General technique for finding minimal polynomial? I always have a lot of trouble with these problems, "find the minimal polynomial of {number} over {field}"
What are the general procedures for solving problems of this format?
Thank you for your help
 A: If$\,$ you know beforehand the degree $n$ of the minimal polynomial of $\beta$, i.e. if you know the field extension degree of $\Bbb Q(\beta)$ over $\Bbb Q$, then $\{1,\beta,\cdots,\beta^{n-1}\}$ is a $\Bbb Q$-basis of $\Bbb Q(\beta)$, and you can look at the $\Bbb Q$-linear transformation of the field given by $z\mapsto \beta z$. You may use the basis I pointed to, to calculate the characteristic polynomial of the above-described linear transformation. You did this in undergraduate Linear Algebra. Cayley and Hamilton say that the matrix satisfies the characteristic polynomial, so that $\beta$ does as well. That’s the minimal polynomial of $\beta$. Believe me, for small values of $n$, it’s eminently computable by hand.
A: I'll assume you're working over $\mathbb Q$.
As an example, take $\alpha = 1+\sqrt{2}$. Notice that $\alpha^2 = 3+2\sqrt{2}$, and so we can write
$$\left[\begin{array}{c} \alpha \\ \alpha^2\end{array}\right] =
\left[\begin{array}{cc} 1 & 1 \\ 3 & 2 \end{array}\right]\left[\begin{array}{c} 1 \\ \sqrt{2}\end{array}\right]$$
Inverting the 2-by-2 matrix on the right gives
$$\left[\begin{array}{cc} -2 & 1 \\ 3 & -1 \end{array}\right]\left[\begin{array}{c} \alpha \\ \alpha^2\end{array}\right]=\left[\begin{array}{c} 1 \\ \sqrt{2}\end{array}\right]$$
Expanding the the first row of the left hand side gives $-2\alpha + \alpha^2 = 1$, i.e.
$$\alpha^2 - 2\alpha-1=0$$
You can do a similar thing with numbers like $\beta = \sqrt{2}+\sqrt{3}$. We note that 
$$\begin{eqnarray*}\beta &=& {\bf 0} + {\bf 1}\sqrt{2}+{\bf 1}\sqrt{3}+{\bf 0}\sqrt{6} \\ \\
\beta^2 &=& {\bf 5} + {\bf 0}\sqrt{2}+{\bf 0}\sqrt{3}+{\bf 2}\sqrt{6} \\ \\
\beta^3 &=& {\bf 0} + {\bf 11}\sqrt{2}+{\bf 9}\sqrt{3}+{\bf 0}\sqrt{6} \\ \\
\beta^4 &=& {\bf 49} + {\bf 0}\sqrt{2}+{\bf 0}\sqrt{3}+{\bf 20}\sqrt{6} 
\end{eqnarray*}$$ 
Putting this into a matrix equation, we get
$$\left[\begin{array}{c} \beta \\ \beta^2 \\ \beta^3 \\ \beta^4 \end{array}\right] = \left[ \begin{array}{cccc} 0 & 1 & 1 & 0 \\
5 & 0 & 0 & 2 \\
0 & 11 & 9 & 0 \\
49 & 0 & 0 & 20 \end{array}\right]\left[\begin{array}{c} 1 \\ \sqrt{2} \\ \sqrt{3} \\ \sqrt{6}\end{array}\right]$$
If you invert the 4-by-4 matrix then you get
$$\left[\begin{array}{cccc} 0 & 10 & 0 & -1 \\ -9/2 & 0 & 1/2 & 0 \\
11/2 & 0 & -1/2 & 0 \\ 0 & -49/2 & 0 & 5/2 \end{array}\right]\left[\begin{array}{c} \beta \\ \beta^2 \\ \beta^3 \\ \beta^4 \end{array}\right]=\left[\begin{array}{c} 1 \\ \sqrt{2} \\ \sqrt{3} \\ \sqrt{6}\end{array}\right]$$
Reading off the first row gives
$$\beta^4-10\beta^2+1=0$$
