Fields having exactly one quadratic extension (up to isomorphism)

Question

Let $F$ be an infinite field such that $F$ has, up to field isomorphisms$^{[1]}$, exactly one extension $K/F$ of degree $2$. Does it imply that $[\overline F : F]<\infty$ ? What happens if $F$ has characteristic $0$?

I don't think that this holds, even if characteristic $0$, but I didn't have an example of a field $F$ of characteristic $0$ ($\implies F$ is infinite) such that $[\overline F : F] = \infty$ and in $F^*$, any product of two non-squares is a square, and there is at least one non-square (see 4) below).

---

My thoughts:

1) The only example of such $F$ I know is $\Bbb R$: any quadratic extension embeds in $\Bbb C$, which has already degree $2$ over $\Bbb R$. My question is to know if other examples exist: if $[\overline F : F]<\infty$ then (by Artin-Schreier theorem) $F$ is a real closed field.

2) Any finite field $\Bbb F_q$ has exactly one extension of degree $n$ (namely $\Bbb F_{q^n}$), up to field isomorphisms, for every $n \geq 1$.

3) It implies that all the quadratic extensions are isomorphic as fields, but this is not sufficient, precisely when $F$ has no quadratic extension, e.g. $\overline F=F$ (or $\bigcup_{n \geq 0} K_n$, with $K_0=\Bbb Q,K_{n+1}=\{x \in \Bbb C \mid x^2 \in K_n\}$).

4) In characteristic different from $2$, any quadratic extension of $F$ is separable and has the form $F(\sqrt a)$ where $\sqrt a \not \in F$. Therefore, as mentioned here, the quadratic extensions of $F$ correpond to $A:=F^* / F^{*,2}$ where $F^{*,2} = \{x^2 \mid \in F^*\}$. Notice that $A$ is a $\Bbb F_2$-vector space, via $[a]_2 \cdot [x]_{F^{*,2}} = [x^a]_{F^{*,2}}$.

So the most interesting case is when we are looking for fields of characteristic $\neq 2$ such that $A=F^* / F^{*,2}$ has order $2$. Equivalently, in $F^*$, any product of two non-squares is a square, and there is at least one non-square (because $x,y$ non squares $\implies x,1/y$ non-squares $\implies x/y = a^2 \in F^{*,2} \implies [x]_{F^{*,2}} = [y]_{F^{*,2}}$).

Let $a$ be a non-square in $F^*$ and let $i = \sqrt a$. Showing that $F(i)$ is algebraically closed is not reasonable. Notice that this condition about $[F^* : F^{*,2}]=2$ is involved in 3. here, which is precisely the situation where $a=-1$ is not a square.

Then I thought to some extension $F$ of $K=\mathrm{Frac}(\Bbb R[x,y]/(x^2+y^2+1))$ since $-1$ is not a square in $K^*$ but is a sum of squares ; we just need $[F^* : F^{*,2}]=2$, and then it's a counterexample, since $F$ won't be a formally real field. I only know how to do $[F^* : F^{*,2}]=1$, see my $K_n$'s in 3).

5) I tried $F = \overline{\Bbb F_2}(t)$, because $t^{1/n}$ has degree $n$ for any $n$, so $[\overline F :F]=\infty$. I think that any extension $F\left(\sqrt{P(t)/Q(t)}\right)$ is isomorphic to $F(\sqrt t) = \overline{\Bbb F_2}(\sqrt t)$ when $P,Q \in \overline{\Bbb F_2}[t]$ i.e. $P/Q \in F$, because we have $\sqrt{a+b}=\sqrt a + \sqrt b$ in the sense that $x^2=a,y^2=b \implies (x+y)^2 = a+b$, so that $\sqrt{P(t)/Q(t)}$ is just a rational fraction in $\sqrt t$, i.e. belongs to $F(\sqrt t)$. However, in characteristic $2$, it is not clear that all quadratic extensions arise as $F(\sqrt a)$ for some non-square $a \in F$.

6) In characteristic 0 (at least $\neq 2$), it is not clear that $F(\sqrt{t+1}) \not \cong F(\sqrt t)$. If there was a field isomorphism, then there is $u \in F(\sqrt t)$ such that $u^2=t+1$, hence there are $a,b \in F[t]$ such that $(a(\sqrt t)/b(\sqrt t))^2 = t+1$, which yields $a(x)^2=(x^2+1)b(x)^2$ as polynomials in $F[x]$...

---

$^{[1]}$ I'm only interested in fields isomorphisms, not in "field extensions" isomorphisms (i.e. not in $F$-algebras isomorphisms – these are equivalent to saying that $f : K \stackrel{\cong}{\to} K'$ commutes with the embeddings $i : F \to K$ and $i' : F \to K'$).

Answer 1

After some searches, I can answer negatively. Actually we can state more:

There are infinite fields with exactly one extension of degree $n$, for every $n \geq 1$.

(Notice that a finite field satisfies this condition, but an algebraically closed field doesn't work since has at most one extension of degree $n \geq 1$ (actually $0$ if $n \geq 2$)).

Such fields $F$ satisfy in particular $[\overline F : F] = \infty$, since there are elements of arbitrarily large degree over $F$.

A concrete example is:

The field of formal Laurent series over $\Bbb C$, denoted by $F = \Bbb C((T))$.

(Its algebraic closure is the field of Puiseux series, but we don't need that to state $[\overline F : F] = \infty$. By the way, $F$ has characteristic 0). This example has actually been discussed here. This example doesn't require the axiom of choice (the other one below does require it)...

---

More generally, we can look at quasi-finite fields (which also require the field to be perfect), or the even stronger notion of pseudo-finite field. An exemple is given here:

Let $Q$ be the set of all prime powers, and let $\mathcal U$ be a non-principal ultrafilter on $Q$ (i.e. for all $q \in Q$, there is $A \in \mathcal U$ such that $q \not \in A$). Then the field $$F' = \left(\prod_{q \in Q} \mathbb F_q\right)/\mathcal U$$ is a pseudo-finite field ; in particular it has exactly one extension of degree $n$, for every $n \geq 1$.

Answer 2

Finite fields still give loads of examples. Let $\kappa$ be any finite field, and let $q$ be any prime (not necessarily distinct from the characteristic). Now take the union $\mathcal K$ of all fields $K_n$ for which $[K_n:\kappa]=q^n$. Then $\mathcal K$ is infinite, and will have a unique extension of every degree prime to $q$, in particular only one quadratic extension, if $q\ne2$.

EDIT — Addition:
On looking more closely at your question, I see that you make some guesses about characteristic two. There the quadratic-extension picture is simultaneously simpler and more complicated. When you adjoin the square root of something, you’re making an inseparable extension of degree $p$, and the truth of the matter is that when the base is perfect, as $\overline{\Bbb F_2}$ is, and your field is finitely generated over the base, and the transcendence degree is only one, then there is precisely one inseparable extension of degree $p$, indeed only one purely inseparable extension of degree $p^m$ for each $m$. In other words, starting with $k=\overline{\Bbb F_2}(t)$, no matter what rational function you adjoin the square root of, you get the same extension, $k^{1/2}$
On the other hand, there are infinitely many nonisomorphic quadratic separable extensions of $k=\overline{\Bbb F_2}(t)$, for instance the ones gotten by adjoining the roots of $X^2+t^mX+t$.

Answer 3

No, it does not. Let $ F $ be an algebraic extension of $ \mathbf Q $, maximal with respect to the property of not containing any of $ \sqrt{2}, \sqrt[3]{2}, \zeta_3 \sqrt[3]{2}, \zeta_3^2 \sqrt[3]{2} $. (The existence of such an $ F $ is guaranteed by Zorn's lemma.) Then, $ F $ has precisely one quadratic extension: $ F(\sqrt{2}) $; however, $ X^{3^k} - 2 $ remains irreducible in $ F[X] $ for all $ k \geq 1 $, thus $ [\bar{\mathbf Q} : F] $ is not finite.

Answer 4

EDIT: This is currently incorrect. I'll work on salvaging it.

Using the axiom of choice, let $s_1, s_2,\ldots$ be a sequence of numbers that are a subset of a transcendence basis of $\mathbb{C}$. Let $F$ be a maximal field extension of $\mathbb{Q}$ that doesn't contain any of the $s_i$ (Zorn's Lemma or transfinite recursion gives this). Notice that $F(s_i)$ and $F(s_j)$ are isomorphic. Now consider $K=F(s_1^2,s_2^2,\ldots)$. We now have that $K(s_i)$ and $K(s_j)$ are isomorphic and are degree $2$ extensions of $K$. It follows that $[\mathbb{C}:K]$ is infinite. One way to see this is to take the first $k$ elements of the sequence and notice that $[K(s_1,\ldots,s_k):K]=2^k$.

The issue, as mentioned in the comments, is that I need to show that $K[\sqrt{s_i^2+s_j^2}]$ doesn't mess things up.