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Let $m,n$ be coprime square-free integers, $m\equiv 3(\mathrm{mod}4)$ and $n\equiv 2(\mathrm{mod}4)$. For the biquadratic extension $L=\mathbb{Q}(\alpha ,\beta )$, with ${\alpha }^2=m$ and ${\beta }^2=n$, an integral basis is $1,\alpha ,\beta ,(\beta +\alpha \beta )/2$. To prove spanning, write an algebraic integer as $(a+b\alpha +c\beta +d\alpha \beta )/2$. The field traces to the three quadratic subfields and the ring of integers of a quadratic field formula force $a,b,c,d\in \mathbb{Z}$. The field norm to $\mathbb{Q}(\alpha )$ has coefficients $(a^2+m{b}^2-n{c}^2-mn{d}^2)/4$ and $(ab-ncd)/2$. Their integrality gives $ab\equiv 0(\mathrm{mod}2)$ and $a^2-b^2-2(c^2+d^2)\equiv 0(\mathrm{mod}4)$. Thus $a,b$ are even and $c\equiv d(\mathrm{mod}2)$, exactly the asserted integer span. Conversely, the half-sum $w$ has $w^2=n(1+m)/4+(n/2)\alpha \in \mathbb{Z}[\alpha ]$, so integral dependence is transitive proves it is an algebraic integer. Rational linear independence is immediate from its nonzero $\alpha \beta$ coefficient.

Put $\alpha =\sqrt{m_1}$, $\beta =\sqrt{m_2}$ and $F=\mathbb{Q}(\alpha )$. The hypotheses make $L/\mathbb{Q}$ a biquadratic extension. To check the degree of a field extension is four, if $\beta =x+y\alpha$ with rational $x,y$, squaring gives $xy=0$. The case $y=0$ would make $m_2$ a rational square; the case $x=0$ would make $m_2/m_1$ a rational square. Neither can occur for the specified distinct coprime square-free integers. Hence $[L:F]=2$.

The nonidentity field automorphism of $L/F$ fixes $\alpha$ and sends $\beta$ to $-\beta$. Thus, for $\gamma =(r+s\alpha )+(t+u\alpha )\beta$, the field norm and field trace areIf $\gamma$ is an algebraic integer, so is its conjugate, and their sum and product are algebraic integers. Conversely, put $T={\mathrm{Tr}}_{L/F}(\gamma )$ and $N=N_{L/F}(\gamma )$. The identityshows that $\gamma$ is integral over $\mathbb{Z}[T,N]$. If $T,N$ are algebraic integers, the ring generated by algebraic integers $\mathbb{Z}[T,N]$ is integral over $\mathbb{Z}$. Since integral dependence is transitive, $\gamma$ is an algebraic integer. This proves the trace and norm integrality criterion for a quadratic extension, including elements already in $F$.

For the biquadratic extension $L=K(\sqrt{a},\sqrt{b})$, the assumptions say that the square classes of $a$ and $b$ are independent. In particular $\sqrt{b}\notin K(\sqrt{a})$: writing $\sqrt{b}=u+v\sqrt{a}$ and comparing the coefficient of $\sqrt{a}$ after squaring would force either $b$ or $ab$ to be a square in $K$. Hence

The field is the splitting field of $(X^2-a)(X^2-b)$ and the characteristic is not two, so it is Galois. Its four automorphisms independently choose the signs of $\sqrt{a}$ and $\sqrt{b}$, givingBy the Galois correspondence, the intermediate-field lattice isEach of the three middle fields has degree two over $K$, and distinct middle fields intersect in $K$.

The Cyclotomic character of a finite cyclotomic extension and the Chinese remainder theorem give the displayed Galois group. It has three order-two subgroups and three order-four subgroups; the latter are kernels of its three nonzero group homomorphisms to $C_2$. By the Fundamental theorem of Galois theory, its proper intermediate fields strictly above $\mathbb{Q}$ are the three quadratic fieldsand the three quartic fieldsThe first quartic has cyclic Galois group of order four and unique quadratic subfield $\mathbb{Q}(\sqrt{5})$; the second is a biquadratic extension, so they differ. The last is real whereas the others are not, by the maximal real subfield of a cyclotomic field formula. Also $\sqrt{-3}+\sqrt{5}$ generates the second quartic: if $s=\sqrt{-3}+\sqrt{5}$, then $\sqrt{-3}=(s-8/s)/2$ and $\sqrt{5}=(s+8/s)/2$.