Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Can someone please give me a tip how to show that for distinct prime $p,q$ that $\mathbb{Q}(\sqrt{p},\sqrt[3]{q})=\mathbb{Q}(\sqrt{p}\sqrt[3]{q})$ and how to find the minimal polynomial of $\sqrt{p}\sqrt[3]{q}$ over $\mathbb{Q}$ ?

share|improve this question

1 Answer 1

up vote 4 down vote accepted

Clearly $\mathbb Q(\sqrt p\sqrt[3]q)\subseteq \mathbb Q(\sqrt p,\sqrt[3] q)$. Also, if $\alpha = \sqrt p\sqrt[3]q$, then we see that $\alpha^3=pq\sqrt p$ and also $\frac{\alpha^3}{pq}=\sqrt p\in \mathbb Q(\sqrt p\sqrt[3]q)$. Similarly, $\frac{\alpha^4}{p^2q}=\sqrt[3]q\in\mathbb Q(\sqrt p\sqrt[3]q)$. We conclude $\mathbb Q(\sqrt p\sqrt[3]q)\supseteq \mathbb Q(\sqrt p,\sqrt[3] q)$.

For the degree of the extension, note that clearly $[\mathbb Q(\sqrt p):\mathbb Q]=2$ and $[\mathbb Q(\sqrt[3] q):\mathbb Q]=3$, hence $[\mathbb Q(\sqrt p,\sqrt[3] q):\mathbb Q]$ is

  • either 2 or 6 from the tower $\mathbb Q(\sqrt p,\sqrt[3] q)\rightarrow \mathbb Q(\sqrt p)\rightarrow \mathbb Q$
  • either 3 or 6 from the tower $\mathbb Q(\sqrt p,\sqrt[3] q)\rightarrow \mathbb Q(\sqrt[3] q)\rightarrow \mathbb Q$

Hence $[\mathbb Q(\sqrt p,\sqrt[3] q):\mathbb Q]=6$ and we conclude that the minimal polynomial of $\sqrt p\sqrt[3]q$ is of degree $6$, hence it is the obvious candidate $X^6-p^3q^2$.

Bonus questions: Where did I use that $p,q$ are prime? Where did I use that $p\ne q$?

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.