4

How to prove that there are seven non isomorphic quadratic field extensions of $\mathbb{Q}_2$?

Approach: I already proved that an unit in $\mathbb{Z}_2$ is congruent with $1,3,5,7 \mod 8$. (maybe that's useful)

bob
  • 1,256

2 Answers2

9

There are infinitely many non-isomorphic field extensions of $\Bbb Q_2$.

There are, however, exactly seven non-isomorphic quadratic extensions of $\Bbb Q_2$.

Each of these is $\Bbb Q_2(\sqrt a)$ where $a\in\Bbb Q_2$ is not a square, and $\Bbb Q_2(\sqrt a)=\Bbb Q_2(\sqrt b)$ iff $a/b$ is a square. So there is one extension for each coset of $(\Bbb Q_2^\times)^2$ in $\Bbb Q_2^\times$ except for $(\Bbb Q_2^\times)^2$ itself.

So we need to prove that $|\Bbb Q_2^\times:(\Bbb Q_2^\times)^2|=8$. This is in all the textbooks. In the end it boils down to each $c\in\Bbb Z_2$ with $c\equiv1\pmod8$ being a square in $\Bbb Q_2$ (this is a Hensel lemma argument),

Angina Seng
  • 158,341
  • 1
    You could try Mahler's book on $p$-adic numbers and functions. He has a whole chapter on quadratic extensions of $\Bbb Q_p$. – Angina Seng Apr 22 '17 at 15:27
  • how can you prove precisely that all these extensions are non isomorphic? – bob Apr 25 '17 at 14:21
  • @bob To show $K(\sqrt a)\cong K(\sqrt b)$ as $K$-extensions implies $a/b$ is a square in $K$, note this implies $b=(r+s\sqrt a)^2$ which is only possible if $r=0$ or $s=0$. Then $b$ is a square or $a$ times a square. – Angina Seng Apr 25 '17 at 15:16
  • Okay thanks, I know that the field extensions are $\mathbb{Q}_2(\sqrt{a})$ with $a=-1,2,-2,3,-3,6,-6$. So for example $\mathbb{Q}_2(\sqrt{2}) \not \cong \mathbb{Q}_2(\sqrt{6})$ since otherwise there is a square $\alpha^2$ such that $6 = 2 \alpha^2$ or $3= \alpha^2$. Hence $\alpha \in \mathbb{Z}_p^{\times}$ and thus $\alpha^2 \equiv 1 \mod 8$. Contradiction. Is this correct? – bob Apr 25 '17 at 15:46
  • @bob That's exactly the sort of reasoning one needs. – Angina Seng Apr 25 '17 at 15:48
  • @LordSharktheUnknown, why do we need to prove $|\mathbb{Q}_2^{\times}: (\mathbb{Q}_2^{\times})^2|=8$ ? How this will help? – MAS Jun 28 '19 at 15:48
5

Perhaps a more accessible reference would be Serre's "Course in Arithmetics", chapter II, §3, thm. 2, which gives the structure of $\mathbf Q_p^*$. But your question is actually a general one, and it admits a geneal answer. More precisely: for a given prime $p$, what is the number of extensions of $\mathbf Q_p$ which are cyclic of degree $p$ ? Since $\mathbf Q_p$ does not contain the $p$-th roots of unity when $p$ is odd, Kummer's theory (as used by @Lord Shark the Unknown when $p=2$) is no longer available. You need a (much) more powerful machinery, which is local class field theory. Consider the composite K of all the above extensions, which is no other than the maximal abelian extension of $\mathbf Q_p$ whose Galois group G has exponent $p$. Then CFT tells you that G is canonically isomorphic (not simply dual) to the multiplicative group $\mathbf Q_p^*/(\mathbf Q_p^*)^{p}$. This latter group is finite of order $p^2$ for $p$ odd (Serre, op. cit.), i.e. it is isomorphic to $\mathbf Z/p\mathbf Z \times \mathbf Z/p\mathbf Z$. It only remains to compute the number of subgroups of order $p$, which is easily seen to be $p+1$. Note the difference between the two cases $p$ odd and even (Serre, op. cit.)

NB. Our argument here takes place in a chosen algebraic closure of $\mathbf Q_p$. This explain why we don't care to distinguish between isomorphic extensions and equal extensions, contrary to @Lord Shark the Unknown.

nguyen quang do
  • 14,578