ and $1-\sqrt{3}$. The next problem is to find an uniformizer for each extension generated by the 3rd root of those. $\endgroup$ – reuns Jun 5 at 18:04
  • $\begingroup$ What you’ve written looks perfectly correct to me. Maybe I can add some clarification and simplification, but I won’t have time till (maybe) three hours from now. $\endgroup$ – Lubin Jun 5 at 22:51

First, you needn’t have worried about what parameter you used: $\sqrt{-6}$ is just as good as $\sqrt3$. Indeed, if $\mathfrak o$ is a complete discrete valuation ring with fraction field $K$ and (additive) valuation $v:K^\times\to\Bbb Z$, and if$f(X)\in\mathfrak o[X]$ is an Eisenstein polynomial with a root $\alpha$, then $\alpha$ is a local parameter for the d.v.r. $\mathfrak o[\alpha]$. Since both $X^2+6$ and $X^2-3$ are Eisenstein for $\Bbb Z_3$, a root of either is good as a local parameter in $\Bbb Q_3(\sqrt{-6}\,)$.

Next, it may help for you to think of $K^\times/(K^\times)^3$ as $K^\times\otimes(\Bbb Z/3\Bbb Z)$. Whether or not, you were quite correct to see that all the contribution to $K^\times/(K^\times)^3$ comes from $1+\mathfrak m$. Here, of course, I’m using $K=\Bbb Q_3(\sqrt{-6}\,)$ and $\mathfrak m=\text{max}(\Bbb Z_3[\sqrt{-6}\,])=\sqrt{-6}\cdot\Bbb Z_3[\sqrt{-6}\,]$.

Now here’s something most useful: the multiplicative group $1+\mathfrak m$ is a $\Bbb Z_3$-module, via exponentiation. That is, for $z\in\Bbb Z_3$ and $\alpha\in\mathfrak m$, the expression $(1+\alpha)^z$ is well-defined, and all the rules that you know for $\Bbb Z$-exponents are valid. How’s it defined? Take any $3$-adically convergent sequence of positive integers with limit $z$, say $n_i\to z$. Then $\bigl\lbrace(1+\alpha)^{n_i}\bigr\rbrace$ is also $3$-adically convergent. I’ll leave it to you to prove that. Of course you see that the statement is true no matter what the $3$-adically complete local ring $\mathfrak o$ you’re dealing with. Note that the exponents are from $\Bbb Z_3$, nothing bigger.

Well: now that you know that $1+\mathfrak m$ is a $\Bbb Z_3$-module, what can you say about its structure? You know that it has no torsion, so it’s a free $\Bbb Z_3$-module. Of what rank? I think you can convince yourself pretty easily that the rank is equal to $[K:\Bbb Q_3]=2$; I’ll leave that to you, too.

Now it’s perfectly clear that $\bigl|(1+\mathfrak m)/(1+\mathfrak m)^3\bigr|=9$, the cardinality of a two-dimensional vector space over the field $\Bbb F_3$. Your enumeration of the elements is quite right, too.

Please don’t hesitate to ask for clarification or expansion of the above.

  • $\begingroup$ When you wrote $K^{\times}\otimes(\mathbb{Z}/3\mathbb{Z})$, did you mean $\mathcal{O}_K^\times/(\mathcal{O}_K^\times)^3\times(\mathbb{Z}/3\mathbb{Z})$? Also, you tacitly left out any discussion of the 2nd roots of unity, but I suppose that's fine because both 1 and -1 are cubes in $\mathbb{Q}_3$. Also it seems a bit leftfield to me to invoke this strange $\mathbb{Z}_3$ exponentiation action. But I suppose it works! I just need to convince myself that it's well-defined, and derive the ranks, as you say. Thanks! $\endgroup$ – Heiro Jun 6 at 15:03
  • $\begingroup$ To the question in your first sentence: By no means! The tensor product is something quite else than the direct product, and in this case its action is (among other things) to kill all torsion prime to $3$. That’s why the $2$-torsion elements of $K^\times$ did not make an appearance in my answer. $\endgroup$ – Lubin Jun 7 at 1:00
  • $\begingroup$ And while we’re at it, the exponentiation by elements of $\Bbb Z_p$ is really not leftfield. It’s not so far removed from the wonderful properties of the logarithm, defined on all of $1+\mathfrak m$, no matter what the ramification index over $\Bbb Q_p$, but with values in $K^+$ (the values may be nonintegral). $\endgroup$ – Lubin Jun 7 at 15:55

Your problem being local, why do you complicate it by bringing it back to a global one ? I'll keep your notation $K_v = \mathbf Q_3 (\sqrt {-6})$ and work locally.

For any $p$-adic local field $K$ of degree $n$ over $\mathbf Q_p$, the quotient $K^*/{K^*}^p$ can be viewed (if written additively) as an $\mathbf F_p$ - vector space, of dimension $n+2$ (resp. $n+1$) according as $K$ contains or not a primitive $p$-th root of $1$ (this is a matter of Herbrand quotients, see e.g. Serre's "Local Fields", chap.14, prop.10 and ex.3). Here your $K_v$ is a quadratic totally ramified extension of $\mathbf Q_3$, not containing $\mu_3$ (because $(-3)(-6)=2.3^2$ is not a square in $\mathbf Q_3$), hence the above dimension is $3$, and we only need to find an $\mathbf F_3$-basis. A first natural vector, coming from an uniformizer, is $\sqrt {-6}$ (or $\sqrt 3$ if you want). It remains only to exhibit two linearly independent vectors in $U_1/{U_1}^3$, where $U_1$ is the group of prinipal units. I found the pair $1\pm \sqrt {-6}$ (but you must check, I am prone to calculation errors).

  • $\begingroup$ Yes, I edit that. When I think that I checked that (-3)(-6) is not a square ! $\endgroup$ – nguyen quang do Jun 7 at 6:26

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