<s>[INST] Minimal polynomial with $f(\alpha)=0$ I'm learning this stuff for the first time so please bear with me.
I'm given $\alpha=\sqrt{3}+\sqrt{2}i$ and asked to find the minimal polynomial in $\mathbb{Q}[x]$ which has $\alpha$ as a root.
I'm ok with this part, quite sure I can find such a polynomial like so:
$\alpha = \sqrt{3}+\sqrt{2}i$
$\iff \alpha - \sqrt{3} = \sqrt{2}i$
squaring both sides,
$\iff \alpha^2 -2\sqrt{3}\alpha + 3 = -2$
$\iff \alpha^2+5 = 2\sqrt{3}\alpha$
again squaring both sides,
$\iff \alpha^4+10\alpha^2+25 = 12\alpha^2$
$\iff \alpha^4-2\alpha^2+25 = 0$
Therefore $f(x)=x^4-2x^2+25$ should have $f(\alpha)=0$.
My question is how can I be sure this polynomial is minimal, i.e., can I be sure there isn't some lower degree polynomial out there with $\alpha$ as a root?
 [/INST] It suffices to prove that $1,\alpha,\alpha^2,\alpha^3$ are linearly independent over $\mathbb Q$. Writing them in the basis $1,\sqrt{3},\sqrt{2}i,\sqrt{6}i$ of $\mathbb Q[\sqrt{3},\sqrt{2}i]$ we get:
$$
\begin{pmatrix}
1 \\ \alpha \\ \alpha^2 \\ \alpha^3
\end{pmatrix}
=
\begin{pmatrix}
1 & 0 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 1 & 0 & 0 & 2 \\ 0 & -3 & 7 & 0
\end{pmatrix}
\begin{pmatrix}
1 \\ \sqrt{3} \\ \sqrt{2}i \\ \sqrt{6}i
\end{pmatrix}
$$
The key point is that the matrix is invertible.
You can prove this by row reduction or by computing its determinant and checking that it is not zero.
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