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Number theory-question again.

Let $K$ be the biquadratic field $K=\mathbb{Q}[\sqrt{m},\sqrt{n}]$ where $m,n$ are distinct squarefree integers. Let $\mathcal{O}_K$ denote the ring of integers of $K$. If $\alpha \in K$, prove that $\alpha \in \mathcal{O}_K$ if and only if the trace and the norm of $\alpha$ over $\mathbb{Q}(\sqrt{m})$ are algebraic integers.

I've been stuck on this one for quite some time now. The left-right direction is easy, but I'm stuck on the other direction. My "intuition" tells me that I should somehow use that an element of a quadratic extension is integral if and only if its norm and its trace are.

I'd like subtle hints in the right direction. Thanks.

Edit: Some minor corrections for clarity.

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After you introduce $\alpha$, you have $\alpha\in\mathcal{O}_k$ (subindex is lower case k). Is that supposed to be $K$, the ring of integers of $K$? Also: either you never use $k$, or you never use $\mathcal{O}_K$ (depending on whether there was a typo on what you are trying to prove or not). –  Arturo Magidin Feb 13 '11 at 5:56
    
I fixed the title, which had the same typo. –  Arturo Magidin Feb 13 '11 at 6:04
    
Was a typo. I ment $\mathcal{O}_K$. –  Fredrik Meyer Feb 13 '11 at 6:05
    
Your intuition is right; just apply it to $\mathbb{Q}(\sqrt{m})$ instead of $\mathbb{Q}$. –  Arturo Magidin Feb 13 '11 at 6:38

1 Answer 1

up vote 1 down vote accepted

Two hints:

  • If $R\subseteq S\subseteq T$ are domains, and $S$ is integral over $R$, and $T$ is integral over $S$, then $T$ is integral over $R$.

  • What is $[\mathbb{Q}(\sqrt{m})(\alpha):\mathbb{Q}(\sqrt{m})]$ ?

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I'm not completely sure if I follow, but $[\mathbb{Q}(\sqrt{m})(\alpha):\mathbb{Q}(\sqrt{m})]$ is either 1 or 2. If it is 1, then $\alpha \in \mathbb{Q}(\sqrt{m})$, so by transitivty of norms/traces we can conclude $\alpha \in \mathcal{O}_K$ (since $\alpha$ lived in a quadratic extension). Similarly, if the index is 2, then the norm and the trace are just the coefficients of the minimal polynomial over $\mathbb{Q}(\sqrt{m})$, and so $\alpha$ is integral over $\mathbb{Q}(\sqrt{m})$. By transitivity of integrality (?) we can conclude $\alpha$ is integral. –  Fredrik Meyer Feb 13 '11 at 22:16
    
Does the above reasoning hold? The question mark is part I'm not sure of. –  Fredrik Meyer Feb 13 '11 at 22:16
    
@Fredrik: Yes, it's okay. The question mark is precisely my first hint: if $\alpha$ is integral over $\mathcal{O}_{\mathbb{Q}(\sqrt{m})}$, and $\mathcal{O}_{\mathbb{Q}(\sqrt{m})}$ is integral over $\mathbb{Z}$, then $\alpha$ is integral over $\mathbb{Q}$. Alternatively: take the minimal polynomial over $\mathbb{Q}(\sqrt{m})$, and multiply by its conjugate; the result will be a polynomial over $\mathbb{Q}$, monic, satisfied by $\alpha$, and the coefficients are algebraic integers, hence integers. So $\alpha$ satisfies a monic polynomial over $\mathbb{Z}$. –  Arturo Magidin Feb 13 '11 at 22:32

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