An element $\theta$ in a Galois extension $L$ of $\mathbb{Q}$ is said to give an integral normal basis if the ring of algebraic integers in $L$ is $\sum \mathbb{Z}\sigma(\theta)$ with $\sigma$ running through Gal$(L / \mathbb{Q})$.

  1. Let $n$ be a square-free positive integer and $\theta$ a primitive $n$ th root of unity. Show $\theta$ gives an integral normal basis for $\mathbb{Q}(\theta)$.

    The ring of algebraic integers of $\mathbb{Q}(\theta)$ is $\mathbb{Z}(\theta)$ and $\sum \mathbb{Z}\sigma(\theta)$ is effectively $\sum_{i=0}^{n-1} \mathbb{Z}(\theta^i) $.

  2. Let $\mathbb{Q} \subseteq K \subseteq \mathbb{Q}(\theta)$. Show $T_{\mathbb{Q}(\theta)/K}(\theta)$ gives an integral normal basis for $K$.

    Here $T_{\mathbb{Q}(\theta)/K}(\theta)= \sum \sigma(\theta)$, then $\sum \mathbb{Z}\sigma(\sum \sigma(\theta) )= \sum \mathbb{Z}\sum \sigma^2(\theta ) $...

    How get $O_{K}$?

  3. Let $K$ be an abelian extension of $\mathbb{Q}$ which is tamely ramified at every prime; that is for each prime $p$, $p$ does not divide the ramification number $e(P/p)$ of $p$ in $K$. Show that $K$ has an integral normal basis.

    NO IDEA!

Any help is appreciated!


1 Answer 1


I'm rusty, so take everything that follows with a grain of salt.

  1. I'll write $G_n$ for $\mathrm{Gal}(\mathbf Q(\zeta_n)/\mathbf Q)$. You seem to think that $G_n$ is $\mathbf Z/n\mathbf Z$, where $\zeta_n$ is a primitive $n$th root of unity; it's not, it's actually $(\mathbf Z/n\mathbf Z)^\times$. You can use the fact that $\mathbf Q(\zeta_{2n})=\mathbf Q(\zeta_n)$ when $n$ is odd to reduce to the case where $n$ is square-free and odd.

    I think you can make something work in a way similar to the proof that the ring of integers of $\mathbf Q(\zeta_n)$ is $\mathbf Z[\zeta_n]$: start with the case where $n$ is the power of a prime. Here, $n$ is square-free, so let's look at the case where $n$ is an odd prime $p$. The conjugates of $\zeta_p$ are $\zeta_p,\zeta_p^2,\ldots,\zeta_p^{p-1}$, and they clearly constitute a $\mathbf Z$-basis of $\mathbf Z[\zeta_p]$, so what you said works.

    Now if $n$ is odd and square-free, write $n=p_1\cdots p_k$; since the discriminant of $\mathbf Q(\zeta_{p_i})$ is a power of $p_i$, the discriminants are pairwise coprime, and so $$ \mathbf Q(\zeta_n) = \mathbf Q(\zeta_{p_1})\cdots\mathbf Q(\zeta_{p_k}),\\ \mathbf Z[\zeta_n] = \mathbf Z[\zeta_{p_1}]\otimes\cdots\otimes\mathbf Z[\zeta_{p_k}],\\ G_n = G_{p_1}\times\cdots\times G_{p_k}. $$

    So by induction, all you need is the fact that if $r$ and $s$ are coprime, and $\zeta_r$ and $\zeta_s$ generate an integral normal basis for $\mathbf Z[\zeta_r]$ and $\mathbf Z[\zeta_s]$ respectively, then $\zeta_r\zeta_s$ (which is a primitive $rs$-th root of unity) generates an integral normal basis for $\mathbf Z[\zeta_r\zeta_s]=\mathbf Z[\zeta_r]\otimes\mathbf Z[\zeta_s]$. Since you have an isomorphism $G_r\times G_s\simeq G_{rs}$, you can conclude as follows: $$ \sum_{\sigma\in G_r}\mathbf Z\sigma(\zeta_r) \otimes \sum_{\sigma'\in G_s}\mathbf Z\sigma'(\zeta_s) = \sum_{\tau\in G_{rs}} \mathbf Z\tau(\zeta_r\zeta_s). $$

  2. I'll write $G=\mathrm{Gal}(\mathbf Q(\theta)/\mathbf Q)$ and $H=\mathrm{Gal}(\mathbf Q(\theta)/K)$. Take an element $x\in\mathcal O_K$. Then $x\in\mathcal O_{\mathbf Q(\theta)}$, so there exist integers $m_\sigma$ such that $x=\sum_{\sigma\in G}m_\sigma\sigma(\theta)$. Now since $x$ is in $\mathcal O_K$, you have $\tau(x)=x$ for all $\tau\in H$; in other words, \begin{align} \sum_{\sigma\in G}m_\sigma\sigma(\theta) &= \sum_{\sigma\in G}m_\sigma\tau\sigma(\theta) \\ &= \sum_{\sigma\in G}m_{\tau^{-1}\sigma}\sigma(\theta) \end{align} by the change of variables $\sigma\mapsto\tau^{-1}\sigma$. Since $\theta$ generates $\mathbf Q(\theta)$, you have $$ \sum_{\sigma\in G}m_\sigma\sigma = \sum_{\sigma\in G}m_{\tau^{-1}\sigma}\sigma $$ and Dedekind's lemma on the independence of automorphisms shows that $m_\sigma=m_{\tau^{-1}\sigma}$ for all $\sigma\in G$ and $\tau\in H$.

    Now the idea is to group the terms in $x=\sum m_\sigma\sigma(\tau)$ into equivalence classes modulo $H$. So let $\Gamma\subset G$ be a lift of $G/H$; then \begin{align} x &= \sum_{\sigma\in G}m_\sigma\sigma(\theta) \\ &= \sum_{\sigma\in\Gamma}\sum_{\tau\in H}m_{\sigma\tau}(\sigma\tau)(\theta) \\ &= \sum_{\sigma\in\Gamma}\sigma\left(\sum_{\tau\in H}m_{\sigma\tau}\tau(\theta)\right). \end{align} I'm going to assume that $K/\mathbf Q$ is Galois, otherwise I don't see how to conclude. If it is Galois, then $H$ is normal in $G$, so any $\sigma\tau$ can be written as a $\tau'^{-1}\sigma$, hence $m_{\sigma\tau} = m_{\tau'^{-1}\sigma} = m_\sigma$, so \begin{align} x &= \sum_{\sigma\in\Gamma}m_\sigma\sigma\left(\sum_{\tau\in H}\tau(\theta)\right) \\ &= \sum_{\sigma\in\Gamma} m_\sigma\sigma\left(\mathrm{Tr}_{\mathbf Q(\theta)/K}(\theta)\right). \end{align} Notice that $\mathrm{Gal}(K/\mathbf Q) = G/H$ and you're done.

  3. The problem apparently wants you to use the Kronecker-Weber theorem: $K$ is abelian, so it's a subfield of a cyclotomic field $\mathbf Q(\zeta_n)$. Given 1. and 2., all you have to show is that $n$ is square-free.


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