3

Given a field $F$, how many non-isomorphic quzdratic extension of $F$ are there ? I don't know if there is a general answer, for instance there is only one for $F=\mathbf{R}$, viz. $\mathbf{C}$, and no one for $F=\mathbf{C}$, for it is algebraically closed.

There may be a more precise answer for quadratic extension of number fields. For $F=\mathbf{Q}$, there are only two, every real extension being isomorphic and of the form $\mathbf{Q}(\sqrt{d})$, and every complex extension being a $\mathbf{Q}(\sqrt{-d})$, with $d$ a positive integer.

What about $p$-adic fields ? If $F$ is a finite extension of degree $d$ of $\mathbf{Q}_p$, how many quadratic extension $E/F$ are there ? Krasner seems to have given an explicit formula for coutning such extensions of degree $n$, but I hope there is an easier way to reach the answer for $n=2$.

If $x$ is in $E \backslash F$, $E = F(x)$. We can suppose $x = \sqrt{a}$ for an $a \in F$, what can be seen as in the real or rational case by factorizing the minimal polynomial (of degree 2) of $x$. We can also suppose it squarefree, for each square factor do not change the extension over $F$. For $\pi$ a uniformizer, every element can be written $x = \pi^r u$ with $u \in O_F$ an integer of $F$, hence quadratic extensions are of the form $E = F(\sqrt{u})$ or $E = F(\sqrt{\pi u})$, with $u$ squarefree unit.

Hence quadratic extensions are parametrized by squarefree units, that is $O_F/O_F^2$, that is the kernel of $x \mapsto x^2$, and hence of cardinal 1 or 2 ? ($x^2=1$ only having one or two solutions in a field) Am I right ?

Any clue or idea would be welcome ;)

Best regards,

user26857
  • 52,094
Desiderius Severus
  • 1,500

2 Answers2

6

By Kummer theory, there is a bijection between the set of separable extensions of F of degree at most 2 and the quotient $F^*/(F^*)^2$, which is not necessarily finite. This quotient is visibly a vector space over the field $F_2$, the addition of vectors being just the multiplication law in the group. Let N be its dimension ; if N is finite, then the number of quadratic extensions which you ask for is $2^N -1$.

Examples:

  • if F is finite , N = 1 always, because F* is cyclic

  • if F is of degree n over $Q_p$, N = n + 1 if F does not contain a primitive p-th root of unity, n + 2 otherwise

  • if F is a number field, N is infinite
nguyen quang do
  • 14,578
  • For finite $F$ of characteristic $2$, $N=0$, but Kummer theory breaks down anyway and there is still one quadratic extension. – Torsten Schoeneberg Jul 13 '23 at 23:12
  • Also, in the case of $F\vert\mathbb Q_p$, you seem to be describing extensions of degree $p$ instead of quadratic ones. (Which happens to be the same iff $p=2$.) In other words, shouldn't we have $N=2$ for odd $p$ (giving us three distinct quadratic extensions, regardless of $n$), while $N=n+2$ for $p=2$? – Torsten Schoeneberg Jul 14 '23 at 00:11
4

Instead of $O_F/O_F^2$, the question is more about $F^\times/(F^\times)^2$, which works for all fields (of characteristic $\ne 2$). For example $(\Bbb C^\times)^2=\Bbb C^\times$, $(\Bbb R^\times)^2=(0,\infty)$ is of index $2$ in $\Bbb R^\times$, $(\Bbb Q^\times)^2$ is of infinite index in $\Bbb Q^\times$.

user26857
  • 52,094
Hagen von Eitzen
  • 374,180