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

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?

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!
• I got $\alpha^{15}=2^7 \sqrt{2} \cdot 3^5 \cdot 5^3 \Rightarrow \sqrt{2} \in M$ and $\alpha^6=2^3 \cdot 3^2 \cdot 5 \sqrt[5]{5} \Rightarrow \sqrt[5]{5} \in M$, so $L \subset M$ and $\mathbb{Q} \subset \mathbb{Q}(\sqrt{2},\sqrt[3]{3},\sqrt[5]{5})$ is a simple extension. – Gurterz Dec 9 '19 at 18:40