Question

If $\mathit{K}$is an extension field of $\mathit{Q}$such that $\left[K:Q\right]\mathbf{=}\mathbf{2}$prove that $\mathit{K}\mathbf{=}\mathit{Q}\sqrt{\mathbf{d}}$ for some square free integer $\mathit{d}$.

Short answer

$K=Q\sqrt{d}$

Step-by-step solution

Definition of extension field.

Let $K$is an extension field of $F$. Then $K$is said to be algebraic extension of $F$is every element of $K$is algebraic over $F$.

Showing that K=Qd .

Let $\alpha \epsilon K-Q$

Since $1,\alpha ,{\alpha }^2$are linearly dependent over $Q$.This implies

$$\begin{gathered} a{\alpha }^2+b\alpha +c1=0 \\ a{\alpha }^2+b\alpha +c=0 \end{gathered}$$

Here $a,b,c\in Q$and since $1,\alpha ,{\alpha }^2$are linearly dependent over $Q$. Therefore not all of $a,b,c\in Q$ is zero.

By taking product with nonzero integer assume that$a,b,c\in Q$.

Now if a = 0 then

$b\alpha +c=0$

Since $a,b,c\in Z$are linearly dependent and as. So either $a=0$and $b$orc is nonzero.

This is not possible. Therefore$a\ne 0$

Hence$Q\sqrt{d}=Q\left(\sqrt{b^2-4ac}\right)$

Since $1,\alpha$are linearly dependent over $Q$.so,

$\left[Q\sqrt{d}:Q\right]=2$

This implies

$K=Q\sqrt{d}$