QUESTION: Show that, $\sqrt{3}+\sqrt[3]{7}$ is algebraic over $\mathbb{Q}$ with degree $6$.

I'm allowed to use this definition: We say that $a \in K$ is algebraic of degree $n$ over $F$ if the minimal polynomial of $a$ over $F$ has degree $n$, i.e., $\deg(Irr_{F}(a))(x)=n$.

MY ATTEMPT: Defining $\alpha:=\sqrt{3}+\sqrt[3]{7}$ we are going to obtain a polinomial $p(x)$ such that $p(\alpha)=0$. Let's start: \begin{align*} \alpha = \sqrt{3}+\sqrt[3]{7} &\implies \alpha -\sqrt{3}=\sqrt[3]{7}\\ &\implies(\alpha -\sqrt{3})^3=7\\ &\implies\alpha^3-3\alpha^2 \sqrt{3}+9\alpha -3\sqrt{3}=7\\ &\implies (\alpha^3 +9\alpha -7)^2=3(3\alpha^2+3)^2\\ &\implies \alpha^6+9\alpha^4-14\alpha^3+27\alpha^2-126\alpha+22=0 \end{align*}

Therefore, $\alpha$ is a root of $p(x)= x^6+9x^4-14x^3+27x^2-126x+22$, where $p(x)\in \mathbb{Q}[x]$ is monic polynomial.

MY DOUBT: Now, it is necessary to show that $p(x)$ is irreducible over $\mathbb{Q}$ in order to conclude this exercise. However here is my problem:

  1. I can't use Eiseinstein criterion, because doesn't work, once there is not any p prime that is suitable to show irreducibility.

  2. If I show all roots by use of De Moivre formula's is not enough. Once we have this result: If a polynomial is irreducible over $F$ then there is not any root of this polinomial over $F$. But, we do not have the opposite implication as a result! So, is not enough use De Moivre formula's.

Would someone help me with this part?

  • 2
    $\begingroup$ I think you can follow the approach explained here. $\endgroup$ Nov 14, 2020 at 4:28
  • 3
    $\begingroup$ At the moment I like this approach the most. $\endgroup$ Nov 14, 2020 at 4:32
  • 1
    $\begingroup$ I think there is a subtraction error in the last step: the polynomial should be $x^6-9x^4-14x^3+27x^2-126x+22$. $\endgroup$
    – robjohn
    Nov 14, 2020 at 14:53
  • $\begingroup$ @JyrkiLahtonen: in cases where that technique works it is arguably the best. The proofs behind the theorems used in that approach are easy compared to using any direct theorems for checking irreducibilty. $\endgroup$
    – Paramanand Singh
    Dec 3, 2020 at 13:53
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    $\begingroup$ +1 for the question. This is how one can ask a good question. Very clear about the attempt and the doubt/block. So much better than just "I am stuck and don't know what to do". $\endgroup$
    – Paramanand Singh
    Dec 3, 2020 at 13:57

1 Answer 1


Following Jyrki Lahtonen suggestion, I will use the approach described here.

First, I prove that $\mathbb{Q}(\alpha)$ contains $\mathbb{Q}(\sqrt{3})$:

Notice that: $\alpha = \sqrt{3} + \sqrt[3]{7} \implies \alpha - \sqrt{3} = \sqrt[3]{7} \implies (\alpha - \sqrt{3})^{3} = (\sqrt[3]{7})^{3} \implies \alpha^{3} - 3\alpha^{2}\sqrt{3} + 9\alpha - 3\sqrt[3]{3} = 7 \implies \alpha^3 + 9\alpha - 7 = 3\alpha^2 \sqrt{3} - 3\sqrt{3} \implies \alpha^3 + 9\alpha - 7 = \sqrt{3} (3\alpha^2 - 3).$ Therefore $$\sqrt{3} = \frac{\alpha^3 + 9\alpha - 7}{(3\alpha^2 - 3)} (*)$$ and thus we conclude that $\mathbb{Q}(\alpha)$ contains $\mathbb{Q}(\sqrt{3})$.

Then, I prove that $\mathbb{Q}(\alpha)$ contains $\mathbb{Q}(\sqrt[3]{7})$:

Notice that $\sqrt[3]{7} = \alpha - \sqrt{3}$ and that $\mathbb{Q}(\alpha)$ contains both $\alpha$ and $\sqrt{3}$ (by the Equation ($*$) ). Thus, $\mathbb{Q}(\alpha)$ contains $\sqrt[3]{7}$ and therefore $\mathbb{Q}(\alpha)$ contains $\mathbb{Q}(\sqrt[3]{7})$.

Since $[\mathbb{Q}(\sqrt{3}) : \mathbb{Q}] = 2$ and $[\mathbb{Q}(\sqrt[3]{7}): \mathbb{Q}] = 3$, we obtain that the $[\mathbb{Q}(\alpha): \mathbb{Q}] $ is a multiple of 6:

Notice that:

$[\mathbb{Q}(\alpha): \mathbb{Q}] = [\mathbb{Q}(\alpha): \mathbb{Q}(\sqrt{3})] [\mathbb{Q}(\sqrt{3}) : \mathbb{Q}] = 2 [\mathbb{Q}(\alpha): \mathbb{Q}(\sqrt{3})] $ $[\mathbb{Q}(\alpha): \mathbb{Q}] = [\mathbb{Q}(\alpha): \mathbb{Q}(\sqrt[3]{7})] [\mathbb{Q}(\sqrt[3]{7}) : \mathbb{Q}] = 3 [\mathbb{Q}(\alpha): \mathbb{Q}(\sqrt[3]{7})]$

and therefore, $[\mathbb{Q}(\alpha): \mathbb{Q}] $ is a multiple of both 2 and 3 (i.e. a multiple of 6).

As you found a monic polynomial $p$ with degree 6, we conclude.

  • 2
    $\begingroup$ Very clever argument (thanks to Jyrki Lahtonen too). +1 $\endgroup$
    – Paramanand Singh
    Dec 3, 2020 at 13:50

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