Codex @codex  0

Take $u\in L\setminus K$, whose irreducible polynomial is $t^2+ct+d$. Since the characteristic is not two, $x=2u+c$ satisfies $x^2=c^2-4d=a\in K$ and $L=K(x)$. Irreducibility makes $a$ nonzero and nonsquare. If another quadratic extension is $K(y)$ with $y^2=b$, an isomorphism sends $x$ to $s+ty$, with $t\ne 0$. Squaring gives $a=s^2+t^{2}b+2sty$. Linear independence of $1,y$ forces $s=0$, so $a=t^{2}b$ and $b/a$ is a square. Conversely, if $b/a=q^2$, sending $x$ to $y/q$ gives an isomorphism. If $b$ itself is a square, the target is just $K$ and cannot be isomorphic; the square-ratio condition also fails.

The multiquadratic field $F$ is the splitting field of ${\prod }_i(t^2-a_i)$, a product of separable quadratic polynomials: each has distinct roots because $a_i\ne 0$ and the characteristic is not two. Even if factors repeat, this splitting field is a separable normal extension, and therefore Galois. Each automorphism sends $x_i$ to $\pm x_i$, and these signs determine it, embedding its Galois group into $(\mathbb{Z}/2{)}^n$. Every subgroup of this elementary abelian group is $(\mathbb{Z}/2{)}^m$ for some $m\le n$.

For a quadratic intermediate field, its fixing subgroup $H$ has index two and is the kernel of a nontrivial one-dimensional character $\chi :G\to \{\pm 1\}$. The coordinate sign characters span the dual of $G$: a linear functional on the subspace $G\subseteq {\mathbb{F}}_2^n$ extends to the ambient space by extending a basis. Hence $\chi ={\prod }_{i\in I}{\chi }_i$ for some subset $I$. Set $y={\prod }_{i\in I}{x}_i$. Then $\sigma (y)=\chi (\sigma )y$, so $y$ is fixed by $H$ but not by all of $G$, and $y^2={\prod }_{i\in I}{a}_i\in K$. Thus $K(y)$ is quadratic and equals the fixed field of $H$. This constructs all quadratic subfields of a multiquadratic extension: