Jtdylktuy

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?

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

You should instead consider $mathbb{Q}subsetmathbb{Q}(sqrt{5})subsetmathbb{Q}(sqrt{2},sqrt{5})$. 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
$$
[mathbb{Q}(sqrt{2},sqrt{5}):mathbb{Q}]=
[mathbb{Q}(sqrt{2},sqrt{5}):mathbb{Q}(sqrt{5})]
[mathbb{Q}(sqrt{5}):mathbb{Q}]=2cdot2=4
$$

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 $mathbb{Q}(sqrt{2}+sqrt{5})subsetmathbb{Q}(sqrt{2},sqrt{5})$. On the other hand
$$
frac{1}{sqrt{5}+sqrt{2}}=frac{sqrt{5}-sqrt{2}}{3}inmathbb{Q}(sqrt{2}+sqrt{5})
$$
so also $sqrt{5}-sqrt{2}inmathbb{Q}(sqrt{2}+sqrt{5})$. 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
$$
(a+bsqrt{5},)^2=2
$$
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?