Can someone tell me if this is right:
I would like to find the minimal polynomial of
(i) $\sqrt[4]{2}i$ over $\mathbb{Q}$
(ii) $\sqrt[4]{2}i$ over $\mathbb{R}$
(i):
$\sqrt[4]{2}i$ is a root of $f(x) = x^4-2$. This is already monic, so to show that this is a minimal polynomial I only need to show that it is irreducible. Edit: To do that, I can use the Eisenstein: $p=2$ does not divide $a_1 = 1$ and $p^2$ does not divide $a_0 = -2$, therefore it is irreducible over $\mathbb{Q}$.
(ii):
This time, $\sqrt[4]{2}i$ is a root of $f(x) = x^2+\sqrt{2}$. For polynomials of degree two it's enough to check if they have a root. The only roots this one has are complex therefore it is irreducible over $\mathbb{R}$ and therefore the minimal polynomial.
So my more general question is: is the way to find a minimal polynomial of an element $a$ in general to find a polynomial $f$ such that $f(a) = 0$ and then to norm $f$ and then to show that $f$ is irreducible?
Many thanks for your help!
Best Answer
In general, you can find the minimum polynomial for an algebraic number $\alpha$ by determining the smallest power of $\alpha$ for which $\{1,\alpha,\alpha^2,\ldots,\alpha^n\}$ is linearly dependent over $\mathbb{Q}$.
For example, let $\alpha=\sqrt{2} + \sqrt{3}$. Then \begin{align*} \alpha^2 &= 5+2\sqrt{6} \\\ \alpha^3 &= 11\sqrt{2}+9\sqrt{3} \\\ \alpha^4 &= 49+20\sqrt{6} \end{align*} Since $\{1,\sqrt{2},\sqrt{3},\sqrt{6}\}$ are linearly independent over $\mathbb{Q}$, we can think of the powers of $\alpha$ as vectors: $$ 1 = \begin{bmatrix}1\\\ 0\\\ 0\\\ 0\end{bmatrix}, \qquad \alpha = \begin{bmatrix}0\\\ 1\\\ 1\\\ 0\end{bmatrix}, \qquad \alpha^2 = \begin{bmatrix}5\\\ 0\\\ 0\\\ 2\end{bmatrix}, \qquad \alpha^3 = \begin{bmatrix}0\\\ 11\\\ 9\\\ 0\end{bmatrix}, \qquad \alpha^4 = \begin{bmatrix}49\\\ 0\\\ 0\\\ 20\end{bmatrix} $$ As you can see, $\{1,\alpha,\alpha^2,\alpha^3\}$ is linearly independent, so $\alpha$ is not the root of any cubic polynomial. However, $\{1,\alpha,\alpha^2,\alpha^3,\alpha^4\}$ is linearly dependent, with $$ \alpha^4 - 10\alpha^2 + 1 \;=\; 0 $$ It follows that $x^4-10x^2+1$ is the minimum polynomial for $\alpha$.
This technique depends on being able to recognize a useful set of linearly independent elements like $\{1,\sqrt{2},\sqrt{3},\sqrt{6}\}$. Depending on how much Galois theory you know, it may be hard to prove that elements like this are linearly independent over $\mathbb{Q}$. In this case, you can use this technique to guess the minimum polynomial, and then prove that you are correct by proving that the polynomial you found is irreducible.
For example, if $\alpha = i\sqrt[4]{2}$, then $$ \alpha^2 = -\sqrt{2},\qquad \alpha^3 = -i\sqrt[4]{8},\qquad \alpha^4 = 2. $$ It seems clear that $ \{1,\alpha,\alpha^2,\alpha^3\} $ is linearly independent, while $\{1,\alpha,\alpha^2,\alpha^3,\alpha^4\}$ satisfies $\alpha^4-2 = 0$. Thus $x^4-2$ should indeed be the minimum polynomial over $ \mathbb{Q} $, though the easiest way to prove it is to show that $x^4 -2$ is irreducible. This can be done using Eisenstein's criterion, or by checking that $ x^4-2 $ has no roots and does not factor into quadratics modulo $4$.