TITLE: On the ring of integers of biquadratic number fields
QUESTION [25 upvotes]: 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?

REPLY [10 votes]: 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.)