proving that $\mathbb{Q}(\sqrt{5}, \sqrt{6}) = \mathbb{Q}(\sqrt{5}+ \sqrt{6}) $ Here is an extract from my Galois Theory notes proving that $\mathbb{Q}(\sqrt{5}, \sqrt{6}) = \mathbb{Q}(\sqrt{5}+ \sqrt{6}) $
My question is after rearranging equation (1) has my lecturer omitted an $\alpha$ that should be on the denominator? I think this may be a typo but solutions to the general case in one of our assignments have been released and it is also missing the $\alpha$ in the denominator.
If it is not a typo then how does this equation hold
thanks in advance

 A: Yes, but correcting the typo leaves the proof correct since $\,\sqrt 5 =  \dfrac{1}2(\alpha - \alpha^{-1})\in\Bbb Q(\alpha)$
Simpler:  $\ \alpha^{-1} = \dfrac{6-5}{\sqrt 6 + \sqrt 5} = \sqrt{6}-\sqrt 5\ \Rightarrow\ \begin{align}\alpha+\alpha^{-1}=2\sqrt 6\\ \alpha-\alpha^{-1} = 2\sqrt{5}\end{align}$ $\,\Rightarrow\, \sqrt 6,\sqrt5 \in \Bbb Q(\alpha)$
Remark $\ $ This exploits the key idea that lies at the heart of the proof of the Primitive Element Theorem. Namely, if a field $F$ has two $F$-linear independent combinations of $\rm\, \sqrt{a},\ \sqrt{b}\, $ then
we can solve for $\rm\, \sqrt{a},\ \sqrt{b}\, $ in $F.$ 
See here for further discussion. 
A: Yes, it's a typo. In the general case of $\mathbb{Q}(\sqrt{a},\sqrt{b})$ where $a$ and $b$ are squarefree integers, we have, after
$$
\alpha=\sqrt{a}+\sqrt{b},
$$
that
$$
(\alpha-\sqrt{a})^2=b
$$
so
$$
\alpha^2-2\alpha\sqrt{a}+a=b
$$
hence
$$
\sqrt{a}=\frac{\alpha^2+a-b}{2\alpha}\in\mathbb{Q}(\alpha).
$$
Thus also
$$
\sqrt{b}=\alpha-\sqrt{a}\in\mathbb{Q}(\alpha)
$$
(no need to repeat the proof).
In order to find the minimal polynomial of $\alpha$, we have to assume $a\ne b$ (otherwise the result would be trivial). Then $\sqrt{b}\notin\mathbb{Q}(\sqrt{a})$, because
$$
(x+y\sqrt{a})^2=b
$$
would mean $x^2+ay^2+2xy\sqrt{a}=b$, forcing $\sqrt{a}$ to be rational, unless $x=0$ or $y=0$; if $x=0$ we'd have $ay^2=b$ which is impossible as $a$ and $b$ are squarefree and unequal; if $y=0$, we'd have $x^2=b$, likewise impossible.
Then the degree of $\mathbb{Q}(\alpha)$ over $\mathbb{Q}$ is $4$. Then, squaring
$$
2\alpha\sqrt{a}=\alpha^2+a-b
$$
we find an expression of degree $4$ in $\alpha$ that gives us a monic polynomial of degree $4$ having $\alpha$ as a root. This is the minimal polynomial.
A: Here is another argument that $\Bbb Q(\sqrt5+\sqrt6\,)=\Bbb Q(\sqrt5,\sqrt6\,)$, but like Dubuque’s second argument, it is not an answer to your question. On the other hand, I think that Suárez-Álvarez will agree that it is well-enough motivated and reasonably natural.
$K=\Bbb Q(\sqrt5,\sqrt6\,)$ is certainly quartic over $\Bbb Q$, in fact biquadratic, which means that there’s another quadratic subfield of $K$ beyond $\Bbb Q(\sqrt5\,)$ and $\Bbb Q(\sqrt6\,)$, namely $\Bbb Q(\sqrt{30}\,)$. And these are the only quadratic subfields of $K$. So if $\Bbb Q(\sqrt5+\sqrt6\,)$ is unequal to $K$, it must be equal to $\Bbb Q(\sqrt{30}\,)$, a field in which $\{1,\sqrt{30}\,\}$ is a good $\Bbb Q$-basis. This would mean that $\sqrt5+\sqrt6=a + b\sqrt{30}$ for rational numbers $a$ and $b$. But $\{1,\sqrt5,\sqrt6,\sqrt{30}\,\}$ is a $\Bbb Q$-basis for $K$, and the preceding equation contradicts linear independence of these four numbers.
