Determine the degree of the field extension
($1$st task) $\mathbb{Q}(\sqrt[3]{5},\sqrt[4]{6},\sqrt[5]{7})/\mathbb{Q}$,
($2$nd task) $\mathbb{Q}(\sqrt{3}, \sqrt[10]{75})/\mathbb{Q}$,
($3$rd task) $N/\mathbb{Q}$, where $N$ is the splitting field of the polynomial $X^5-3\in \mathbb{Q}[X]$.
Studying for my upcoming algebra exam, I found this exercise from an old algebra exam.
So far I have:
- The first task
For the $\sqrt[3]{5}$ we have $\alpha = \sqrt[3]{5} \rightarrow \alpha^3-5=0$ is the minimal polynomial, so we have $f(t)=t^3-5$. The Eisenstein criteria is fulfilled with $p=5$, so the degree is $\deg(f(t))=3$.
So for $\sqrt[4]{6}$ the degree should be $4$ and for $\sqrt[5]{7}$ the degree should be $5$. But I am not sure how to proceed from here, because normally you should factorize the given numbers which is not possible here.
- The second task
$\sqrt[2]{3}$ so the minimal polynomial should be $\alpha^2-3=0$ and the degree is $2$. The term $\sqrt[5]{75}$ can be factorized into $\sqrt[5]{75}=\sqrt[5]{25} \cdot \sqrt[5]{3}=\sqrt[5]{5} \cdot \sqrt[5]{5} \cdot \sqrt[5]{3}$, but I am unsure how to proceed to determine degree from here.
- The third task
For the third task I unfortunately have no idea how to start solving it. Maybe someone could give me hint?
Edit Thanks for the quick answer. For the third task the splitting field is $\mathbb Q(\zeta_5,\sqrt[5]3)$. So I think the degree of $\mathbb{Q}[\sqrt[5]{3}]$ over $\mathbb{Q}$ is 5 because the polynomial is irreducible in $\mathbb{Q}$. And the solutions for the polynomial are $\alpha=\sqrt[5]{3}$ times the $5$th roots of unity $\zeta_5=e^{\frac{2\pi i}{5}}$, that is $\alpha, \alpha \zeta_5, \alpha \zeta_5^2, \alpha \zeta_5^3$ and $\alpha \zeta_5^4$. Then the degree of $\mathbb{Q}[\zeta_5]$ over $\mathbb{Q}$ is 4. So the degree of the splitting field should be $[\mathbb{Q}[\alpha, \zeta_5]:\mathbb{Q}]=20$, because $4$ and $5$ have no common divisor.
Is this right?
Thanks in advance!
For the first task: You already noted that \begin{align*} [\Bbb Q(\sqrt[3]{5}):\Bbb Q]&=3\\ [\Bbb Q(\sqrt[4]{6}):\Bbb Q]&=4\\ [\Bbb Q(\sqrt[5]{7}):\Bbb Q]&=5 \end{align*} Since these degrees are pairwise coprime the corresponding field extensions are pairwise linearly disjoint and we get as you did in task three: \begin{align*} [\Bbb Q(\sqrt[3]{5},\sqrt[4]{6},\sqrt[5]{7}):\Bbb Q]&=[\Bbb Q(\sqrt[3]{5}):\Bbb Q][\Bbb Q(\sqrt[4]{6}):\Bbb Q][\Bbb Q(\sqrt[5]{7}):\Bbb Q]=3\cdot 4\cdot 5 = 60 \end{align*}
For the second task: We can do this actually a bit quicker than the way I hinted in the comment. Note that $$(\sqrt[10]{75})^5=5\cdot\sqrt{3}$$ Hence $$\Bbb Q(\sqrt 3,\sqrt[10]{75})=\Bbb Q(\sqrt[10]{75})$$ And $\Bbb Q(\sqrt[10]{75})$ has degree $10$ over $\Bbb Q$ since $t^{10}-75$ is irreducible.