Showing $\mathbb{Q}(\sqrt[]{m})\neq \mathbb{Q}(\sqrt[]{m},\sqrt[]{n})$

Question

Let $m,n\in \mathbb{Z}$ be square free integers such that $m\neq n$

Show $\mathbb{Q}(\sqrt[]{m})\neq \mathbb{Q}(\sqrt[]{m},\sqrt[]{n})$

My approach to doing this is showing that $\sqrt[]{m}\notin \mathbb{Q}(\sqrt[]{n})$

I am trying to do this currently by supposing by contradiction that we could write $\sqrt[]{m}$ as

$$\frac{a+b\sqrt[]{n}}{c+d\sqrt[]{n} }$$

By rationalizing the denominator we get

$$c\sqrt[]{m}+d\sqrt[]{mn}-b\sqrt[]{n}=a$$

I have other results, but the important thing is if I can show that the sum on the left is not a rational, I am done

I know that all the roots are distinct (even if the middle one reduces a little)

Is there a result about finite rational sums of square roots that could help show that this result is not possible? (Since $m\neq n$ we know all the roots are different square free integers)

Answer 1

If $$\sqrt{m}=a+b\sqrt{n}$$ with $a$ and $b$ rational then

$$m=a^2+b^2n+2ab\sqrt{n}$$ this is only possible if $ab=0$, and $b\neq 0$ from the first equation, so $a=0$ and we have $$\sqrt{m}=b\sqrt{n}$$ so $mn$ is a perfect square.

Answer 2

I think you’ve made things too difficult for yourself by taking a quotient of two expressions of form $A+B\sqrt n$. After all, you have $$ \frac{a+b\sqrt n}{c+d\sqrt n}=\frac{(a+b\sqrt n)(c-d\sqrt n)}{c^2-nd^2}=A+B\sqrt n\,, $$ for rational numbers $A$ and $B$ that you can figure out as easily as I can.

Then you just want to show that for rational numbers $A$ and $B$, it is not possible to have $(A+B\sqrt n)^2=m$, and I think you should be able to do that without much trouble.