Exception in the characterization of equality of quadratic extensions when the field is of characteristic $2$.

I've been brushing up on Galois Theory with the book Galois Theory Through Exercises and I have come across one such exercise that claims what I believe is false when a field is of characteristic $2$. The problem goes as follows:

2.5 (a) Let $L\supseteq K$ be a field extension and let $\alpha \in L\setminus K$, $\alpha^2 \in K$. Show that $$K(\alpha) = \{a + b\alpha, \text{ where }a,b\in K\}.$$ (b) Let $K(\sqrt{a})$ and $K(\sqrt{b})$ where $a,b\in K$, $ab \neq 0$, be two field extensions of $K$. Show that $K(\sqrt{a}) = K(\sqrt{b})$ if and only if $ab$ is a square in $K$ (that is, there is a $c\in K$ such that $ab = c^2$).

Part (a) is easy enough and so is proving that $K(\sqrt{a}) = K(\sqrt{b})$ when $ab = c^2$ for some $c\in K$. Now, if $K(\sqrt{a}) = K(\sqrt{b})$ then we distinguish two cases:

• If $K = K(\sqrt{a}) = K(\sqrt{b})$ then let $u,v\in K$ such that $u = \sqrt{a}$ and $v = \sqrt{b}$. Then $ab = (uv)^2$, as we wanted to show.
• If $K \subsetneq K(\sqrt{a}) = K(\sqrt{b})$ then we have that $\sqrt{b}\in K(\sqrt{a})$, so using part (a) there are $c,d\in K$ such that $\sqrt{b} = d + c\sqrt{a}$. This means that $$b = d^2 + c^2a + 2cd\sqrt{a}$$ so $2cd\sqrt{a} = b - d^2 - c^2a$. Since the RHS is in $K$, so must the LHS. But $\sqrt{a}\notin K$ so $2cd = 0$.

If $c = 0$ then $\sqrt{b} = d \in K$ which contradicts our assumption that $K\subsetneq K(\sqrt{b})$.

If $d = 0$ then $\sqrt{b} = c\sqrt{a}$ so $\sqrt{ab} = ac$ and, therefore, $ab = (ac)^2$, as we wanted to prove.

However, if $\text{char}(K) = 2$ then we cannot deduce that $c$ or $d$ is $0$. And I think that the claim is false. Indeed, let $a \in K\setminus K^2$ and let $b = a + 1$. Then $\sqrt{b} = \sqrt{a} + 1$ so $K(\sqrt{a}) = K(\sqrt{b})$. However, $\sqrt{ab} = a + \sqrt{a}\notin K$.

Am I missing something here or should the question state that $\text{char}(K) \neq 2$

Your proposed counterexample works provided your field $K$ of characteristic $2$ is such that not every element of $K$ is a square in $K$.

For example, choosing $K=F_2$ won't work.

More generally, if $K$ is a field of characteristic $2$, the map $x\mapsto x^2$ is an injective homomorphism, so in the case where $K$ is finite, every element of $K$ is a square in $K$.

But for your proposed counterexample to work, it's not enough to simply assume that the field $K$ of characteristic $2$ is infinite, since if $K$ is an algebraic extension of $F_2$, every element of $K$ lies in a finite subfield of $K$, hence once again, every element of $K$ is a square in $K$.

Thus, a qualifying $K$ must be a transcendental extension of $F_2$.

As an example of a $K$ that works, you can choose $K=F_2(x)$, and then you can let $a=x$ and $b=x+1$, exactly as you proposed.