Let \(F\) be a field of characteristic \(\neq 2\). Let \(D_1\) and \(D_2\) be elements of \(F\), neither of which is a square of \(F\). Prove that \(F(\sqrt{D_1}, \sqrt{D_2})\) is of degree 4 over \(F\) if \(D_1D_2\) is not a square in \(F\) and is of degree 2 over \(F\) otherwise (when \(F(\sqrt{D_1}, \sqrt{D_2})\) is of degree 4 over \(F\) it is called a biquadratic extension of \(F\)).
Remark: note that the condition \(\text{chr}(F) \neq 2\) is required, to exclude e.g. \(\mathbb{F}_2\), all of whose elements are squares.
Let \(F' = F(\sqrt{D_1D_2})\) and \(F'' = F(\sqrt{D_1},\sqrt{D_2})\). Consider the chain \(F \subseteq_{n} F' \subseteq_{m} F''\), where \(n\) and \(m\) denote the respective degrees. Note the second inclusion is valid since each element of \(F'\) can be written in the form \(a+b\sqrt{D_1D_2}\) for \(a,b \in F\). Furthermore, \(m > 1\) because \(\sqrt{D_1}\) cannot be expressed in that form (otherwise \(\sqrt{D_2}^{-1} \in F\) which is a contradiction).
\(D_1D_2\) not being a square in \(F\) means \(n = 2\), forcing \(nm = 4\) as \(nm \le 4\) and \(m > 1\). On the other hand, \(nm \neq 2\) forces \(nm = 4\) (as \(nm = 1\) is not a valid possibility), so \(n = 2\) and \(D_1D_2\) is not a square in \(F\).
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