Quadratic Fields

We can now say a bit more about the relationship between quadratic fields and cyclotomic fields.

Let \(\omega = e^{2\pi /p}\) for an odd prime \(p\). Recall \(disc(\omega)={\pm p^{p-2}}\) where the sign is positive if and only if \(p=1 (mod 4)\). Using the definition of the discriminant, we have

\[|\sigma_i(\omega^j)| = p^{(p-3)/ 2}\sqrt{\pm p}\]

where the \(\sigma_i\) are the embeddings of \(\mathbb{Q}[\omega]\) in \(\mathbb{C}\). But each embedding simply maps each \(\omega^i\) to some other \(\omega^j\), thus we may compute \(\sqrt{\pm p}\) using field operations on the powers of \(\omega\). In other words, \(\sqrt{\pm p} \in \mathbb{Q}[\omega]\), with the sign positive if and only if \(p = 1 (mod 4)\).

For example, for \(p = 3\) the above equation becomes

\[{| \array { 1 & \omega \\ 1 & \omega^2 } |} = \sqrt{-3}\]

which can be rewritten \(\sqrt{-3} = 2\omega + 1\).

Similarly for \(p = 5\) we obtain \(\sqrt{5} = \omega^2 - \omega^4 + \omega^3 - \omega = 1 -2 \omega -2\omega^4\).

The \(8\)th cyclotomic field contains \(\sqrt{2}\) because in this case we have \(\omega = \sqrt{2}/2 + i\sqrt{2}/2\), and hence \(\sqrt{2} = \omega + \omega^{-1}\).

If the \(q\)th cyclotomic field contains \(\mathbb{Q}[\sqrt{p}]\), the \(4q\)th cyclotomic field contains \(\mathbb{Q}[\sqrt{- p}]\) because it must contain the fourth root of unity \(i\) along with \(\sqrt{p}\).

Now consider any squarefree \(m = p_1 ... p_r\). For each \(p_i\) take the cyclotomic field containing \(\sqrt{p}\). Then take the smallest cyclotomic field \(K\) containing all these fields. Then \(K\) contains \(\mathbb{Q}[\sqrt{m}]\). Set \(d = disc(\mathbb{A}\cap\mathbb{Q}[\sqrt{m}])\). It can be easily verified that the desired \(K\) is in fact the \(d\)th cyclotomic field.

Kronecker and Weber proved that every abelian extension of \(\mathbb{Q}\) (normal with abelian Galois group) is contained in a cyclotomic field. Hilbert and others studied abelian extensions of general number fields, and their results are known as class field theory.

Ben Lynn blynn@cs.stanford.edu 💡