Quadratic extension of field with two square classes

Question

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.

Answer 1

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.

Answer 2

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.