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I am trying to solve a problem that goes as follows.

Show that the polynomial $f(X) = X^4 - 16X^2 + 4$ is irreducible over $\mathbb{Q}$. Let $\alpha$ be a root of $f(X)$ in some field extension of $\mathbb{Q}$. Determine the minimal polynomials of $\alpha^2$ and of $\alpha^3 - 14 \alpha$ over $\mathbb{Q}$.

I have shown that $f(X)$ is irreducible over $\mathbb{Q}$ and found the minimal polynomial of $\alpha^2$. I am now stuck on the very last part of this problem, to find the minimal polynomial of $\alpha^3 - 14\alpha$. I know that $0 = \alpha^4 - 16 \alpha^2 + 4$ but I can not proceed from here. To find the minimal polynomial of $\alpha^2$ was easy enough, just set $X := \alpha^2$ in the equation and simplify. However, with $\alpha^3 - 14\alpha$ I can't seem to find a way to use substitution in this way. Any hints?

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    $\begingroup$ Not sure what you mean by "setting $X=\alpha^2$ and simplify". The minimal polynomial of $\alpha^2$ can be read off, as $x^2-16x+4$. $\endgroup$
    – lulu
    Mar 1, 2021 at 16:30
  • $\begingroup$ @lulu What I meant by that was that you can set $X = \alpha^2$ to then notice that $\alpha^2$ satisfies $0 = X^2 - 16X + 4$. One can also check that it is irreducible in $\mathbb{Q}$ just to be sure. $\endgroup$
    – Ello
    Mar 1, 2021 at 16:35
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    $\begingroup$ No need...if it factored, $\alpha$ would be rational. $\endgroup$
    – lulu
    Mar 1, 2021 at 16:37

2 Answers 2

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We can simplify the expression $\alpha^3 - 14 \alpha$. This occurs because $\alpha^2$ satisfies $\alpha^4 = 16\alpha^2 - 4$, therefore $$\alpha^3 = 16\alpha - 4\alpha^{-1} $$

and so $\theta = \alpha^3 - 14\alpha = 2\alpha - 4\alpha^{-1}$. (note that $\alpha \neq 0$ so is invertible)

Let us square this term : $$ \theta^2 = 4\alpha^2 - 16\alpha^{-2} -16 = \frac{4\alpha^4 - 16}{\alpha^2} - 16 = 4\frac{\alpha^4 - 4}{\alpha^2} -16 $$

but $\alpha^4 - 4 = 16 \alpha^2$! So substitute that in and get $4 \times 16 - 16 = 48 = \theta^2$.

Thus, the minimal polynomial should be $x^2-48$ as long as you see that is irreducible.


We are lucky that such a manipulation worked out. In general, however, we'd either have to visit higher powers, or need good manipulative intuition to find the minimal polynomial. In some cases, knowing a root can be helpful, so you can directly compute the expression for that root.

For example, the polynomial given is in fact the minimal polynomial of $\sqrt 3 + \sqrt 5$, so one can try the manipulation with this as $\alpha$ to see that $\alpha^3 - 14 \alpha = 4\sqrt 3$ (in case that no intelligent techniques work, this is brute force).

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    $\begingroup$ Thank you for the acceptance, looking forward to seeing your improvement in this course over the next few months. $\endgroup$ Mar 1, 2021 at 17:22
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To find the minimal polynomial of $\gamma=\alpha^3 - 14\alpha$, we may start by looking at the powers of $\gamma$.

Luckily, $\gamma^2=48$ (WA), and we're done. (But there is one crucial point to prove. Find it!)

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