This is Daniel A. Marcus, Number Fields, Exercise 2.29

If anyone can help with this problem, I'd greatly appreciate it.

Let $K$ be the biquadratic field $\mathbb Q[\sqrt{m}, \sqrt{n}] = \{a + b\sqrt{m} + c\sqrt{n} + d\sqrt{mn}: a,b,c,d \in \mathbb Q\}$, where $m$ and $n$ are distinct squarefree integers. Suppose $m$ and $n$ are relatively prime. Find an integral basis and the discriminant of $\mathcal{O}_K$ in each of the cases:

(a) $m,n \equiv 1 \pmod 4$.

(b) $m \equiv 1 \pmod 4, n \not\equiv 1 \pmod 4$.

So, I understand how to do this problem with $K$ being a quadratic field $\mathbb Q[\sqrt{m}]$. I would assume $m \equiv 1 \pmod 4$ so that $\mathcal{O}_K$ has integral basis $\{1, \frac{(1+\sqrt{m})}{2} \}$. But how would I go about it with a biquadratic field?


Here's how we go about doing (b). First, a lemma:

Lemma 1: For $\alpha \in K$, we have that $\text{Tr}_{K/\Bbb{Q}(\alpha)}(\alpha)$ and $N_{K/\Bbb{Q}(\alpha)}(\alpha)$ being algebraic integers iff $\alpha$ is integral over $\Bbb{Z}$.

Proof: Suppose that $\alpha \in \mathcal{O}_K$. Then $\alpha$ is certainly integral over $\mathcal{O}_{\Bbb{Q}(\sqrt{m})}$ and thus its minimal $m_\alpha(t)$ polynomial has coefficients in $\mathcal{O}_{\Bbb{Q}(\sqrt{m})}$. Since $\text{Tr}_{K/\Bbb{Q}(\sqrt{m})}(\alpha) = (-1) \times \text{(coefficient of $t$})$ and $N_{K/\Bbb{Q}(\sqrt{m})}(\alpha)$ is the constant term in $m_\alpha(t)$, these are both in $\mathcal{O}_{\Bbb{Q}(\sqrt{m})}$ and thus are algebraic integers. Conversely if $T = \text{Tr}_{K/\Bbb{Q}(\sqrt{m})}(\alpha)$ and $N = N_{K/\Bbb{Q}(\sqrt{m})}(\alpha)$ are algebraic integers, $\Bbb{Z}[N,T]$ is a finitely -generated $\Bbb{Z}$ - module. Also, $\alpha$ is integral over $\mathcal{O}_{\Bbb{Q}(\sqrt{m})}$ and so $\Bbb{Z}[N,T][\alpha]$ is a finitely - generated $\Bbb{Z}[N,T]$ - module. Then $\Bbb{Z}[N,T][\alpha]$ is a finitely - generated $\Bbb{Z}$ - module and so by Proposition 5.1(c) of Atiyah - Macdonald, $\alpha$ is integral over $\Bbb{Z}$.

Proposition 1: Suppose $m \equiv 1\mod{4}$, $n \equiv k \equiv 2$ or $3 \mod{4}$. Then an integral basis for $\mathcal{O}_K$ is

$$\left\{ 1, \frac{1 + \sqrt{m}}{2},\sqrt{n},\frac{\sqrt{n} + \sqrt{k}}{2} \right\}$$

where $k = mn/(m,n)^2$.

Proof: Write down the minimal polynomial for $\alpha$ over $\Bbb{Q}(m)$. Then by the lemma we see that if $\alpha \in \mathcal{O}_K$ is of the form

$$\alpha = \frac{a + b\sqrt{m} + c \sqrt{n} + d\sqrt{k}}{2}$$

with $a,b,c,d\in\Bbb{Z}$ and $a \equiv b\mod{2}$, $c\equiv d\mod{2}$. We can rewrite $\alpha$ as

$$\begin{eqnarray*} \alpha &=& \frac{a +b\sqrt{m} - b + b + c\sqrt{n} - d\sqrt{n} +d\sqrt{n} + d\sqrt{k} }{2}\\ &=& \frac{ a-b}{2} + b\left(\frac{1 + \sqrt{m}}{2}\right) +\frac{c-d}{2}\left(\sqrt{n}\right) + d\left(\frac{\sqrt{n} + \sqrt{k}}{2}\right)\end{eqnarray*}$$

Since $\frac{a-b}{2}$ and $\frac{c-d}{2}$ are arbitrary integers this concludes the proof of the proposition. Now that you have an integral basis, you only need to calculate the determinant of a $4 \times 4$ matrix to get the discriminant of $\mathcal{O}_K$.

Note to user: Since this is a homework problem I have left out some details. Among the details you need to fill in are:

  1. How did I get that $\alpha$ must be of the form prescribed above?

  2. How does my calculation in the proposition show that what I claimed is an integral basis for $\mathcal{O}_K$?

You are urged to fill them in and follow my method to do (a).

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fpqc's answer is absolutely fine! But I want to point out another solution which works in this case.

In what follows, $K, L \subseteq \overline{\mathbf Q}$ are linearly disjoint number fields, so we have a canonical isomorphism $K \otimes_{\mathbf Q} L \cong KL$.

Claim: Under these hypotheses, the discriminant $\mathfrak d_{KL}$ divides $\mathfrak d_K^s\mathfrak d_L^r$, where $s=[L:\mathbf Q]$ and $r=[K:\mathbf Q]$. Moreover, if the discriminants $\mathfrak d_K$ and $\mathfrak d_L$ are relatively prime in $\mathbf Z$, then $\mathfrak d_{KL} = \mathfrak d_K^s\mathfrak d_L^r$, and the canonical injection of $\mathbf Z$-modules $\mathcal O_K \otimes_{\mathbf Z} \mathcal O_L \hookrightarrow \mathcal O_{KL}$ is an isomorphism (in particular, bases for $\mathcal O_K$ and $\mathcal O_L$ give a basis of $\mathcal O_{KL}$ in the obvious way).

Proof: (With A. Fiori) Let $M$ be the image of $\mathcal O_K \otimes_{\mathbf Z} \mathcal O_L$ in $\mathcal O_{KL}$. Then $M$ is an order of $\mathcal O_{KL}$. If $x=ab \in \mathcal O_{KL}$ is the image of a pure tensor $a \otimes b$, then the trace form on $KL$ acts on $x$ by $T_{KL}(ab) = T_{K}(a) T_{L}(b)$ (to prove this, reduce to the case where $K$ and $L$ are Galois and use the canonical isomorphism $\text{Gal}(KL/\mathbf Q) \simeq \text{Gal}(K/\mathbf Q) \times \text{Gal}(L/\mathbf Q)$ provided by linear disjointness). In other words, $T_{KL}$ acts on $M$ as $T_K \otimes_\mathbf Z T_L$. By picking a basis $\{a_1, \dots, a_r\}$ for $\mathcal O_K$ and $\{b_1, \dots, b_s\}$ for $\mathcal O_L$, it follows that the matrix for $T_{KL}$ on $M$ is the Kronecker product of the matrices for $T_K$ and $T_L$, and thus its determinant is $\mathfrak d_K^s\mathfrak d_L^r$. It follows that $\mathfrak d_{KL}$ divides $\mathfrak d_K^s\mathfrak d_L^r$. By considering the towers $\mathbf Q \subseteq K \subseteq KL$ and $\mathbf Q \subseteq L \subseteq KL$, we know also that $\mathfrak d_{KL}$ is divisible by $\mathfrak d_K^s$ and $\mathfrak d_L^r$, hence the assertion that $\mathfrak d_{KL} = \mathfrak d_K^s\mathfrak d_L^r$ if $\mathfrak d_K$ and $\mathfrak d_L$ are coprime. It follows in this case that $M=\mathcal O_{KL}$.

(This applies in both cases (a) and (b), since the discriminants are $m$ and $n$ in the first case, and $m$ and $4n$ in the second, which are relatively prime because $m$ and $n$ are, and because $m$ is odd.)

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  • $\begingroup$ Dear Bruno, Thank you for your answer. I had committed for the additional bounty intending it to go to BenjaLim. In fact I was surprised when it came up that I had to wait 23 hours. Naively, I never expected another answer. To be honest, your answer is over my head. But that is irrelevant. I feel an ethical responsibility, in that I would never explicitly or implicitly ask anyone to do anything for naught. And even if I could, I would be reluctant to differentiate between such answers. That being the case, I will leave it up to you: I will post another 50 point bounty intended to you if you... $\endgroup$ – user12802 Feb 18 '13 at 3:34
  • $\begingroup$ like. Please do not hesitate to tell me what you think. Meantime, I have to go to bed and will check first thing in the AM. Hope I don't get any more answers. P.S. Love the cartoon. Best regards, Andrew $\endgroup$ – user12802 Feb 18 '13 at 3:37
  • $\begingroup$ Dear @Andrew : I wrote this answer knowing that you had already awarded the bounty, and I certainly don't hold you responsible for it. You did more than your part already by offering the bounty and bringing this question to my attention - so thank you! Best regards, $\endgroup$ – Bruno Joyal Feb 18 '13 at 3:58
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    $\begingroup$ @Bruno +1 For your solution, a real thermonuclear weapon! Do continue to post answers on math.se as I can't express how much I learned from this answer! $\endgroup$ – user38268 Feb 18 '13 at 13:09
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    $\begingroup$ @AlexYoucis You should post what you have in mind! :) $\endgroup$ – Bruno Joyal Nov 4 '13 at 8:33

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