Find the minimal polynomial using Galois theory One exercise in chapter 14 of Dummit and Foote is to find the minimal polynomial of $\alpha = \sqrt{2} + \sqrt{3}$.  I know this can be done by computing higher powers of $\alpha$ and then finding a nontrivial relation among them, but I think the point of this question is to use Galois theory.  
I think the solution to this question goes something like:


*

*Note that $\mathbb{Q}(\sqrt{2} + \sqrt{3}) \subseteq \mathbb{Q}(\sqrt{2}, \sqrt{3})$.  

*The field extension $\mathbb{Q}(\sqrt{2}, \sqrt{3})$ is Galois over $\mathbb{Q}$.  For any $\sigma \in \text{Gal}(\mathbb{Q}(\sqrt{2}, \sqrt{3})/ \mathbb{Q})$, the only options for $\sigma(\sqrt{2} + \sqrt{3})$ are 
\begin{equation*}
\pm \sqrt{2} \pm \sqrt{3}
\end{equation*}

*Now form the polynomial:
\begin{equation*}
m(x): = \big( x - [\sqrt{2} + \sqrt{3}] \big)\big( x - [-\sqrt{2} + \sqrt{3}] \big)\big( x - [\sqrt{2} - \sqrt{3}] \big)\big( x - [-\sqrt{2} - \sqrt{3}] \big)
\end{equation*}

*This polynomial is the minimal polynomial for $\sqrt{2} + \sqrt{3}$ over $\mathbb{Q}$, because:
(a) It is monic.  This is obvious.
(b) It is irreducible.  This can be seen from noting that $\mathbb{Q}(\sqrt{2}, \sqrt{3}) = \mathbb{Q}(\sqrt{2} + \sqrt{3})$, and the fact that this is a degree $4$ polynomial and $[\mathbb{Q}(\sqrt{2}, \sqrt{3}): \mathbb{Q}] = 4$.
(c) Has $\sqrt{2} + \sqrt{3}$ as a root.
(d) Has coefficients in $\mathbb{Q}$.  I guess this can be seen by just multiplying out the terms of $m(x)$.

I don't have any questions about the correctness of the assertions in any of the above steps.  My questions are more related to why these steps are done in the first place.



*

*The assertion in Step 1 is obviously true, but why would anyone ever think to do this step in the first place?  $\mathbb{Q}(\sqrt{2} + \sqrt{3})$ is also contained in the Galois extension $\mathbb{Q}(\sqrt{2}, \sqrt{3}, \sqrt{5})$, so why not work with this extension instead?

*I understand that $\sigma(\sqrt{2} + \sqrt{3}) = \pm \sqrt{2} \pm \sqrt{3}$.  But why should knowing this fact lead me to want to multiply together all the $(x - \sigma(\alpha))$?   This cannot just be an arbitrary trick...there clearly must be some connection here that I am failing to see.

*Is there are more elegant way (than just multiplying out the polynomial) to see that $m(x)$ has coefficients in $\mathbb{Q}$?
 A: The reason to choose $F=\mathbb Q(\sqrt2,\sqrt 3)$ is that there is no non-trivial automorphism of this field which fixes $\sqrt{2}+\sqrt{3}$. That is not true for $\mathbb Q(\sqrt 2,\sqrt 3,\sqrt 5)$.
If $p(x)$ is a rational polynomial, then for any automorphism $\sigma$ of $F$ and any $u\in F$, $\sigma(p(u))=p(\sigma(u))$. This is because $\sigma(r)=r$ for any rational $r$. 
So if $\sqrt{2}+\sqrt{3}$ is a root, then so must be $\sigma(\sqrt{2}+\sqrt{3})$. So any rational polynomial with root $\sqrt{2}+\sqrt{3}$ has to have $\sigma(\sqrt{2}+\sqrt{3})$ for all $\sigma$.
Finally, the coefficients of $m(x)=\prod_{\sigma\in G}(x-\sigma(\alpha))$ are fixed by any automorphism of $\sigma'\in G$. In the above field, at least, the only elements fixed by all automorphisms of the $F$ are the rationals, so the coefficients must be rational.'
So $m(x)$ has all the roots it has to have, and it has rational coefficients, so it is the minimal polynomial.

So what we have is that 


*

*$m(x)$ has no repeated roots, by the consideration in the first paragraph, that there is no non-trivial automorphism of $F$ which fixes $\alpha$.

*Any polynomial with rational coefficients which has $\alpha$ as a root has $m(x)$ as a factor.


From this, we see that $m(x)$ is irreducible - if it factors non-trivially, then there is a rational polynomial of smaller degree that has $\alpha$ as a root, but that must be divisible by $m(x)$.

So the two key properties we have is that:


*

*If $\sigma(\alpha)=\alpha$, then $\sigma=1$.

*For $x\in F$, if $\sigma(x)=x$ for all $\sigma\in G$, then $x\in\mathbb Q$.


It's worth considering examples where this is not the case.
Taking $F=\mathbb Q(\sqrt[4]{2})$ and $\alpha=\sqrt[4]{2}$, there is one non-trivial automorphism, $\sqrt[4]{2}\mapsto -\sqrt[4]{2}$. But (2) isn't true here, since $\sqrt{2}$ is fixed by both automorphisms. So:
$$(x-\sqrt[4]2)(x+\sqrt[4]2)=x^2-\sqrt{2}$$ is not a minimal polynomial fo $\alpha$ over $\mathbb Q$.
The case where (1) fails is actually your $\mathbb Q(\sqrt{2},\sqrt{3},\sqrt{5})$ example. You get repeated factors in this case.

One final note. There is a way to remove repeated roots. If $m(x)$ has repeated roots, you can remove them by computing:
$$m_1(x)=\frac{m(x)}{\gcd(m(x),m'(x))}$$
This has problems when your fields have finite characteristic, but over the rationals, this will always give you the minimal polynomial even when $\sigma(\alpha)=\alpha$ for some non-trivial case.
A: The other roots of the minimal polynomials are the conjugates (images under $\sigma\in G$), that's what makes $\prod_{\sigma\in G}(X-\sigma(\alpha))$ such a good guess.
Incidentally, you could try the same with $\Bbb Q(\sqrt 2,\sqrt 3,\sqrt 5)$, but that would - as you may easily find - produce the square of the minimal polynomial.
