Computing minimal polynomial An example in chapter 14.2 in Dummit and Foote computes the minimal polynomial for $\sqrt{2}+\sqrt{3}$ over $\mathbb{Q}$. We consider the field $\mathbb{Q}(\sqrt{2}+\sqrt{3})$, and note that this is the same field as $\mathbb{Q}(\sqrt{2},\sqrt{3})$. Now, the other roots of the minimal polynomial for $\sqrt{2}+\sqrt{3}$ over $\mathbb{Q}$ are the distinct conjugates of $\sqrt{2}+\sqrt{3}$ under the Galois group. The distincr conjugates are $\pm\sqrt{2}\pm\sqrt{3}$. The minimal polynomial is therefore $[x-(\sqrt{2}+\sqrt{3})][x-(\sqrt{2}-\sqrt{3})][x-(-\sqrt{2}+\sqrt{3})][x-(-\sqrt{2}-\sqrt{3})]$. The next sentence states that this is "quickly computed" to be the polynomial $x^4-10x^2+1$. My question is how is this quickly computed? I'm not sure of a way to do it other than tediously expanding it. Further, how do we know that $x^4-10x^2+1$ is, in fact, irreducible? 
 A: if $x  = \sqrt 2 + \sqrt 3,$ then $x^2 = 5 + 2 \sqrt 6.$ So $(x^2 - 5)^2 = 24,$ and $x^4 - 10 x^2 + 25 = 24,$ and
$$ x^4 - 10 x^2 + 1 = 0. $$
By the Gauss lemma, without any rational roots, the possible rational factorings are
$$ (x^2 + ax +1)(x^2 - ax + 1),  $$
$$ (x^2 + ax -1)(x^2 - ax - 1).  $$
Neither one gives integer $a.$ Either $2 - a^2 = -10$ or $-2 - a^2 = -10$
A: *

*The expansion isn't too tedious if you take advantage of some differences of squares. Start from $$ [x-(\sqrt2+\sqrt3)][x+(\sqrt2+\sqrt3)][x-(\sqrt2-\sqrt3)][x+(\sqrt2-\sqrt3)] $$ 

*We already know how to break this polynomial into linear factors, and none have coefficients in $\mathbb Q$. It's not too hard to see that any pair won't combine give a quadratic factor with coefficients in $\mathbb Q$: let the four factors be $y_1, y_2, y_3, y_4$. For any pair $i,j$, $(x-y_i)(x-y_j) = x^2 -(y_i+y_j)x + y_iy_j$. For our four factors, $y_i+y_j$ is in $\mathbb Q$ if and only if $y_i = -y_j$, in which case $y_iy_j = -y_i^2$ is not in $\mathbb Q$. Since there aren't any linear or quadratic factors, it's irreducible.
A: 
Further, how do we know that $x^4-10x^2+1$ is, in fact, irreducible? 

Other answers have addressed the expansion, so I'll tackle this question. 
Irreducibility can be shown using the following (important) theorem:

Let $f$ be a separable polynomial over a field $F$ with $E$ denoting its splitting field.   Then $f$ is irreducible over $F \iff$ the action of $\text{Aut}(E/F)$ on the roots of $f$ is transitive$^\dagger$.

In this case, we are taking $f \in \mathbb{Q}[x]$ as the minimal polynomial for $\sqrt{2} + \sqrt{3}$.  Bearing in mind the tower $\mathbb{Q} \subset \mathbb{Q}(\sqrt{2} + \sqrt{3}) \subseteq \mathbb{Q}(\sqrt{2}, \sqrt{3})$, the automorphisms of $E$ are simply the automorphisms of $\mathbb{Q}(\sqrt{2}, \sqrt{3})$ restricted to $E$ (the latter two fields are actually equal but that would require further proof).  Now note that, because $f$ has precisely all the Galois conjugates of $\sqrt{2} + \sqrt{3}$ as its roots per its construction, the action is clearly transitive!  So the theorem guarantees irreducibility.

$^\dagger$ For a proof, see Theorem 2.9(b) here.
