# What is $\mathbb{Q}_3(\sqrt{-6})^{\times}/\left(\mathbb{Q}_3(\sqrt{-6})^{\times}\right)^3$?

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.

• $(1+x)^3 = 1+3x+3x^2+x^3$ 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. – reuns Jun 5 at 18:04
• 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. – Lubin Jun 5 at 22:51

## 2 Answers

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.

• 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! – Heiro Jun 6 at 15:03
• 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. – Lubin Jun 7 at 1:00
• 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). – 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).

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