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 (http://planetmath.org/ExtensionField) 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 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 $\mathbb{Z}$ (http://planetmath.org/PowerBasisOverMathbbZ). Note that $3$ splits completely in both $\mathbb{Q}(\sqrt{m})$ and $\mathbb{Q}(\sqrt{n})$ and thus in $L$. Therefore, $3\mathcal{O}_{L}=P_{1}P_{2}P_{3}P_{4}$ for distinct prime ideals (http://planetmath.org/PrimeIdeal) $P_{1}$, $P_{2}$, $P_{3}$, 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   .