$\mathbb{Q} \subset \mathbb{Q}(\sqrt{2},\sqrt[3]{3},\sqrt[5]{5})$ (simple extension)

Question

Let $\mathbb{Q} \subset \mathbb{Q}(\sqrt{2},\sqrt[3]{3},\sqrt[5]{5})$.

How to show that this is a simple extension without the primitive element theorem?

My idea was to show that $\mathbb{Q}(\sqrt2, \sqrt[3]{3}, \sqrt[5]{5})=\mathbb{Q}(\sqrt{2} \cdot \sqrt[3]{3} \cdot \sqrt[5]{5})$.

The field $L:= \mathbb{Q}(\sqrt2, \sqrt[3]{3}, \sqrt[5]{5})$ contains $\sqrt 2, \sqrt[3]{3}$ and $\sqrt[5]{5}$.

So $\mathbb{Q}(\sqrt 2 + \sqrt[3]{3} + \sqrt[5]{5}) \subset L$.

To show that $L \subset \mathbb{Q}(\sqrt 2 + \sqrt[3]{3} + \sqrt[5]{5})=:M$, I tried to indicate that $\sqrt2, \sqrt[3]{3}, \sqrt[5]{5} \in M$.

$\sqrt 2 + \sqrt[3]{3} + \sqrt[5]{5} \in M \Rightarrow (\sqrt 2 + \sqrt[3]{3} + \sqrt[5]{5})^2 \in M$

But now I don't know how to continue.

How to show that this is a simple extension?

Answer 1

I like your original idea of showing that $\mathbb{Q}(\sqrt2, \sqrt[3]{3}, \sqrt[5]{5})=\mathbb{Q}(\sqrt{2} \cdot \sqrt[3]{3} \cdot \sqrt[5]{5})$. Now, let $L$ be the field $\mathbb{Q}(\sqrt2, \sqrt[3]{3}, \sqrt[5]{5})$ and let $M$ be the field $\mathbb{Q}(\sqrt{2} \cdot \sqrt[3]{3} \cdot \sqrt[5]{5})$.

If I understand correctly, I think you already know how to prove $M\subseteq L$, so I will focus on proving that $L\subseteq M$. First, let $\alpha$ be $\sqrt{2} \cdot \sqrt[3]{3} \cdot \sqrt[5]{5}$. It is easy to see that $\alpha \in M$, by definition. Now, take $\alpha$ to the $10^{\text{th}}$ power. This will get rid of the square and fifth roots, so we get: $$\alpha^{10}=2^5\cdot 3^3\sqrt[3]{3}\cdot 5^2$$ From here, one can see that $\sqrt[3]{3}\in M$, because you just need to divide $\alpha^{10}$ by an integer to get $\sqrt[3]{3}$.

I think the above demonstrates the basic idea behind the rest of the proof, so I will leave it as an exercise to you to show that $\sqrt 2$ and $\sqrt[5]{5}$ are in $M$ as well. Good luck!