I was reading a paper and the author had an irreducible quadratic polynomial $f(x)$, with (non-real) root $\alpha$. He stated that if $p$ ramified or stayed prime in $\mathbb{Q}(\alpha)$ then $f(x)$ would be irreducible in $\mathbb{Q}_p$. So I thought that if we let $\alpha = a + b\sqrt{-d}$, then if $p \not = 2$ ramified then $p|d$, and thus the root would not be in $\mathbb{Q}_p$ as $\sqrt p \not \in \mathbb{Q}_p$. Therefore $f(x)$ would be irreducible over $\mathbb{Q}_p$.
My question is, first of all, is my reasoning correct and if not, what would be a correct argument. Secondly I wanted to know if this would be true in general. That is if we have an irreducible polynomial $f(x)$ with a (non-real) root $\alpha$, then if $p$ ramified or stayed prime in $\mathbb{Q}(\alpha)$ then $f(x)$ would stay irreducible in $\mathbb{Q}_p$. Obviously my argument doesn't extend because in general not having a root does not imply irreducible.
Not quite right, as $p$ may ramify in $\Bbb Q(\alpha)$ without $\sqrt p$ being in the picture. Consider $\Bbb Q(\sqrt{-3}\,)$, ramifying at $3$ only. But it’s $\sqrt{-3}$ here that isn’t in $\Bbb Q_3$, not $\sqrt3$.
As to your second question, the condition “nonreal” has nothing to do with the case. But for polynomials of degree greater than $2$, the picture can get complicated.
Consider $x^3-3\in\Bbb Q_2[x]$. Let’s use strong Hensel here, which says that if something factors in characteristic $p$ with factors that are relatively prime, then it factors upstairs. Over $\Bbb F_2$, we have $x^3-3=x^3+1=(x+1)(x^2+x+1)$, two factors that are relatively prime there. So you get a lifting in $\Bbb Q_3[x]$, namely $x^3-3=(x-\alpha)(x^2+\alpha x+\alpha^2)$, where $\alpha$ is the unique cube root of $3$ in $\Bbb Q_2$. Both these factors are irreducible.
Something moderately deep is going on here, and you’ll see what it is when you study algebraic number theory a little more deeply.