# Minimal polynomial of $\alpha^3 - 14 \alpha$ over $\mathbb{Q}$

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?

• Not sure what you mean by "setting $X=\alpha^2$ and simplify". The minimal polynomial of $\alpha^2$ can be read off, as $x^2-16x+4$.
– lulu
Mar 1, 2021 at 16:30
• @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.
– Ello
Mar 1, 2021 at 16:35
• No need...if it factored, $\alpha$ would be rational.
– lulu
Mar 1, 2021 at 16:37

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).

• Thank you for the acceptance, looking forward to seeing your improvement in this course over the next few months. Mar 1, 2021 at 17:22

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!)