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?

  • 4
    $\begingroup$ The inclusion $\mathbb{Q}[\sqrt{5}] \subseteq \mathbb{Q}[\sqrt{2}]$ is not true. $\endgroup$ – Dante Grevino Dec 2 '18 at 17:57
  • 2
    $\begingroup$ Your conclusion is also false (in fact you've written down a $\Bbb Q$-basis for $\Bbb Q(\sqrt{2}, \sqrt{5})$) $\endgroup$ – Edward Evans Dec 2 '18 at 18:08
  • $\begingroup$ @ÍgjøgnumMeg I see. I apologize for my elementary knowledge on this. How do I approach such a problem then? $\endgroup$ – user482939 Dec 2 '18 at 18:14
  • 2
    $\begingroup$ @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; $\{1,\sqrt{2}\}$. $\endgroup$ – 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 $\mathbb{Q}(\sqrt2, \sqrt5) = \mathbb{Q}(\sqrt2 + \sqrt5)$.


$Let$ $x=√2+√5$
$x-√5=√2$ $(x-√5)^2=(√2)^2$
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?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy