Find the smallest normal extension (upto isomorphism) of $\Bbb{Q}(2^{1/4}, 3^{1/4})$ in $\overline{\Bbb{Q}}.$

Question

I have taken the polynomial $p(x)=(x^2+\sqrt{2})(x^2+\sqrt{3})$ over $\Bbb{Q}(2^{1/4}, 3^{1/4})$.
Now, $p(x)$ doesn't have any root in $\Bbb{Q}(2^{1/4}, 3^{1/4})$, more explicitly roots of $p(x)$ are $+i\sqrt{2}$, $-i\sqrt{2}$, $+i\sqrt{3}$, $-i\sqrt{3}$.
Then it is easy to see $\Bbb{Q}(2^{1/4}, 3^{1/4},+i\sqrt{2}, -i\sqrt{2}, +i\sqrt{3}, -i\sqrt{3})=\Bbb{Q}(2^{1/4}, 3^{1/4}, i)$, hence $\Bbb{Q}(2^{1/4}, 3^{1/4}, i)$ is splitting field of $p(x)\implies\Bbb{Q}(2^{1/4}, 3^{1/4}, i)$ is normal extension of $\Bbb{Q}(2^{1/4}, 3^{1/4})$. And $[\Bbb{Q}(2^{1/4}, 3^{1/4}, i):\Bbb{Q}(2^{1/4}, 3^{1/4})]=2$ since $i$ is a root of the irreducible polynomial $x^2+1$ over $\Bbb{Q}(2^{1/4}, 3^{1/4})$.
MY CLAIM: $\Bbb{Q}(2^{1/4}, 3^{1/4}, i)$ is the smallest(upto isomorphism) normal extension of $\Bbb{Q}(2^{1/4}, 3^{1/4})$. But I can't prove the smallest(upto isomorphsm) thing.
Can anybody show that this extension is smallest one(upto isomorphism)?
Thanks for assistance in advance.

Answer 1

The degree of $\mathbb Q(2^{1/4}, 3^{1/4}, i)$ over ${\mathbb Q}(2^{1/4}, 3^{1/4})$ is $2$. So there are no smaller proper extensions.