Square classes of $\mathbb{Q}_2^\times$?

Question

Does anyone know how to easily compute them? We know that a number is a square modulo $2^k$ if and only if it's a square modulo $8$. This gives a bunch of integers that represent square classes. I also know that

$$\mathbb{Q}_2^\times \cong \mathbb{Z}\times (1+2\mathbb{Z}_2),$$

but I can't figure out how to find the square classes in $1+2\mathbb{Z}_2$. Is there some really simple solution for this? I can't seem to figure out how to apply Hensel's lemma here.

Maybe my little knowledge is a dangerous thing, but why isn't the answer simply everything of the form $2^k (1 + 8z)$, where $k \in \mathbb Z$ and $z \in {\mathbb Z}_2$? — Greg Martin, Jan 11 '12 at 19:28

score 4 · Accepted Answer · answered Jan 11 '12 at 19:33

4

By your observations, a square element of $1+2\mathbb{Z}_2$ must actually live in $1+8\mathbb{Z}_2$. So $$ \mathbb{Q}_2^\times/\mathbb{Q}_2^{\times 2}\approx \mathbb{Z}/2\mathbb{Z}\times \frac{1+2\mathbb{Z}_2}{1+8\mathbb{Z}_2}\approx \mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}. $$ The first factor of $\mathbb{Z}/2\mathbb{Z}$ corresponds to choosing even/odd-ness of the power of 2 dividing the element of $\mathbb{Q}_2^\times$, the latter two any choice of coset represenatives for odd squares mod 8. So one possible enumeration of representatives for these 8 classes are the classes of $$ \{\pm 1,\pm 2,\pm 5,\pm 10\}. $$

answered Jan 11 '12 at 19:33

Cam McLeman

6,333

Cam, that is the choice on page 43 of Cassels, Rational Quadratic Forms. There was some poor slob a few months ago here on MSE who wanted to know the primes for which some indefinite ternary, diagonal such as $3 x^2 + 7 y^2 - 15 z^2,$ was isotropic. I typed it all in but I do not believe it took. Oh, also he wouldn't buy Cassels, he "had no access" to it. http://math.stackexchange.com/questions/89138/isotropy-over-p-adic-numbers – Will Jagy Jan 11 '12 at 20:20
@WillJagy: Indeed, I'm at the moment the unique upvote on that answer. :)
And it appears that this particular class of representatives is a heavy favorite -- I quickly checked, and both Gouvea and Serre use it as well!
– Cam McLeman Jan 11 '12 at 23:14
1

Thanks, Cam, I do like to educate the OP, but that takes time, and often they want just enough to do the homework, but quickly. I see, both answers there were mine, after the OP said the first answer wasn't good enough. Sigh. – Will Jagy Jan 11 '12 at 23:51

Square classes of $\mathbb{Q}_2^\times$?

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