Question

Find a basis of $\mathbf{Q}\left(\sqrt{2}+\sqrt{3}\right)\mathbf{}\mathbf{over}\mathbf{}\mathbf{Q}\left(\sqrt{3}\right)$ over .

Short answer

$\left\{1,\sqrt{2}\right\}$is a basis of this.

Step-by-step solution

Definition of basis.

A set of element in vector space is called a basis or a set of basis vectors, if the vectors are linearly independent and every vector in vector space is linear combination of this set.

Step 2:Showing that {1,2} is a basis.

As $Q\subseteq Q\sqrt{3}\subseteq Q\left(\sqrt{2}+\sqrt{3}\right)$ and $Q\left(\sqrt{3}:Q\right)=2$

Now

$\left[\left(Q\sqrt{2},\sqrt{3}\right):Q\sqrt{3}\right]\le 2$

As $\sqrt{2}$satisfy $x^2-2=0$ 0ver $Q\sqrt{3}$

Let $f\left(x\right)=x^2-2$

Suppose on contrary $f\left(x\right)$is reducible over $Q\sqrt{3}$. This implies

$\sqrt{2}\in Q\sqrt{3}$

This gives

$\sqrt{2}=a+\sqrt{3}b:a,b\in Q$

Square both the side

$$\begin{gathered} {\left(\sqrt{2}\right)}^2={\left(a+\sqrt{3}b\right)}^2 \\ 2=a^2+3b^2+2\sqrt{3}ab \end{gathered}$$

If $ab\ne 0$then $\sqrt{3}\in Q$

This is not possible as $\sqrt{3}$ is irrational. Therefore $ab=0$then this Implies

Either a = 0,b = 0

Now if $a=0$then $\sqrt{2}=\sqrt{3}b$

This means $b=\frac{\sqrt{2}}{\sqrt{3}}\in Q$

And if $b=0$ then$\sqrt{2}=a\notin Q$

In both cases it creates a contradicition. Therefore $x^2-2$is irreducible over $Q\sqrt{3}$

Thus

$\left[\left(Q\sqrt{2},\sqrt{3}\right):Q\sqrt{3}\right]=2$

And $\left\{1,\sqrt{2}\right\}$ is the basis for$Q\left(\sqrt{2}+\sqrt{3}\right)$ over$Q\sqrt{3}$ .