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}.

Post a comment or leave a trackback: Trackback URL.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: