# Find a basis for $\mathbb{Q}(\sqrt{2}+\sqrt{5})$ over $\mathbb{Q}(\sqrt{5})$.

Find a basis for $$\mathbb{Q}(\sqrt{2}+\sqrt{5})$$ over $$\mathbb{Q}(\sqrt{5})$$.

I used the following facts:

1. $$\mathbb{Q}(\sqrt{2}+\sqrt{5}) = \mathbb{Q}(\sqrt{2},\sqrt{5})$$, so a tower of fields is $$\mathbb{Q}(\sqrt{5})\subset \mathbb{Q}(\sqrt{2}) \subset \mathbb{Q}(\sqrt{2},\sqrt{5})$$.
2. Minimal polynomial for $$\sqrt{2}$$ over $$\mathbb{Q}(\sqrt{5})$$ is $$x^{2}-2$$, so $$[\mathbb{Q}(\sqrt{2}):\mathbb{Q}(\sqrt{5}]=2$$.
3. Simple extension of $$\mathbb{Q}(\sqrt{2})$$ is $$\mathbb{Q}(\sqrt{2})(\sqrt{5})$$.
4. Minimal polynomial for $$\sqrt{5}$$ over $$\mathbb{Q}(\sqrt{2})$$ is $$x^{2}-5$$. Then $$[\mathbb{Q}\sqrt{2},\sqrt{5}):\mathbb{Q}(\sqrt{2})]=2$$.

Therefore, $$[\mathbb{Q}(\sqrt{2}+\sqrt{5}):\mathbb{Q}(\sqrt{5})]=4$$ and a basis is $$\left \{ 1,\sqrt{2},\sqrt{5},\sqrt{10}\right \}$$.

Are these facts valid though?

• The inclusion $\mathbb{Q}[\sqrt{5}] \subseteq \mathbb{Q}[\sqrt{2}]$ is not true. – Dante Grevino Dec 2 '18 at 17:57
• Your conclusion is also false (in fact you've written down a $\Bbb Q$-basis for $\Bbb Q(\sqrt{2}, \sqrt{5})$) – Edward Evans Dec 2 '18 at 18:08
• @ÍgjøgnumMeg I see. I apologize for my elementary knowledge on this. How do I approach such a problem then? – user482939 Dec 2 '18 at 18:14
• @numericalorange As I see it, you need to show that $\mathbb{Q}(\sqrt{2}+\sqrt{5})$ is not the same as $\mathbb{Q}(\sqrt{5})$, but is the same as $\mathbb{Q}(\sqrt{2},\sqrt{5})$. Then the obvious basis will work; $\{1,\sqrt{2}\}$. – BlarglFlarg Dec 2 '18 at 18:16

There is no such tower of subfields as $$\mathbb{Q}(\sqrt{5})\subset\mathbb{Q}(\sqrt{2})\subset\mathbb{Q}(\sqrt{2},\sqrt{5})$$, because the leftmost inclusion doesn't hold.

You should instead consider $$\mathbb{Q}\subset\mathbb{Q}(\sqrt{5})\subset\mathbb{Q}(\sqrt{2},\sqrt{5})$$. However, you claim that the minimal polynomial of $$\sqrt{2}$$ over $$\mathbb{Q}(\sqrt{5})$$ is $$x^2-2$$ without proving it. Once you prove it, together with the fact that $$x^2-5$$ is the minimal polynomial of $$\sqrt{5}$$ over $$\mathbb{Q}$$, you get $$[\mathbb{Q}(\sqrt{2},\sqrt{5}):\mathbb{Q}]= [\mathbb{Q}(\sqrt{2},\sqrt{5}):\mathbb{Q}(\sqrt{5})] [\mathbb{Q}(\sqrt{5}):\mathbb{Q}]=2\cdot2=4$$

So your proof jumps over two facts:

1. $$\mathbb{Q}(\sqrt{2}+\sqrt{5})=\mathbb{Q}(\sqrt{2},\sqrt{5})$$;
2. $$x^2-2$$ is irreducible over $$\mathbb{Q}(\sqrt{5})$$.

Both facts can be easily proved.

Clearly $$\mathbb{Q}(\sqrt{2}+\sqrt{5})\subset\mathbb{Q}(\sqrt{2},\sqrt{5})$$. On the other hand $$\frac{1}{\sqrt{5}+\sqrt{2}}=\frac{\sqrt{5}-\sqrt{2}}{3}\in\mathbb{Q}(\sqrt{2}+\sqrt{5})$$ so also $$\sqrt{5}-\sqrt{2}\in\mathbb{Q}(\sqrt{2}+\sqrt{5})$$. Hence $$\sqrt{2}$$ and $$\sqrt{5}$$ both belong to $$\mathbb{Q}(\sqrt{2}+\sqrt{5})$$. This proves the reverse inclusion.

The polynomial $$x^2-2$$ has no root in $$\mathbb{Q}(\sqrt{5})$$, hence it is irreducible (having degree $$2$$). Indeed, if $$(a+b\sqrt{5}\,)^2=2$$ we obtain $$a^2+5b^2=2$$ and $$2ab=0$$. This is a contradiction.

Therefore $$\{1,\sqrt{2}\}$$ is a basis for $$\mathbb{Q}(\sqrt{2},\sqrt{5})$$ over $$\mathbb{Q}(\sqrt{5})$$, which in turn has basis $$\{1,\sqrt{5}\}$$ over $$\mathbb{Q}$$.

The standard proof of the dimension theorem tells us that $$\{1,\sqrt{2},\sqrt{5},\sqrt{2}\sqrt{5}\}=\{1,\sqrt{2},\sqrt{5},\sqrt{10}\}$$ is a basis for $$\mathbb{Q}(\sqrt{2},\sqrt{5})$$ over $$\mathbb{Q}$$.

Clearly $$\mathbb{Q}(\sqrt2 + \sqrt5) \subseteq \mathbb{Q}(\sqrt2, \sqrt5)$$. Since $$\mathbb{Q}(\sqrt2 + \sqrt5)$$ is a field, the inverse of $$\sqrt2 + \sqrt5$$ must be in $$\mathbb{Q}(\sqrt2 + \sqrt5)$$ i.e.

$$\frac{1}{\sqrt2 + \sqrt5} = \frac{\sqrt5 - \sqrt2}{3} \in \mathbb{Q}(\sqrt2 + \sqrt5)$$

Then $$\sqrt5 = \frac{3}{2}(\frac{\sqrt5 - \sqrt2}{3}) + \frac{1}{2}(\sqrt2 + \sqrt5) \in \mathbb{Q}(\sqrt2 + \sqrt5)$$ and similarly you can check that $$\sqrt2 \in \mathbb{Q}(\sqrt2 + \sqrt5)$$. Thus, $$\mathbb{Q}(\sqrt2, \sqrt5) \subseteq \mathbb{Q}(\sqrt2 + \sqrt5)$$ and hence $$\mathbb{Q}(\sqrt2, \sqrt5) = \mathbb{Q}(\sqrt2 + \sqrt5)$$.

Now you know that $$\{ 1, \sqrt2 \}$$ is a basis for $$\mathbb{Q}(\sqrt2)$$ over $$\mathbb{Q}$$ and $$\{ 1, \sqrt5 \}$$ is a basis for $$\mathbb{Q}(\sqrt5)$$ over $$\mathbb{Q}$$. Then, as you stated, $$x^2 -2$$ is the minimal polynomial for $$\sqrt2$$ over $$\mathbb{Q}(\sqrt5)$$ and hence $$\mathbb{Q}(\sqrt2, \sqrt5)$$ is an extension field of degree 2 over $$\mathbb{Q}(\sqrt5)$$. Hence a basis for $$\mathbb{Q}(\sqrt2, \sqrt5)$$ over $$\mathbb{Q}(\sqrt5)$$ is $$\{ 1, \sqrt2 \}$$ since $$\sqrt2 \notin \mathbb{Q}(\sqrt5)$$ and it is clear that it is also a basis for $$\mathbb{Q}(\sqrt2 + \sqrt5)$$ over $$\mathbb{Q}(\sqrt5)$$ since $$\mathbb{Q}(\sqrt2, \sqrt5) = \mathbb{Q}(\sqrt2 + \sqrt5)$$.

$$Let$$ $$x=√2+√5$$
$$x-√5=√2$$ $$(x-√5)^2=(√2)^2$$
$$x^2-2√5x+5=2$$
$$x^2-2√5x+3=0$$
This polynomial belongs to Q(√5) therefore degree of √2+√5 over Q(√5) is 2. That means basis for Q(√2+√5) over Q(√5) is {1, √2+√5 }. Am I right?