Let $K$ be a field of characteristic 0, $f \in K[X]$ and $F$ the splitting field of $f$. Is it true that if Gal$(F|K)$ acts transitively on the roots of $f$, then $f$ ir irreducible? From all the results I have seen, this seems to be true. However, this one example is making me confused.
Take $f = (x^2 + 1)^2$, $K = \mathbb{Q}$. Then $F = \mathbb{Q}(i)$ and $G = \{Id_{F}, \sigma\}$, where $\sigma(i) = -i$. The set of roots of $f$ is $\{i, -i\}$, and it seems to me that $G$ is acting transitively on this set.
Am I missing something very basic?
The answer to your question is yes it will be irreducible. To prove this pick a subset of the galois group on which it does act transitively and set $g(x)$ to be the polynomial with precisely these galois conjugates as its roots. It is quite easy to show that this polynomial both divides $f(x)$ and is in $\mathbb{Q}(x)$ (to see this note that all of the coefficients are symmetric polynomials in these galois conjugates so the action of any element of the galois group will fix these and are in $\mathbb{Q}$). The reason your example causes some problems is that when we talk about the polynomial for which a given field is a splitting field we tend to not pointlessly adjoin multiple copies since they give the same field.