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? 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.
 A: A partial answer (i.e. one direction, and not quite computation-free) is that if $L/\mathbb{Q}$ is a $\mathbb{Z}/4$ extension with Galois group generated by $g$, $K$ is the fixed field of $g^2$, and $\beta$ is a field generator for $L/\mathbb{Q}$ that is the square root of an element of $K$, then $g(\beta)/\beta$ is in $K$ and has norm $-1$.  For, as argued in the OP, $g^2(\beta)=-\beta$, thus 
$$g^2(g(\beta)/\beta)=g(g^2(\beta))/g^2(\beta)=g(-\beta)/(-\beta) = g(\beta)/\beta$$
so $g(\beta)/\beta\in K$, and its conjugate over $\mathbb{Q}$ must be its image $g^2(\beta)/g(\beta)=-\beta/g(\beta)$ under $g$; thus its $K/\mathbb{Q}$-norm is the product
$$-\frac{\beta}{g(\beta)}\cdot\frac{g(\beta)}{\beta}=-1$$
A: I'll refer to my notes for some precise statements and definitions: https://dl.dropboxusercontent.com/u/27883775/math%20notes/ant.pdf
Define $I_K^S$ to be the group of ideals in $K$ generated by prime ideals not in $S$ (or not dividing primes in $S$). Define $P_K(1,\mathfrak m)$ to be the group of principal ideals that are "1 modulo $\mathfrak m$." (10.3)
By class field theory (10.4.1), if $L/K$ is abelian, $\mathfrak m$ is a modulus containing the set of primes of $K$ ramifying in $L$, and $S$ the set of primes dividing $\mathfrak m$, there is an isomorphism $\psi_{L/K}:I^S_K/(P_K(1,\mathfrak m)\cdot \text{Nm}_{L/K}(I_L^S))\xrightarrow{\cong} G(L/K)$, and this is compatible with field extension $M/L$ to form a commutative square, the right side of the square being the natural projection $G(M/K)\to G(L/K)$. 
The case of $\mathbb Q$:
The above may seem complicated, but it's much easier to understand in the case of $\mathbb Q$, because Kronecker-Weber encapsulates much of the statement above.
Every abelian extension $L/\mathbb Q$ corresponds to $(N,m)$ where $N\subseteq (\mathbb Z/m)^{\times}$ satisfies $(\mathbb Z/m)^{\times}/N\cong G(L/\mathbb Q)$.
Indeed, $I_{\mathbb Q}^S/P_{\mathbb Q}(1,m\infty)=(\mathbb Z/m)^{\times}$, and the norm group $\text{Nm}_{L/\mathbb Q}(I_L^S)$ can be thought of as a subgroup of $(\mathbb Z/m)^{\times}$. We can all make this very explicit if we want (and avoid CFT): $L=\mathbb Q(\sum_{j\in N}\zeta_m^j)$, and the isomorphism is given by $j\mapsto (\zeta_m\mapsto \zeta_m^j)$. From this it is clear, for example, that $-1\in N$ iff $L$ is real.
Note that if $m\mid m'$ and $N'$ projects to $N$, then
$(N,m)$ and $(N',m')$ correspond to the same extension. For quadratic $L$, the smallest $m$ we can take is the discriminant. (It is the same as the smallest $m$ such that $L\subseteq \mathbb Q(\zeta_m)$.)
The Problem
Let $L/\mathbb Q$ be quadratic. The following are equivalent.
(1) $L/\mathbb Q$ occurs inside a $\mathbb Z/4$ field.
(2) There is $(N,m)$ corresponding to $L$ and $N_1\subseteq N$ such that $(\mathbb Z/m)^{\times}/N_1\cong \mathbb Z/4$.
(3) $L$ is real and the discriminant of $L$ is divisible only by primes $p\equiv 1,2\pmod 4$.
(4) $-1\in \text{Nm}_{L/\mathbb Q}(L)$.
(1)$\iff$(2) follows from the preceding discussion.
(3)$\implies$(2): The argument is elementary but the group theory is messy. Choose $m$ such that $8\mid m$ and the discriminant $d$ divides $m$, and write $m=\prod_i p_i^{e_i}$.  Let $h_i:G_i\to \mathbb Z/2$ be the unique nontrivial map. 
Let $h_i:G_i\to \mathbb Z/2$ be a nontrivial map (the only place we have to be careful is $p_i=2$, here use the map $G_i\cong \mathbb Z/2\oplus  (\mathbb Z/2)^{e_i-2}\to  (\mathbb Z/2)^{e_i-2}\to \mathbb Z/2$).
One can check $N=\ker \sum h_i$ corresponds to the unique real quadratic extension with discriminant $d$. (The image to have in mind is a checkerboard.)
Because all $p_i\equiv 1,2\pmod 4$, there exist surjective maps $h_i':G_i\to \mathbb Z/4$ and we can let $N_1=\ker \sum h_i'$. This shows (2).
(3)$\implies$(2): Suppose by way of contradiction $N_1$ works. Write $m=\prod_i p_i^{e_i}$ and keep the notation from above.
First suppose $L$ is not real. Then $-1\not\in N$. But then $-1$ would have order 4 in $G/N_1$, contradiction.
Suppose $k$ is such that $p_k\ne 4k+1,2$. 
Now $G_k/G_k\cap N\cong \mathbb Z/2$ (as opposed to $\{1\}$) because otherwise $(N',m/p_k)$ would also represent $m$, where $N'$ is the projection of $N$ to $\prod_{i\ne k}G_i$.
Now consider $g\in \prod G_i$ such that $g_i=1$ for each $i\ne k$ and $g_k$ is a generator of $G_k$. Then $g\not\in G_k\cap N$ so $g\not\in N$. This means $g$ has order 4 in $G/N_1$. But $\mathbb Z/4$ is not a quotient of $G_k$, contradiction.
For (3)$\iff$(4), note $x^2-my^2$ is solvable over every $\mathbb Q_p$ iff $m$ is only divisible by 2 and $4k+1$ primes. Then use the Hasse norm theorem: for cyclic $L/K$, an element is a global norm iff it is a local norm over every place.
The moral reason why you can detect the possibility of a $\mathbb Z/4$ extension with norms is the correspondence between norm groups and abelian extensions; however, (4) is not immediate from (2) because of the technicality with norms of elements vs. norms of ideals. We've already remarked $-1\in N$ iff $N$ is real, so whether $-1\in N$ only measures part of the failure to be part of a $\mathbb Z/4$-extension. 
Possibly there's something more elementary you can do with quadratic forms.
@Ben: $\frac{g(\beta)}{\beta}$ reminds me of cohomology, although I don't know if that has anything to do with it.
A: 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).
A: In Artin-Tate's "Class Field Theory", at the very end of chapter 10 on "Grunwald's Theorem", you can find  the  purely algebraic (as opposed to arithmetic) following characterization:  Let p be a given prime, k any field containing a primitive $p^r$ - primitive root of unity w, K/k a cyclic extension of degree divisible by p . Then there exists an extension L/K such that L/k is cyclic and L/K is cyclic of degree $p^r$ iff w is a norm from K . The proof is "conceptual",  it is essentially a good exercise in Galois cohomology .
A: Embedding Problem
Yes, it is possible to give a summarized argument , but to understand it requires a certain familiarity with the cohomological machinery. Let me try to give a (not too) sketchy survey. I apologize in advance for my poor typing, I’m kind of « TEX illitterate ». Throughout we’ll consider only finite groups, although profinite groups (which appear in Galois theory) could also be dealt with. Let K/k be a Galois extension with Galois group G. Let E be a group extension of G by a given group A, i.e. A is normal in E and G is isomorphic to E/A. Two such extensions E and E’ are called  equivalent if there exists an isomorphism between E and E’ which induces the identity on A and G. The so called « embedding problem » (K /k, [E]) is two-fold :
1) On the side of group extensions, characterize the equivalence classes [E]. If A is abelian, it is naturally a G-module, and [E] corresponds to a cohomology class e in $H^2$(G, A). This is explained in any textbook on group cohomology, but the underlying reason is simple : for any section of sets j: G -> E, j(xy) is a priori different from j(x)j(y), and the "deviation" yields a 2-cocycle with values in A. Note that the trivial cohomology class corresponds to a split extension. If moreover G is cyclic, it is known that $H^2$(G, A) is isomorphic to the quotient $A^G$/N(A) , where the numerator denotes the fixed points under G and the denominator the image of the norm map attached to G. If in addition A is cyclic, it follows that $H^2$(G, A) itself is cyclic.
2) On the side of field extensions, construct a Galois extension L/K/k such that Gal(L/K) is isomorphic to A and [E] = [Gal(L/k)]. If A is abelian, the embedding problem will be denoted (K/k, e), and there is a well established theory starting from a citerion of Klaus Hoechsmann, J. reine angew. Math, 229 (1968), 81-106 : let  $G_k$ be the absolute Galois group of k and  consider the inflation-restriction-transgression exact seqence (i.e. the beginning of the Hochschild-Serre spectral sequence)  1 -> $H^1$(G, A)  -> $H^1$($G_k$, A) -> $H^1$($G_K$, A)$^{G_k}$  -> $H^2$(G, A) ; then the embedding problem (K/k, e) is solvable iff e lies in the image of the transgression map (= the rightmost one). 
Let us apply this theory to our situation, where G is cyclic and k contains the group A = $W_q$ of q-th roots of 1, where q is a power of a prime p. An elementary cohomological reduction step allows us to assume that G is a p-group. From Kummer theory (for this, suppose that p is not the characteristic of k), we know that K*/$K*^q$ is isomorphic to Hom($G_K$, $W_q$) = $H^1$($G_K$, $W_q$). Besides, it is a classical exercise to show that the field L = K($a^{1/q}$) is Galois over k iff a mod $K*^q$ belongs to (K*/$K*^q$)$^G$. Manipulating the q-th power map in K*, one gets a canonical exact sequence  (K*/$K*^q$)$^G$  -> $H^2$(G, $W_q$) -> $H^2$(G, K*) , where the leftmost map can be identified with the transgression. Hence its image is the kernel of the righmost map,which can be identified (since G is cyclic) with the kernel of $W_q$/N($W_q$) -> K*/N(K*), and we are done.
Of course, in your initial problem, you don’t need the whole machinery here. It suffices to follow the map road and, at each step, give an elementary solution.
