A biquadratic field (or biquadratic number field) is a biquadratic extension of $\mathbb{Q}$ . To discuss these fields more easily, the set $S$ will be defined to be the set of all squarefree integers not equal to $1$ . Thus, any biquadratic field is of the form $\mathbb{Q}(\sqrt{m}, \sqrt{n})$ for distinct elements $m$ and $n$ of $S$ .

Let $\displaystyle k=\frac{mn}{(\gcd(m,n))^2}$ . It can easily be verified that $k \in S$ , $k \neq m$ , and $k \neq n$ . Since $\displaystyle \sqrt{k}=\frac{\sqrt{mn}}{\gcd(m,n)} \in \mathbb{Q}(\sqrt{m}, \sqrt{n})$ , the three distinct quadratic subfields of $\mathbb{Q}(\sqrt{m}, \sqrt{n})$ are $\mathbb{Q}(\sqrt{m})$ , $\mathbb{Q}(\sqrt{n})$ , and $\mathbb{Q}(\sqrt{k})$ . Note that $\mathbb{Q}(\sqrt{k})=\mathbb{Q}(\sqrt{mn})$ .

Of the three cyclotomic fields of degree four over $\mathbb{Q}$ , $\mathbb{Q}(\omega_8)$ and $\mathbb{Q}(\omega_{12})$ are biquadratic fields. The quadratic subfields of $\mathbb{Q}(\omega_8)$ are $\mathbb{Q}(\sqrt{2})$ , $\mathbb{Q}(\sqrt{-1})$ , and $\mathbb{Q}(\sqrt{-2})$ ; the quadratic subfields of $\mathbb{Q}(\omega_{12})$ are $\mathbb{Q}(\sqrt{3})$ , $\mathbb{Q}(\sqrt{-1})$ , and $\mathbb{Q}(\sqrt{-3})$ .

Note that the only rational prime $p$ for which $e(P|p)=4$ is possible in a biquadratic field is $p=2$ . (The notation $e(P|p)$ refers to the ramification index of the prime ideal $P$ over $p$ .) This occurs for biquadratic fields $\mathbb{Q}(\sqrt{m}, \sqrt{n})$ in which exactly two of $m$ , $n$ , and $k$ are equivalent to $2 \operatorname{mod} 4$ and the other is equivalent to $3 \operatorname{mod} 4$ . For example, in $\mathbb{Q}(\omega_8)=\mathbb{Q}(\sqrt{2}, \sqrt{-1})$ , we have that $e(P|2)=4$ .

Certain biquadratic fields provide excellent counterexamples to statements that some people might think to be true. For example, the biquadratic field $K=\mathbb{Q}(\sqrt{2}, \sqrt{-3})$ is useful for demonstrating that a subring of a principal ideal domain need not be a principal ideal domain. It can easily be verified that $\mathcal{O}_K$ (the ring of integers of $K$ ) is a principal ideal domain, but $\mathbb{Z}[\sqrt{-6}]$ , which is a subring of $\mathcal{O}_K$ , is not a principal ideal domain. Also, biquadratic fields of the form $L=\mathbb{Q}(\sqrt{m}, \sqrt{n})$ with $m$ and $n$ distinct elements of $S$ such that $m \equiv 1 \operatorname{mod} 3$ and $n \equiv 1 \operatorname{mod} 3$ are useful for demonstrating that rings of integers need not have power bases over $\mathbb{Z}$ . Note that $3$ splits completely in both $\mathbb{Q}(\sqrt{m})$ and $\mathbb{Q}(\sqrt{n})$ and thus in $L$ . Therefore, $3\mathcal{O}_L=P_1P_2P_3P_4$ for distinct prime ideals $P_1$ , $P_2$ , $P_3$ , and $P_4$ of $\mathcal{O}_L$ . Now suppose $\mathcal{O}_L=\mathbb{Z}[\alpha]$ for some $\alpha \in L$ . Then $L=\mathbb{Q}(\alpha)$ , and the minimal polynomial $f$ for $\alpha$ over $\mathbb{Q}$ has degree $4$ . This yields that $f$ , considered as a polynomial over $\mathbb{F}_3$ , is supposed to factor into four distinct monic polynomials of degree $1$ , which is a contradiction.