Degree of extension to the splitting field - where's the mistake? I have this false prove which I can't find the problem in.
Trying to find the field extension degree $[E:\mathbb{Q}]$ where $E$ is the splitting field of the polynomial $f(x)=x^3-5$ over $\mathbb{Q}$.
My false proof is:
The roots of $f$ in E are $a,b,c$, we don't need to state them precisely at the moment.
Since $f$ is irreducible over $\mathbb{Q}$, $f(x)=irr(a,\mathbb{Q})=irr(b,\mathbb{Q})=irr(c,\mathbb{Q})$, and hence $\mathbb{Q}/(x^3-5) \cong \mathbb{Q}(a) \cong \mathbb{Q}(b) \cong \mathbb{Q}(c)$, therefore $b,c\in \mathbb{Q}(a)$, therefore $E=\mathbb{Q}(a,b,c)=\mathbb{Q}(a)$, therefore $[E:\mathbb{Q}]=3$.
This contradicts several other things, which are the reason I believe this prove is false. Can you spot the not?
Thanks,
G.
 A: The error is in "$\mathbb{Q}(a) \cong \mathbb{Q}(b) \cong \mathbb{Q}(c)$ therefore $b,c\in \mathbb{Q}(a)$". The three fields you list are isomorphic but distinct subfields of $E$, which has degree $6$ over $\Bbb Q$. Indeed one can see that $\frac ab,\frac bc, \frac ac$ are all roots of $X^2+X+1$, which is an irreducible quadratic polynomial over $\Bbb Q$, whence neither of these roots can lie in any of the degree $3$ extensions listed, so each of those extensions can contain only one of $a,b,c$.
A: As others said, your mistake is to assume that isomorphism of fields implies equality. To make this absolutely clear consider the following. Exactly one of the roots of $x^3-5$ is real. Therefore exactly one of those three fields consists of real numbers only - the other two do not. Hence they cannot be equal (even though they are isomorphic).
A: Your error is with the misunderstanding of what an isomorphism is and what it does. The two groups $\mathbb{Z}/2\mathbb{Z}$ and the group of permutations of $\{1,2\}$ are isomorphic but they have completely different elements (one consists of classes of integers, the other permutations).
Just because the three fields you have are isomorphic it doesnt mean they are equal.
