Minimal polynomial of $\sqrt{2}+\sqrt{3}$ over $\mathbb Q$ I had an example in the book given as follows:   

Find the minimal polynomial of $\sqrt{2}+\sqrt{3}$ over $\mathbb Q$ . 
Solution:  $~~~~~$$(\sqrt{2}+\sqrt{3})^2=5+2 \sqrt6$
$(\sqrt{2}+\sqrt{3})^4=49+20 \sqrt6$ 
Then $(\sqrt{2}+\sqrt{3})^4-10(\sqrt{2}+\sqrt{3})^2+1=0.$ 
Thus $a=\sqrt{2}+\sqrt{3}$ satisfies $f(x)=x^4-10x^2+1$ over $\mathbb Q$. 
Let $p(x)$ be the minimal polynomial of $\sqrt{2}+\sqrt{3}$ over $\mathbb Q$.Then $(\sqrt{2}-\sqrt{3}),(-\sqrt{2}+\sqrt{3}),(-\sqrt{2}-\sqrt{3})$ are also roots of $p(x).$
  So degree of $p(x)$ is atleast $4$.
But $f(a)=0 $ and $f(x) \in \mathbb Q[x] \implies p(x)$ divides $f(x)$ .
So $f(x)$ is minimal polynomial of $\sqrt{2}+\sqrt{3}.$

But I can't get the step  how are $(\sqrt{2}-\sqrt{3}),(-\sqrt{2}+\sqrt{3}),(-\sqrt{2}-\sqrt{3})$ also roots of $p(x).$ 
Kindly help with this.
 A: That fact comes from Galois Theory. Under the assumption that $K/F$ is Galois, the (monic) minimal polynomial $f$ of an element $\alpha \in K$ (over $F$) is of the form
$$f(x) = \prod_{\sigma \in \text{Gal}(K/F)} (x - \sigma \alpha).$$
This implies that $\sigma \alpha$ are all roots of the minimal polynomial for $\alpha$.
In your case, let $K = \mathbb{Q}(\sqrt{2}, \sqrt{3})$ and $F = \mathbb{Q}$. Then $K/F$ is Galois and there are 4 automorphisms in $\text{Gal}(K/F)$: namely


*

*identity on $K$

*the one that sends $\sqrt{2} \mapsto -\sqrt{2}$ keeping $\sqrt{3}$ fixed: this automorphism sends $\sqrt{2} + \sqrt{3}$ to $-\sqrt{2} + \sqrt{3}$; hence $-\sqrt{2} + \sqrt{3}$ must be a root of $p(x)$.

*the one that keeps $\sqrt{2}$ fixed and sends $\sqrt{3} \mapsto -\sqrt{3}$: like above, we deduce $\sqrt{2} - \sqrt{3}$ is root of minimal polynomial $p(x)$

*the one that negate both roots: finally $- \sqrt{2} - \sqrt{3}$ must also be a root.

A: Good question: that is a big leap in the argument in my opinion.
If we let $K = \mathbb{Q}(\sqrt{2},\sqrt{3})$, then $\sqrt{2} + \sqrt{3}$, $\sqrt{2} - \sqrt{3}$, $-\sqrt{2} + \sqrt{3}$, and $-\sqrt{2} - \sqrt{3}$ all lie in $K$.
In order to prove that they are all roots of $p(x)$, the minimal polynomial of $\sqrt{2} + \sqrt{3}$, we follow An Hoa's idea: find automorphisms $\sigma, \tau$ of $K / F$ such that $\sigma(\sqrt{2}) = -\sqrt{2}$, $\sigma(\sqrt{3}) = \sqrt{3}$, $\tau(\sqrt{2}) = \sqrt{2}$, and $\tau(\sqrt{3}) = -\sqrt{3}$.
The argument is then completed by noting that $\sigma, \tau$ both preserve $p(x)$ (since its coefficients are in $K$), so they preserve the roots of $p(x)$. (If $\alpha$ is a root of $p(x)$, then $\sigma(\alpha)$ is a root of $\sigma(p(x)) = p(x)$ and $\tau(\alpha)$ is a root of $\tau(p(x)) = p(x)$.)
Therefore what we need to show is that these automorphisms $\sigma$ and $\tau$ exist.
$K = \mathbb{Q}(\sqrt{2},\sqrt{3})/\mathbb{Q}$ is the splitting field of $(x^2 - 2)(x^2 - 3)$, so it is a Galois extension.
To argue the extension has degree $4$, further notice that it contains both $\mathbb{Q}(\sqrt{2})$ and $\mathbb{Q}(\sqrt{3})$. These two fields are distinct because one contains a square root of two and one doesn't (all elements of $\mathbb{Q}(\sqrt{3})$ can be written $a + b\sqrt{3}$ and you can show that such a thing squared cannot be $2$). So $K$ has degree divisible by $2$ and strictly greater than $2$ over $\mathbb{Q}$, and it also has degree at most $4$ since it is the splitting field of a degree-4 polynomial, so it has degree exactly $4$.
Now that we know $K$ has degree $4$ over $\mathbb{Q}$ and is Galois, it must have four automorphisms (this is the definition of Galois, that the number of automorphisms equals the degree). These automorphisms permute the roots of $(x^2 - 2)(x^2 - 3)$ and are defined by where they send these roots. They must send roots of $(x^2 - 2)$ to roots of $(x^2 - 2)$ and roots of $(x^2 - 3)$ to roots of $(x^2 - 3)$. But there are only $4$ different ways to do this. Therefore, every possible way of sending $\sqrt{2} \mapsto \pm \sqrt{2}$ and $\sqrt{3} \mapsto \pm \sqrt{3}$ is an automorphism. In particular, $\sigma$ and $\tau$ are automorphisms.
$\square$
A: Assume that $m$ and $n$ are square-free coprime integers. Then, if we prove that $Q(\sqrt m +\sqrt n) = Q(\sqrt m, \sqrt n)$, we can conclude that that $f(x) = x^4 -2(m+n)x +(m-n)^2$ must be the minimal polynomial of $\sqrt m +\sqrt n$ since $Q(\sqrt m +\sqrt n)$ is a four-dimensional $Q$-vector space. But if $\xi = (\sqrt m + \sqrt n)$, then ${\frac {{\xi}^2-(m+n)} {2}}={\sqrt m}{\sqrt n}$. Thus, $({\frac {{\xi}^2-(m+n)} {2}})\xi = m{\sqrt n}+n{\sqrt m}$. Hence, $$({\frac {{\xi}^2-(m+n)} {2}})\xi -n{\xi} = (m-n){\sqrt n},$$ and $$({\frac {{\xi}^2-(m+n)} {2}})\xi -m{\xi} = (n-m){\sqrt n}.$$ Thus, $Q(\sqrt m +\sqrt n)$ contains $\{\sqrt m, \sqrt n \}$, and so $Q(\sqrt m +\sqrt n)\supset Q(\sqrt m ,\sqrt n)$. The reverse set inclusion is immediate. The case $m =2$, and $n = 3$ is of course what inspired this train of thought which is hopefully correct.
