Field Extension must Adjoin a Square Root

Suppose that \( E \) is a field extension of \( \mathbb{ Q } \) of degree 2. Prove that \( E = \mathbb{ Q } \left( \sqrt{ a } \right) \) for some \( a \in \mathbb{ Q } \) such that \( \sqrt{ a } \notin \mathbb{ Q } \)

Since \( E \) is a vector space of degree 2 over \( \mathbb{ Q } \) then it has some basis \( \left( 1, k \right) \). It must be that \( k \notin \mathbb{ Q } \) because if \( k \in \mathbb{ Q } \) then \( \left( 1, k \right) \) would no longer be linearly independent. Moreover we can write \( k ^ 2 \in E \) as a linear combination of the basis vectors so that we have some \( c, d \in \mathbb{ Q } \) such that \( k ^ 2 = -c - dk \) which is to say that \( k ^ 2 + dk + c = 0 \) which means that \( k \) is a solution to the polynomial \( x ^ 2 + dx + c = 0 \).

The quadratic formula determines that \( k \) is of the form \( \frac{- \left( -d \right) \pm \sqrt{ d ^ 2 - 4 \cdot 1 \cdot c } }{2} \), now if \( d ^ 2 - 4c \) is a square, then the entire expression yields a rational, but \( k \) is not rational, so we know that \( d ^ 2 - 4c \) is not a square. Let \( m := d ^ 2 - 4c \in \mathbb{ Q } \) so that \( k = \frac{d}{2} \pm \frac{\sqrt{ m } }{2} \iff \sqrt{ m } = \pm \left( 2k - d \right) \). Recall that \( k \notin \mathbb{ Q } \) so that \( 2k - d \notin \mathbb{ Q } \) so that \( \sqrt{ m } \notin \mathbb{ Q } \)

For any number \( x \notin \mathbb{ Q } \) we know that \( \mathbb{ Q } \left( x \right) = \mathbb{ Q } \left( a x + b \right) \) where \( a, b \in \mathbb{ Q } \) then we know that \( \mathbb{ Q } \left( k \right) = \mathbb{ Q } \left( \sqrt{ m } \right) \) so we've proven the statement true.