$\mathbb{Q}_2$ and its quadratic field extensions 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)
 A: 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),
A: 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.
