Biquadratic field extensions

Let F be a field of characteristic not 2, and let D_1 and D_2 be non-squares in F. Prove that F(\sqrt{D_1}, \sqrt{D_2}) has degree 4 over F if D_1D_1 is nonsquare in F, and degree 2 otherwise. (If the degree is 4, F(\sqrt{D_1},\sqrt{D_2}) is called a biquadratic extension of F.

Suppose D_1D_2 is nonsquare in F. We claim that p(x) = x^2 - D_2 is irreducible over F(\sqrt{D_1}). To see this, suppose to the contrary that (a+b\sqrt{D_1})^2 = D_2. Comparing coefficients, we have a^2 + b^2D_1 - D_2 = 0 and 2ab = 0. If b = 0, then D_2 = a^2, a contradiction. If a = 0, then D_1 = D_2/b^2. But then D_1D_2 = (D_2/b)^2 is square, a contradiction. So F(\sqrt{D_1},\sqrt{D_2}) has degree 2 over F(\sqrt{D_1}), and thus degree 4 over F.

Now suppose that D_1D_2 = \eta^2 is square in F. then (x - \eta/\sqrt{D_1})(x + \eta/\sqrt{D_1}) = x^2 - D_2 is reducible over F(\sqrt{D_1}), and we have F(\sqrt{D_1},\sqrt{D_2}) = F(\sqrt{D_1}) of degree 2 over F.

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