I'm taking a class on Galois Theory in another language and the prof is saying my answer on this is incorrect and I'm wondering why, particularly since sometimes there's a language barrier. Basically the question is this:
Using the primitive element theorem, show that:
$\mathbb{Q} \big[ \sqrt{2}, \sqrt[3]{2} \big] = \mathbb{Q} \big[ \sqrt{2} + \alpha \sqrt[3]{2} \big] \, \, \forall \alpha \in \mathbb{Q} - \{0\}$.
I basically said the primitive elt theorem states there is some alpha satisfying the above, and that by an isomorphism of vector spaces $f_c : x + ty \to x + tcy$ which trivially respects the operations of a vector space, one can take any t and convert it into any alpha by simply multiply by alpha over t....
Where have I gone wrong?
Let's denote $\mathbb{Q}[\sqrt{2},\sqrt[3]{2}]$ by $K$. If you have an isomorphism of fields $\phi: K \rightarrow L$ over $\mathbb{Q}$, then such an isomorphism necessarily acts as the identity map on $\mathbb{Q}$ (simply by working through the definition of field homomorphism; any such map must send $1$ to $1$, then by addition you get that it sends $2$ to $2$, $3$ to $3$ etc., and then by taking quotients $\phi(m/n) = \phi(m)/\phi(n) = m/n$ for $m,n \in \mathbb{Z}$).
Thus if we take $\alpha \in K$, satisfying some irreducible polynomial $f(X) \in \mathbb{Q}[X]$, then our field isomorphism must take $\alpha$ to another root of $f(X)$ (as $\phi(f(X)) = f(X)$ as elements of $\mathbb{Q}[X]$, where here we act on the coefficients, and thus $0 = f(\alpha) = \phi(f(\alpha)) = f(\phi(\alpha))$). The map you've written down above doesn't in general take $\sqrt{2} + t\sqrt[3]{2}$ to another root of its minimal polynomial. If it did, then such a polynomial would have an infinite number of roots in $\overline{\mathbb{Q}}!$
The mistake you've made here is that whilst any homomorphism of fields is determined by what you send the primitive element to, this is subject to the condition that this primitive element is still sent to some root of its minimal polynomial. Not every vector space isomorphism will be a field isomorphism, else all quadratic extensions of $\mathbb{Q}$ would be isomorphic as fields!