Let be a field of characteristic not 2, and let with not square. Prove that for some if and only if is square in . Use this to give a sufficient condition for the extension to be biquadratic over .
Suppose we have for some . Evidently, is a root of and of . Note that the degree of over is 2, and that the degree of over is either 1 or 2. So the degree of over is either 2 or 4.
Suppose the degree of over is 2. Then there exist such that . Comparing coefficients, we have and .
Substituting, we have , which is quadratic in . Letting , we see that the discriminant of is . So is a perfect square trinomial, and in fact . Thus is square in .
Now suppose the degree of over is 4. In this case, is the (unique) monic minimal polynomial of , and we have square in .
Conversely, suppose is square in . Letting and , evidently we have , and thus .
Note that by a very similar argument, we can show that this theorem holds if we replace by .
Now consider the case , which has characteristic 0, and let . Suppose is irreducible over and is square in . By the above result, for some . So . Note that the first field in this chain has degree 4, and that the last has degree at most 4. So in fact is a biquadratic extension of .