How to prove the minimality of a minimal polynomial? I have a problem: find the minimal polynomial of $\sqrt 2+\sqrt 3$ over $\mathbb Q$
I can calculate the answer i.e. $x^4-10x^2+1$ but how can I prove the minimality? Thanks in advance!
 A: Hint; $\sqrt 2+\sqrt 3 \in \mathbb Q(\sqrt 2,\sqrt 3)$, which has degree $4$ over $\mathbb Q$. Therefore, the degree of every element is $1$, $2$, or $4$. Prove that $\sqrt 2+\sqrt 3$ does not have degree $1$ or $2$. (Use that $\sqrt 2$ and $\sqrt 3$ are linearly independent over $\mathbb Q$.)
A: $$(a+b+c)\prod_{cyc}(a+b-c)=\sum_{cyc}(2a^2b^2-c^4).$$
For $a=\sqrt2$, $b=\sqrt3$ and $c=x$ we obtain:
$$2(6+2x^2+3x^2)-4-9-x^4=0,$$
which gives
$$x^4-10x^2+1=0.$$
Now, easy to show that $x^4-10x^2+1$ is irreducible over $\mathbb{Q}$
and $\sqrt2+\sqrt3\notin\mathbb{Q}$.  
A: It is enough to exploit a bit of linear algebra. We have that $\alpha^2-2\in\mathbb{Q}[\alpha]$ is the minimal polynomial of $\sqrt{2}$ over $\mathbb{Q}$ and $\beta^2-3\in\mathbb{Q}[\beta]$ is the minimal polynomial of $\sqrt{3}$. A base of the ring $\mathbb{Q}[\alpha,\beta]/(\alpha^2-2,\beta^2-3)$ as a vector space over $\mathbb{Q}$ is given by $1,\alpha,\beta,\alpha\beta$, and we may represent every polynomial $(\alpha+\beta)^n$ with respect to such a base:
$$ \begin{array}{|c|c|c|c|c|c|}\text{Polynomial}&1&\alpha&\beta&\alpha\beta\\ \hline (\alpha+\beta)^0 & 1 & 0 & 0 & 0 \\ 
\hline (\alpha+\beta)^1 & 0 & 1 & 1 & 0 \\
\hline (\alpha+\beta)^2 & 5 & 0 & 0 & 2 \\ 
\hline (\alpha+\beta)^3 & 0 & 11 & 9 & 0 \\
\hline (\alpha+\beta)^4 & 49 & 0 & 0 & 20 \\
\hline  \end{array}$$
Given five vectors in $\mathbb{Q}^4$, there always is a non-trivial linear combination of them that equals zero.
In the present case the fifth row minus ten times the third row plus the first row gives zero, hence $z^4-10z^2+1$ is a polynomial that vanishes at $\alpha+\beta$.  That is the minimal polynomial of $\sqrt{2}+\sqrt{3}$ over $\mathbb{Q}$ because the matrix
$$ M=\begin{pmatrix}1 & 0 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 5 & 0 & 0 & 2 \\ 0 & 11 & 9 & 0\end{pmatrix}$$
is invertible, hence there isn't any non-trivial polynomial in $\mathbb{Q}_{\leq 3}[x]$ vanishing at $\sqrt{2}+\sqrt{3}$.
A: Let $y = (x-1)$. Then $x = y+1$ and so your polynomial can be expressed as a polynomial in $y$, which is irreducible by Eisenstein's criterion using the prime $2$, and hence your polynomial is also irreducible. Of course, the general proof of linear independence (as mentioned by lhf) is the way to go if you want the general result that the sum of square-roots of primes generates the whole field extension.
