TITLE: Conceptual reason why a quadratic field has $-1$ as a norm if and only if it is a subfield of a $\mathbb{Z}/4$ extension?
QUESTION [19 upvotes]: I have convinced myself in a computation-heavy, ad-hoc way that a quadratic extension $K$ of $\mathbb{Q}$ occurs as the unique quadratic subfield of a $\mathbb{Z}/4$-extension of $\mathbb{Q}$ if and only if $N_{K/\mathbb{Q}}(x)=-1$ for some $x\in K$.  (See below for proof sketch.) There must be a more conceptually illuminating proof?

Morally / conceptually, why is it that a quadratic field occurs inside a cyclic degree 4 field if and only if -1 occurs in it as a norm?

What's the story here? An optimum answer would be a proof that doesn't ask the reader to follow, trust, or replicate any calculations. (As opposed to the following, where I have written "computation shows" about 5 times.)
Computation-heavy, ad-hoc proof sketch: $\Rightarrow$ If $K=\mathbb{Q}(\alpha)$ with $\alpha^2\in \mathbb{Q}, \alpha\notin\mathbb{Q}$, and $\exists x\in K$ with $N_{K/\mathbb{Q}}(x)=-1$, we have $x=a+b\alpha$, $a,b\in\mathbb{Q}$, with $N(x)=a^2-\alpha^2b^2 = -1$. Let $\beta = \sqrt{\alpha^2b-a\alpha}$ and let $L=K(\beta)$. Then, by computation, $\beta$ satisfies the polynomial
$$f=x^4-2\alpha^2bx^2+\alpha^2\in\mathbb{Q}[x]$$
which is irreducible by Eisenstein, thus $\beta^2$ is not a square in $K$, so $\mathbb{Q}(\beta)=L$ is degree $4$. Let
$$\beta'=(a+b\alpha)\beta\in L$$
Computation shows $\beta'$ also satisfies $f$, so $\beta\mapsto\beta'$ induces an automorphism of $L$; call it $g$. More computation shows $g(\beta')=-\beta$, so $g$ has order $4$, and $\beta$ has 4 conjugates in $L$, thus $L/\mathbb{Q}$ is Galois, with $\operatorname{Gal}(L/\mathbb{Q})$ generated by $g$. This proves the "if" direction.
$\Leftarrow$ If $L$ is a $\mathbb{Z}/4$ extension of $\mathbb{Q}$, let $g$ be a generator of its Galois group and let $K$ be the unique quadratic subfield. We can take $K=\mathbb{Q}(\alpha)$ with $\alpha^2\in\mathbb{Q}$ and $L=K(\beta)$ with $\beta^2\in K$. $1,\alpha,\beta,\alpha\beta$ is a basis for $L/\mathbb{Q}$; we can write a matrix for $g$, seen as a $\mathbb{Q}$-linear transformation, with respect to this basis. We must have $g(1)=1$, and $g(\alpha)=-\alpha$ because $g$ cannot act trivially on $K$. Let $g(\beta)=c+d\alpha+a\beta+b\alpha\beta$ with $a,b,c,d\in\mathbb{Q}$. Then, using the fact that $g(\alpha\beta)=g(\alpha)g(\beta)$, the matrix of $g$ is
$$\begin{pmatrix}1& &c&-\alpha^2d\\ &-1&d&-c\\ & &a&-\alpha^2b\\ & &b&-a\end{pmatrix}$$
Also, $g^2(\alpha)=\alpha$ because $K$ is fixed by $g^2$, and $g^2(\beta)=-\beta$ because $-\beta$ is $\beta$'s only conjugate over $K$ and $L$ is not fixed by $g^2$. Thus the matrix of $g^2$ is
$$\begin{pmatrix}1& & & \\ &1& & \\ & &-1& \\ & & &-1\end{pmatrix}$$
It follows by computation that $a^2-\alpha^2b^2=-1$. (It also follows that $c=d=0$, but this isn't needed.) Thus $a+b\alpha\in K$ has norm $-1$. This proves the "only if" direction.

REPLY [3 votes]: Elements of norm $-1$ in $K = {\mathbb Q}(\sqrt{m})$ give a solution of $x^2 - my^2 = -1$; writing this as $x^2 + 1 = my^2$ shjows that $m$ must be a sum of two squares, which is the condition for embeddability of $K$ in some cyclic quartic extension.  This last equivalence was discussed at length in the Jahresberichte of the DMV in the 1930s, with proofs ranging from elementary algebra to class field theory and division algebras (see the 1st chapter of Serre's notes on Galois theory for more ideas).