# On the ring of integers of a compositum of number fields

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?

• – Watson Jul 23 '16 at 12:24

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).

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.)