Question

verify that $\mathbf{[}\mathbf{Q}\left(\sqrt{2},\sqrt{5},\sqrt{10}\right)\mathbf{:}\mathbf{Q}\mathbf{]}\mathbf{=}\mathbf{4}$ .

Short answer

$[\mathrm{Q}(\sqrt{2},\sqrt{5},\sqrt{10}):\mathrm{Q}]=4$

Step-by-step solution

Definition of monic polynomial.

The monic polynomial $p\left(x\right)$ over a field $F$ of least degree such that $p\left(\alpha \right)=0$ for an algebraic element $\alpha$ over a field$F$ .

Step 2:Showing that  [Q2,5,10:Q]=4

As $\sqrt{2},\sqrt{5}\in Q\left(\sqrt{2},\sqrt{5}\right)$. Then $\sqrt{2},\sqrt{5}=\sqrt{10}\in Q\left(\sqrt{2},\sqrt{5}\right)$

Therefore $\sqrt{10}\in Q\left(\sqrt{2},\sqrt{5}\right)$

This implies

$Q\left(\sqrt{2},\sqrt{5},\sqrt{10}\right)\subseteq Q\left(\sqrt{2},\sqrt{5}\right)$

And also role="math" localid="1657964968372" $Q\left(\sqrt{2},\sqrt{5},\right)\subseteq Q\left(\sqrt{2},\sqrt{5},\sqrt{10}\right)$

From this $Q\left(\sqrt{2},\sqrt{5},\sqrt{10}\right)=Q\left(\sqrt{2},\sqrt{5}\right)$

As if are distinct prime number then $Q\left(\sqrt{p},\sqrt{q}\right)=Q\left(\sqrt{q}+\sqrt{p}\right)$ has minimal polynomial of degree 4. The minimal polynomial is ${\left(x^2-p-q\right)}^2-4pq$

Hence

$\left[Q\left(\sqrt{2},\sqrt{5}\right):Q\sqrt{2}\right]=2$

And

$$\begin{gathered} \left[Q\left(\sqrt{2},\sqrt{5}\right):Q\right]=\left[Q\left(\sqrt{2},\sqrt{5}\right):Q\sqrt{2}\right]\left[Q\sqrt{2}:Q\right] \\ =2.2 \\ =4 \end{gathered}$$

Thus

$$\begin{gathered} \left[Q\left(\sqrt{2},\sqrt{5}\right):Q\right]=4 \\ \left[Q\left(\sqrt{2},\sqrt{5}\right):Q\right]=\left[Q\left(\sqrt{2},\sqrt{5}\right)\sqrt{10}\right] \end{gathered}$$

So,

$Q\left(\sqrt{2},\sqrt{5}\sqrt{10}\right)=4$.