I have a question about finding the minimal polynomial over $\mathbb{Q}$ of the element $1 + \sqrt[3]{2} + \sqrt[3]{4}$. The answer is discussed in the MSE post here: Find the minimal polynomial for $\sqrt[3]{2} + \sqrt[3]{4}$ over $\mathbb{Q}$ but I believe that there is another way of determining the minimal polynomial in such a situation and I am not sure if the procedure works.

Let $\zeta_3$ denote a primitive 3rd root of unity. I believe then that the minimal polynomial $m(x)$ should be able to be directly calculated by

$$ m(x) = \prod_{c_1, c_2, c_3 \in \{1, \zeta_3, \zeta_3^2\}}(x + c_1(1) + c_2(\sqrt[3]{2}) + c_3(\sqrt[3]{4})), $$

i.e., let the coefficients $c_1, c_2, c_3$ (there are three coefficients because there are three elements being summed in the element I'm trying to find the minimal polynomial for?) range across all three of the 3rd roots of unity (let them range over the third roots of unity since there are cube roots in the element I'm trying to find the minimal polynomial for?) and then multiply all of the resulting linear polynomials together.

I know from the linked post that the answer is the degree 3 polynomial $x^3 - 6x - 6$, but in order to verify the above product I would need to multiply out 27 linear polynomials.

So, I guess my question is two part - does the above algorithm work in general for determining minimal polynomials over $\mathbb{Q}$, and is there an online application I can use that would allow me to compute such products as the one above? I.e., an online calculator that allows for inputs such as $\zeta_3$, etc.

  • $\begingroup$ One computer algebra system you can use is Gap, there are others listed here. $\endgroup$
    – plop
    Aug 19, 2022 at 0:18
  • $\begingroup$ $X^3-3X^2-3X-1$ $\endgroup$
    – GEdgar
    Aug 19, 2022 at 0:22
  • $\begingroup$ @gedgar you blew my cover! $\endgroup$ Aug 19, 2022 at 0:31

1 Answer 1


Upon further review, your $27$ linear polynomials reduce to three.

First off, the $1$ is not under any root signs, so you can just call it $1$ without multiplying in any roots of unity. That leaves $1+c_2\sqrt[3]{2}+c_3\sqrt[4]{4}$.

Then, with real variables, you can slip in $\sqrt[3]{4}=\sqrt[3]{2}^2$ and thereby you have only the single radical expression $\sqrt[3]{2}$! So that is the only factor that needs an independent cube root of unity factor, and your polynomial product reduces to


where of course $\omega$ is either complex cube root of unity and $\omega^4$ in the third factor reduces to $\omega$. This still requires a little work, but nowhere near having to multiply $27$ factors.

  • $\begingroup$ can you elaborate on how you 'slip in' the equality $\sqrt[3]{4} = \sqrt[3]{2}^2$? After doing so, I get that the element in question becomes $1 + \sqrt[3]{2} + \sqrt[3]{2}^2$, but it still feels like these are 'different' radicals in some sense, and would require their own coefficients $c_1$ and $c_2$. Also, I'm not sure how you came up with $\omega^4$ in the last factor. $\endgroup$
    – Oderus
    Aug 19, 2022 at 1:40

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