# How to show this polynomial is irreducible over $\mathbb{Q}$?

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?

• I think you can follow the approach explained here. Nov 14, 2020 at 4:28
• At the moment I like this approach the most. Nov 14, 2020 at 4:32
• I think there is a subtraction error in the last step: the polynomial should be $x^6-9x^4-14x^3+27x^2-126x+22$.
– robjohn
Nov 14, 2020 at 14:53
• @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. Dec 3, 2020 at 13:53
• +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". Dec 3, 2020 at 13:57

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.

• Very clever argument (thanks to Jyrki Lahtonen too). +1 Dec 3, 2020 at 13:50