Showing that the Field Extension $\mathbb{Q}(T^{1/4})/ \mathbb{Q}(T)$ is not Galois 
Prove that $\mathbb{Q}(T^{1/4})$ is not Galois over $\mathbb{Q}(T)$, where $T$ is an indeterminate.

I am not sure how to proceed due to the indeterminate. It suffices to show that the degree of $\mathbb{Q}(T^{1/4})$ over $\mathbb{Q}$, is larger than the $\vert \text{Aut}(\mathbb{Q}(T^{1/4})/\mathbb{Q}) \vert$. The degree of the field extension is $4$ since the minimal polynomial of $T^{1/4}$ over the base field is $x^4-T$. I am not sure if the following factorization makes sense since $T$ is an indeterminate: $x^4-T=(x^2-T^{1/2})(x^2+T^{1/2})=(x-T^{1/4})(x+T^{1/4})(x-iT^{1/4})(x+iT^{1/4})$; this factorization assumes that $T$ is a positive real number. If this factorization is correct, wouldn't the Galois group have 4 elements generated by the maps $f:i \mapsto i, T^{1/4}\mapsto iT^{1/4}$, $g:i\mapsto -i, T^{1/4}\mapsto T^{1/4}$?
 A: The extension $\Bbb Q(T^{1/4})/\Bbb Q(T)$ is not a Galois extension since the minimal polynomial of $T^{1/4}$, which indeed is $X^4-T$, can not be factorized in the extension; it has roots that are not in the extension:
$$(X^4-T)=(X^2-T^{1/2})(X^2+T^{1/2})=(X-T^{1/4})(X+T^{1/4})(X^2+T^{1/2})$$
But the last factor has no roots in $\Bbb Q(T^{1/4})$.
Note that the splitting field of $X^4-T$ is $\Bbb Q(T^{1/4},i)$.
A: Your factorization perfectly works. $\mathbb{Z}[T]$ is just a factorial ring, so that you may apply what you know about general factorial rings. Its field of fractions is $\mathbb{Q}(T)$. If $K$ is any field, then $i$ denotes an element in an algebraic extension of $K$ with $i^4=1$ and $i^2 \neq 1$.
The automorphism group of $\mathbb{Q}(T^{1/4})$ over $\mathbb{Q}(T)$ is isomorphic to $\mathbb{Z}/2\mathbb{Z}$, since $\pm T^{1/4}$ are the only roots of $X^4 - T$ in $\mathbb{Q}(T^{1/4})$.
The normal closure of $\mathbb{Q}(T^{1/4})$ over $\mathbb{Q}(T)$ is $\mathbb{Q}(T^{1/4},i)$. This is the splitting field of $X^4 - T$, and has the Galois group which you have described.
