Galois Group of any polynomial over $\mathbb{R}$ This problem came up on a homework and I immediately came up with the following: The Galois Group is $Z/2Z$ if the polynomial has a complex root ($\mathbb{C} = \mathbb{R}[i]$ is algebraically closed). 
Am I correct or have I made a trivial error?

This is correct.
By the fundamental theorem of algebra, any polynomial $f(x) \in \mathbb R[x]$ splits completely into linear factors over $\mathbb C$. So the splitting field of $f(x)$ is certainly some intermediate field between $\mathbb R$ and $\mathbb C$.
But $[\mathbb C: \mathbb R] = 2$. So the only such intermediate fields are $\mathbb R$ and $\mathbb C$ themselves!
Therefore, there are two cases:


*

*All roots of $f(x)$ are in $\mathbb R$. If this is true, then the splitting field of $f(x)$ is $\mathbb R$ itself. The degree of the extension is one, and the Galois group is trivial.

*Some root of $f(x)$ is not contained in $\mathbb R$. In this scenario, the splitting field is bigger than $\mathbb R$, and the only possibility is that it is $\mathbb C$. The degree of the extension is two, and the Galois group is therefore of order two, i.e. the Galois group is $\mathbb Z_2$.
