## A class of biquadratic extensions

Let $F$ be a field of characteristic not 2, and let $a,b \in F$ with $b$ not square. Prove that $\sqrt{a + \sqrt{b}} = \sqrt{m} + \sqrt{n}$ for some $m,n \in F$ if and only if $a^2-b$ is square in $F$. Use this to give a sufficient condition for the extension $\mathbb{Q}(\sqrt{a+\sqrt{b}})$ to be biquadratic over $\mathbb{Q}$.

Suppose we have $\sqrt{a + \sqrt{b}} = \sqrt{m} + \sqrt{n} = \zeta$ for some $m,n \in F$. Evidently, $\zeta$ is a root of $p(x) = x^4 - 2ax^2 + (a^2-b)$ and of $q(x) = x^4 - 2(m+n)x^2 + (m-n)^2$. Note that the degree of $F(\sqrt{b})$ over $F$ is 2, and that the degree of $F(\sqrt{a+\sqrt{b}})$ over $F(\sqrt{b})$ is either 1 or 2. So the degree of $F(\sqrt{a+\sqrt{b}})$ over $F$ is either 2 or 4.

Suppose the degree of $\zeta$ over $F$ is 2. Then there exist $r,s \in F$ such that $(r+s\sqrt{b})^2 = a+\sqrt{b}$. Comparing coefficients, we have $r^2+bs^2 = a$ and $2rs = 1$.

Substituting, we have $a^2-b = s^4b^2 + (rs-1)b + r^4$, which is quadratic in $b$. Letting $p(x) = s^4x^2 + (rs-1)x + r^4$, we see that the discriminant of $p$ is $(rs-1)^2 - 4s^4r^4 = 0$. So $p(x)$ is a perfect square trinomial, and in fact $p(b) = (s^2b - \dfrac{1-rs}{2s^2})^2$. Thus $a^2-b$ is square in $F$.

Now suppose the degree of $\zeta$ over $F$ is 4. In this case, $p(x) = q(x)$ is the (unique) monic minimal polynomial of $\zeta$, and we have $a^2-b = (m-n)^2$ square in $F$.

Conversely, suppose $a^2-b = t^2$ is square in $F$. Letting $m = (a+t)/2$ and $n = (a-t)/2$, evidently we have $(\sqrt{m} + \sqrt{n})^2 = a+\sqrt{b}$, and thus $\sqrt{a+\sqrt{b}} = \sqrt{m} + \sqrt{n}$.

Note that by a very similar argument, we can show that this theorem holds if we replace $\sqrt{a+\sqrt{b}} = \sqrt{m} + \sqrt{n}$ by $\sqrt{a - \sqrt{b}} = \sqrt{m} - \sqrt{n}$.

Now consider the case $F = \mathbb{Q}$, which has characteristic 0, and let $a,b \in \mathbb{Q}$. Suppose $p(x) = x^4 - 2ax^2 + (a^2-b)$ is irreducible over $\mathbb{Q}$ and $a^2-b$ is square in $\mathbb{Q}$. By the above result, $\sqrt{a+\sqrt{b}} = \sqrt{m} + \sqrt{n}$ for some $m,n \in \mathbb{Q}$. So $\mathbb{Q}(\sqrt{a+\sqrt{b}}) \subseteq \mathbb{Q}(\sqrt{m} + \sqrt{n}) \subseteq \mathbb{Q}(\sqrt{m},\sqrt{n})$. Note that the first field in this chain has degree 4, and that the last has degree at most 4. So in fact $\mathbb{Q}(\sqrt{a+\sqrt{b}}) = \mathbb{Q}(\sqrt{m},\sqrt{n})$ is a biquadratic extension of $\mathbb{Q}$.

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