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:
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I can't use Eiseinstein criterion, because doesn't work, once there is not any p prime that is suitable to show irreducibility.
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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?
Best Answer
Following Jyrki Lahtonen suggestion, I will use the approach described here.
First, I prove that $\mathbb{Q}(\alpha)$ contains $\mathbb{Q}(\sqrt{3})$:
Then, I prove that $\mathbb{Q}(\alpha)$ 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:
As you found a monic polynomial $p$ with degree 6, we conclude.