TITLE: How to prove the minimality of a minimal polynomial?
QUESTION [8 upvotes]: 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!

REPLY [4 votes]: 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 $\sqrt{2}+\sqrt{3}$.