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.