Finding the degree and basis of $\mathbb{Q}(\sqrt{5}, \sqrt{7})$ I'm having a difficult time finding the degree and basis of $\mathbb{Q}(\sqrt{5}, \sqrt{7})$ over $\mathbb{Q}$.
I know that $\sqrt{7}$ satisfies $f(x) = x^2 - 7$ and is irreducible over $\mathbb{Q}$. So $[\mathbb{Q}(\sqrt{7}) : \mathbb{Q}] = 2$. I also know that $\sqrt{5}$ satisfies $f(x) = x^2 - 5$ which is also irreducible in $\mathbb{Q}$. So $[\mathbb{Q(\sqrt{5})}: \mathbb{Q}] = 2$.
Then by some corollary which I forget the name of $[\mathbb{Q}(\sqrt{5}, \sqrt{7}) : \mathbb{Q}] \leq 4$.
I am unsure of what to do from here.
 A: You know that $[\mathbb{Q}(\sqrt{5}):\mathbb{Q}]=2$. Then $[\mathbb{Q}(\sqrt{5},\sqrt{7}):\mathbb{Q}]=[\mathbb{Q}(\sqrt{5},\sqrt{7}):\mathbb{Q}(\sqrt{5})][\mathbb{Q}(\sqrt{5}):\mathbb{Q}]$. If $\sqrt{7}\in\mathbb{Q}(\sqrt{5})$ then $[\mathbb{Q}(\sqrt{5},\sqrt{7}):\mathbb{Q}(\sqrt{5})]=1$, if not then $[\mathbb{Q}(\sqrt{5},\sqrt{7}):\mathbb{Q}(\sqrt{5})]=2$. You fill in the details.
A: $\mathbb{Q}(\sqrt{5},\sqrt{7})$ as a vector space over $\mathbb{Q}$ is spanned by $1$ and products of the generators $\sqrt{5}$ and $\sqrt{7}$.  Therefore, $\mathbb{Q}$ is spanned by $\{1,\sqrt{5},\sqrt{7},\sqrt{35}\}$.  Notice that I didn't say that this is a basis.  The next step is to show that these elements are linearly independent over $\mathbb{Q}$.
One way to see that these are linearly independent is to find an irreducible polynomial (of degree 4) which has $\sqrt{5}+\sqrt{7}$ as a root.  (This polynomial will have roots of $\pm\sqrt{5}\pm\sqrt{7}$, none of which are in $\mathbb{Q}$ and none of the products of the factors are in $\mathbb{Q}[x]$, either).
A: Some of the "hard part" done for you: It should be clear that
$\Bbb Q(\sqrt{5} + \sqrt{7}) \subseteq \Bbb Q(\sqrt{5},\sqrt{7})$ since $\sqrt{5} + \sqrt{7} \in \Bbb Q(\sqrt{5},\sqrt{7})$.
The other direction is a bit "trickier", note that:
$(\sqrt{5} + \sqrt{7})^3 = 5\sqrt{5} + 15\sqrt{7} + 21\sqrt{5} + 7\sqrt{7} = 26\sqrt{5} + 22\sqrt{7}$
Hence $(\sqrt{5} + \sqrt{7})^3 - 26(\sqrt{5} + \sqrt{7}) = -4\sqrt{7}$, that is, if we set $\alpha = \sqrt{5} + \sqrt{7}$, then:
$\sqrt{7} = -\frac{1}{4}(\alpha^3 - 26\alpha) \in \Bbb Q(\alpha) = \Bbb Q(\sqrt{5} + \sqrt{7})$.
In turn, this means $\sqrt{5} = \sqrt{5} + \sqrt{7} - \sqrt{7} \in \Bbb Q(\sqrt{5} + \sqrt{7})$, so that $\Bbb Q(\sqrt{5},\sqrt{7}) \subseteq \Bbb Q(\sqrt{5} + \sqrt{7})$
So we conclude $\Bbb Q(\sqrt{5},\sqrt{7}) = \Bbb Q(\sqrt{5} + \sqrt{7})$. This is useful.
Since we know the minimal polynomial of $\alpha$ is of degree at most $4$, it's natural to start with:
$\alpha^4 = 284 + 48\sqrt{35}$. Now $\alpha^2 = 12 + 2\sqrt{35}$, so:
$\alpha^4 - 24\alpha^2 = -4$, that is $\alpha$ is a root of $x^4 - 24x + 4$. Proving this is irreducible over $\Bbb Q$ then shows $[\Bbb Q(\sqrt{5},\sqrt{7}):\Bbb Q] = 4$.
One possible basis from this is $\{1,\alpha,\alpha^2,\alpha^3\}$, the linear independence of which is guaranteed by the minimal degree of our quartic. Spanning should be clear, from the degree of the extension, but also: any rational polynomial in $\alpha$ can be reduced to one of degree $< 4$, and from:
$\alpha^4 - 24\alpha^2 + 4 = 0$ we obtain: $\alpha\left(6\alpha - \dfrac{\alpha^3}{4}\right) = 1$, that is:
$\alpha^{-1} = 6\alpha - \dfrac{\alpha^3}{4}$, so any rational function of $\alpha$ (that is, $\Bbb Q(\alpha)$) can be expressed as a linear combination of our basis.
A: Followin @A. Wong's hints, to show $[\mathbb{Q}(\sqrt{5},\sqrt{7}):\mathbb{Q}(\sqrt{5})]=2$, you have to show $x^2-7$ remains irreducible over $\mathbb{Q}(\sqrt{5})$. If not, it has a root in this field, and you can show this implies $\sqrt 5$ is rational.
A: An element $a+b\sqrt5$ has a minimum polynomial $x^2-2ax+a^2-5b^2$. But $\sqrt7$ has minimum polynomial $x^2-7$. These can only be the same if $a=0$ and $5b^2=7$, and this is impossible for rational b. So $\sqrt7=a+b\sqrt5$ is impossible for rational a,b.
