Let \(F\) be a field of characteristic not equal to 2.
Prove that any extension \(K\) of \(F\) of degree 2 is of the form \(F(\sqrt D)\) where \(D\in F\) is not a square in \(F\) and, conversely, that each such extension has degree 2 over \(F\).
Let \(D_1,D_2\in F\) neither of which is a square in \(F\). Prove that \([F(\sqrt{D_1},\sqrt{D_2}):F]=4\) if \(D_1D_2\) is not a square in \(F\) and is of degree 2 otherwise.
<1>4. One has \[ \sqrt{D_2}\in F(\sqrt{D_1}) \quad\Longleftrightarrow\quad D_1D_2\text{ is a square in }F. \]
<1>5. Therefore \[ [F(\sqrt{D_1},\sqrt{D_2}):F] =\begin{cases} 4,&D_1D_2\text{ is not a square in }F,\\ 2,&D_1D_2\text{ is a square in }F. \end{cases} \]
Solution
1. Let \(K/F\) be an extension of degree \(2\). Then \(K=F(\sqrt D)\) for some nonsquare \(D\in F\).
Choose \(\alpha\in K\setminus F\). Since \([K:F]=2\), its minimal polynomial over \(F\) has degree \(2\), say \[ m_\alpha(x)=x^2+bx+c. \] Because \(\operatorname{char}F\ne2\), put \[ \beta=\alpha+\frac b2. \] Then \[ \beta^2=\alpha^2+b\alpha+\frac{b^2}{4}=\frac{b^2}{4}-c=:D\in F. \] Also \(F(\beta)=F(\alpha)=K\). If \(D\) were a square in \(F\), then \(\beta\in F\), hence \(\alpha\in F\), contradiction. Thus \(D\) is not a square and \(K=F(\sqrt D)\).
2. Conversely, if \(D\in F\) is not a square, then \([F(\sqrt D):F]=2\).
The polynomial \(x^2-D\) has no root in \(F\), so as a quadratic it is irreducible. Hence it is the minimal polynomial of \(\sqrt D\) and the extension degree is \(2\).
3. Let \(D_1,D_2\) be nonsquares. Then\[ [F(\sqrt{D_1},\sqrt{D_2}):F] =[F(\sqrt{D_1},\sqrt{D_2}):F(\sqrt{D_1})]\cdot2. \]Thus the degree is \(2\) or \(4\) according as \(\sqrt{D_2}\) does or does not lie in \(F(\sqrt{D_1})\).
Suppose first that \(\sqrt{D_2}=a+b\sqrt{D_1}\) with \(a,b\in F\). Squaring gives \[ D_2=a^2+b^2D_1+2ab\sqrt{D_1}. \] Since \(1,\sqrt{D_1}\) are linearly independent over \(F\), \(2ab=0\). Because \(\operatorname{char}F\ne2\), either \(a=0\) or \(b=0\). The case \(b=0\) would make \(D_2=a^2\) a square, impossible. Hence \(a=0\), so \(D_2=b^2D_1\) and therefore \[ D_1D_2=(bD_1)^2 \] is a square in \(F\).
Conversely, suppose \(D_1D_2=c^2\) for some \(c\in F\). Since \(D_1\ne0\), \[ D_2=\frac{c^2}{D_1^2}D_1, \] so \[ \sqrt{D_2}=\frac{c}{D_1}\sqrt{D_1} \] for a suitable choice of square root inside an algebraic closure. Hence \(F(\sqrt{D_2})\subseteq F(\sqrt{D_1})\).
Combine 3 and 4.