Proving something is not a Normal Extension Let $M = \mathbb{Q}(\sqrt{3}, i\sqrt[4]{5})$ be an extension of $\mathbb{Q}$. Then work out the basis of $M$ over $\mathbb{Q}$ and show that the extension $M/\mathbb{Q}$ is not a normal extension. So this is how I showed it and just wanted to check that it is actually a correct way of going about it:
So I calculated the basis correctly I think, but my concern is regarding the normal extension  bit. So I deduced that $M/\mathbb{Q}$ is separable since its finite and characteristic of $\mathbb{Q}$ is $0$ so to show that M isn't a normal extension, I just showed that $M$ isnt the splitting field of the following  polynomial $f = (x^2 - 3)(x^4 - 5)$. $f$ is separable since it has distinct roots but its splitting field is not equal to $M$, so hence we can deduce that $M/\mathbb{Q}$ is not a normal extension. Is this correct?

You are trying to show the right thing, but be careful. There are of course many polynomlas $M$ is not the splitting field of. And even if a polnomial $f$ happens to split in $M$ we can still multiply it with suitable $g$ such that $fg$ does not split. Therefore, to show that $M$ is normal, it is best to exhibit a polynomial $f$ such that


*

*$f$ is irrducible over $\mathbb Q$

*$f$ has some root in $M$

*$f$ does not have all roots in $M$ (i.e., doesn't cplit completely)


Your choice of an - obviously - reducible polynomial is therefore "suspicious". 
Since $\sqrt 3\in M$ implies $-\sqrt 3\in M$, the polynomial $X^2-3$ is not suitable for this task. However, the other generator $i\sqrt[4]5$ can helkp us: It is certainly a root of $f(X)=X^4-5$. So try to answer the three bullit points for this $f$.
