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?

Solution 1:

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

$$-\frac{\beta}{g(\beta)}\cdot\frac{g(\beta)}{\beta}=-1$$

Solution 2:

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.

Solution 3:

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).