# Kronecker-Weber does not apply

Let $K=\mathbb{Q}(\sqrt{D})\neq \mathbb{Q}$. Show that $K$ has an abelian extension that is not contained in $K(\theta)$ for any root of unity $\theta$.

Hint: Find $u \in K$ such that $K(\sqrt{u})$ is not normal over $\mathbb{Q}$. If $\sqrt{u} \in K(\theta)$ for some root of unity $\theta$ then $K(\theta)$ is abelian over $\mathbb{Q}$ and every subfield is normal over $\mathbb{Q}$.

Suppose that $K$ has a abelian extension $L$ such that $L \subset K(\theta)$ for some root unity $\theta$.

If I put $u=\sqrt{D}$, then $L=K(\sqrt[4]{D})=\mathbb{Q}(\sqrt{D},\sqrt[4]{D} )$ is not normal over $\mathbb{Q}$. If $\sqrt[4]{D} \in K(\theta)$, then $L=K(\sqrt[4]{D}) \leqslant K(\theta)$, but $K(\theta)$ is normal over $\mathbb{Q}$ and any subfield of $K(\theta)$ is normal. This contradicts our assumption.

This is correct?

Thanks!

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Does this work when $D=-1$? –  Gerry Myerson Jul 14 '13 at 6:54
@GerryMyerson I think you mean that, when $D=-1$, then $L=\mathbb Q(\text{i},\sqrt{\text{i}})$, which is normal, since $L=\mathbb Q(\text{i},\sqrt2)$. Is this correct? --A confused student. –  awllower Jul 14 '13 at 7:07

Set $u = a + b \sqrt{D}$. We will show that we can choose $a,b \in \mathbb{Q}$ such that $L = K(\sqrt{u})$ is not normal over $\mathbb{Q}$.

Obviously we should not choose $u$ to be a square in $K$, so we can assume that [$L$:$\mathbb{Q}$]=4.

The conjugates of $\sqrt{u}$ are $\pm \sqrt{u}$ and $\pm \sqrt{u'}$, where $u' = a - b \sqrt{D}$.

If $L$ is normal, then $L$ contains $\sqrt{uu'}= \sqrt{a^2 - D b^2}$. The extension $K_2 = \mathbb{Q}(\sqrt{a^2-Db^2})$ is in general different from $K = \mathbb{Q}(\sqrt{D})$ (they would only be equal for some special choices of $a$, $b$). Hence $L$ contains at least 2 distinct quadratic subextensions, and so the Galois group of $L$ is $C_2 \times C_2$, which means that $L$ contains exactly 3 quadratic subextensions: $K_1 = K$, $K_2 = \mathbb{Q}(\sqrt{a^2-Db^2})$, and a third one which we will call $K_3$.

Next, observe that the norms of $\sqrt{u}$ down to each of the $K_i$ are obtained by multiplying $\sqrt{u}$ by each of its other 3 conjugates (corresponding to the 3 non-identity elements of the Galois group). These norms are:

$$\sqrt{u} \cdot -\sqrt{u} = -u = -(a+b \sqrt{D})$$ $$\sqrt{u} \cdot \sqrt{u'} = \sqrt{a^2 - Db^2}$$ $$\sqrt{u} \cdot -\sqrt{u'} = -\sqrt{a^2 - Db^2}$$

The first norm lies in $K_1$, the last 2 norms both lie in $K_2$, and none lie in $K_3$. This is a contradiction.

(Essentially, what we have proven here is: If $\mathbb{Q}(\sqrt{a+b\sqrt{D}})$ is a normal extension of $\mathbb{Q}$ of degree 4, then its Galois group must be $C_4$.)

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Why $\sqrt{uu'} \in L$? I'm a little lost! The other things I understand. Thanks! –  P. M. O. Jul 14 '13 at 20:37
Because if a normal extension of $\mathbb{Q}$ contains a number, then it also contains all of its conjugates. So $\sqrt{u} \in L$ implies $\sqrt{u'} \in L$. –  Ted Jul 14 '13 at 23:44