I was thinking about how to show whether or not a polynomial irreducible over $\mathbb Q$ could be irreducible over $\mathbb Q(i)$
I am aware of $x^4 + 1$ being irreducible over $\mathbb Q$ but not $\mathbb Q(i)$. However, it also has strictly complex roots. I was wondering if whenever there are only two non real roots and at least one real root, that the polynomial is irreducible over both $\mathbb Q$ and $\mathbb Q(i)$.
Are there any more examples of polynomials irreducible over $\mathbb Q$ but not $\mathbb Q(i)$ that have only two complex roots and at least one real root? I've realised that my original reasoning is actually flawed but I still suspect that maybe there isn't such a polynomial.
Also, a related question:
I am aware of many tests for irreducibility over $\mathbb Q$, but what I can I do other than check the roots, to see that a polynomial is or isn't irreducible over $\mathbb Q(i).$ I am aware that Eisenstein's criterion may be generalised to integral domains, but what if the constant of the polynomial is $2$? In that case Eisenstein wouldn't apply, so is there anything else that I could do?
Let $f$ be irreducible over $\Bbb Q$ but not over $\Bbb Q(i)$. Then over $\Bbb Q(i)$ we have the factorisation $f=g\overline g$ where the coefficients of $\overline g$ are the complex conjugates of those of $g$. The zeroes of $\overline g$ the complex conjugates of those of $g$. A real zero of $f$ is then both a zero of $g$ and of $\overline g$ and so a double zero of $f$, which we'd assumed irreducible...