Quadratic extension of field with two square classes Let $K$ be a field which has at most two square classes, that is $|K^*/(K^*)^2|\le 2$, and let $L$ be a quadratic field extension of $K$. Is it true that $L$ also has at most two square classes?
The examples that come to my mind are finite fields, real closed and algebraically closed fields. For all of these this is true.
 A: Let $K$ be a field with $\text{char}(K)\ne 2$ such that $[K^*:(K^*)^2]\le 2$, and suppose $L$ is a quadratic extension of $K$.

Claim:$\;[L^*:(L^*)^2]\le 2$.

Proof:

Since $L$ is a quadratic extension of $K$, and $\text{char}(K)\ne 2$, it follows that $L=K(\sqrt{n})$, for some $n\in K^*{\setminus}(K^*)^2$.

From $n\in K^*{\setminus}(K^*)^2$, we get $[K^*:(K^*)^2]\ge 2$, hence $[K^*:(K^*)^2]=2$.

First, a lemma . . .

Lemma $(1)$:

*

*If $u,v\in K^*{\setminus}(K^*)^2$, then $uv\in (K^*)^2$.$\\[4pt]$

*If $uv\in K^*{\setminus}(K^*)^2$, then exactly one of $u,v$ is in $(K^*)^2$.$\\[4pt]$

*If $u\in (K^*)^2$ and $v\in K^*{\setminus}(K^*)^2$, then $uv\in K^*{\setminus}(K^*)^2$.

Proof of lemma $(1)$:$\;$Immediate consquence of $[K^*:(K^*)^2]=2$.

Another lemma . . .

Lemma $(2)$:$\;K^*{\setminus}(K^*)^2=\{g^2n{\,\mid\,}g\in K^*\}$.

Proof of lemma $(2)$:

If $g\in K^*$, then $g^2\in (K^*)^2$, hence, since $n\in K^*{\setminus}(K^*)^2$, we get $g^2n \in K^*{\setminus}(K^*)^2$.

Conversely, suppose $x\in K^*{\setminus}(K^*)^2$.

From $n\in K^*{\setminus}(K^*)^2$, we get ${\Large{\frac{1}{n}}}\in K^*{\setminus}(K^*)^2$, hence
\begin{align*}
&
x{\,\cdot\,}\frac{1}{n}\in (K^*)^2
\\[4pt]
\implies\;&
\frac{x}{n}=g^2\;\,\text{for some $g\in K^*$}
\\[4pt]
\implies\;&
x=g^2n
\\[4pt]
\end{align*}
which completes the proof of lemma $(2)$

Next let $N:L^*\to K^*$ be given by
$$
N(u)=a^2-nb^2
$$
where $u=a+b\sqrt{n}$, with $a,b\in K$, not both zero.

An easily verified, well known result, is the following . . .

Lemma $(3)$:$\;N(uv)=N(u)N(v)$, for all $u,v\in L^*$.

Proof of lemma $(3)$:

Just write $u=a+b\sqrt{n}$ and $v=c+d\sqrt{n}$, and then expand both sides.

Next, the key lemma . . .

Lemma $(4)$:$\;(L^*)^2=\{u\in L^*{\,\mid\,}N(u)\in (K^*)^2\}$.

Proof of lemma $(4)$:

First suppose $u\in (L^*)^2$.
\begin{align*}
\text{Then}\;\;&
u\in (L^*)^2
\\[4pt]
\implies\;&
u=v^2\;\,\text{for some $v\in L^*$}
\\[4pt]
\implies\;&
N(u)=N(v^2)
\\[4pt]
\implies\;&
N(u)=N(v)^2
\\[4pt]
\implies\;&
N(u)\in (K^*)^2
\\[4pt]
\end{align*}
Conversely, suppose $u\in L^*$ is such that $N(u)\in (K^*)^2$.

Our goal is to show $u\in (L^*)^2$.

Write $u=a+b\sqrt{n}$, with $a,b\in K$, not both zero.

Consider two cases . . .

Case $(1)$:$\;b=0$.

Then $u=a$, with $a\in K^*$.

If $a\in (K^*)^2$, then $a\in (L^*)^2$.

If $a\in K^*{\setminus}(K^*)^2$, then $a=g^2n$ for some $g\in K^*$, 
hence $a=(g\sqrt{n})^2\in (L^*)^2$.

Either way, we have $u\in (L^*)^2$, which resolves case $(1)$.

Case $(2)$:$\;b\ne 0$.
\begin{align*}
\text{Then}\;\;&
N(u)\in (K^*)^2
\\[4pt]
\implies\;&
a^2-nb^2=g^2\;\,\text{for some $g\in K^*$}
\\[4pt]
\implies\;&
a^2-g^2=nb^2
\\[4pt]
\implies\;&
4(a^2-g^2)=4nb^2
\\[4pt]
\implies\;&
\Bigl(2(a+g)\Bigr)
\Bigl(2(a-g)\Bigr)
=
(2b)^2n
\\[4pt]
\implies\;&
\Bigl(2(a+g)\Bigr)
\Bigl(2(a-g)\Bigr)
\in
 K^*{\setminus}(K^*)^2
\\[4pt]
\end{align*}
hence exactly one of $2(a+g),\,2(a-g)$ is in $(K^*)^2$.

Without loss of generality, we can assume $2(a-g)\in (K^*)^2$, else replace $g$ by $-g$.

Then we can write $2(a-g)=h^2$ for some $h\in K^*$.

Now let $v\in L^*$ be given by $v=c+d\sqrt{n}$, where $c=h/2$ and $d=b/h$.

Then we have
\begin{align*}
v^2
&=\;
(c+d\sqrt{n})^2
\\[4pt]
&=\;
\Bigl(
\frac{h}{2}
+
\frac{b}{h}
\sqrt{n}
\Bigr)^2
\\[4pt]
&=\;
\Bigl(
\frac{h^2}{4}
+
\frac{nb^2}{h^2}
\Bigr)
+
b\sqrt{n}
\\[4pt]
&=\;
\Bigl(
\frac{h^2}{4}
+
\frac{a^2-g^2}{h^2}
\Bigr)
+
b\sqrt{n}
\\[4pt]
&=\;
\Bigl(
\frac{2(a-g)}{4}
+
\frac{a^2-g^2}{2(a-g)}
\Bigr)
+
b\sqrt{n}
\\[4pt]
&=\;
\Bigl(
\frac{a-g}{2}
+
\frac{a+g}{2}
\Bigr)
+
b\sqrt{n}
\\[4pt]
&=\;
a+b\sqrt{n}
\\[4pt]
&=\;
u
\\[4pt]
\end{align*}
so $u\in (L^*)^2$, which resolves case $(2)$, and completes the proof of lemma $(4)$.

Returning to the proof of the main claim, suppose $u,v\in L^*{\setminus}(L^*)^2$.
\begin{align*}
\text{Then}\;\;&
u,v\in L^*{\setminus}(L^*)^2
\\[4pt]
\implies\;&
N(u),N(v)\in K^*{\setminus}(K^*)^2
\\[4pt]
\implies\;&
N(u){\,\cdot\,}N(v)\in (K^*)^2
\\[4pt]
\implies\;&
N(uv)\in (K^*)^2
\\[4pt]
\implies\;&
uv\in (L^*)^2
\\[4pt]
\end{align*}
hence $\;[L^*:(L^*)^2]\le 2$, as was to be shown.
A: 
It is true when $char(K)=2$.


*

*If $K$ is algebraic over $\Bbb{F}_2$ then so is $L$, therefore $L^{*2}=L^*$ and we are done.


*Otherwise take $a\in K$ transcendental over $\Bbb{F}_2$ and let $E= \Bbb{F}_2(a^{1/2^\infty})\cap K$, so that $E=\Bbb{F}_2(a^{1/2^n})$ with $n \in \Bbb{N}\cup \infty$.
The kernel of $E^*/E^{*2}\to K^*/K^{*2}$ is trivial, so it must be that $|E^*/E^{*2}|\le 2$.
If $n <\infty$ then $E^{*2},a^{1/2^n}E^{*2}, (1+a^{1/2^n})E^{*2}$ are 3 distinct elements of $E^*/E^{*2}$, a contradiction.
Whence $n=\infty$, which means that $a$ is a square, so every transcendental element of $K$ is a square, and since algebraic elements are squares as well, we get that $K^*= K^{*2}$.
Next $L=K(b)$, $K(b)/K$ must be separable so that $K(b^2)=K(b)$ and hence $L^{*2} = K(b)^{*2}=K^2(b^2)^{*}= K(b^2)^*=K(b)^*=L^*$, qed.
