$f$ is irreducible, $f$ and $g$ share a root, then $f$ divides $g$. I think I may be missing something obvious here. Say $f,g \in \mathbb{Q}[t]$ , and $f$ is an irreducible element of $\mathbb{Q}[t]$. If $\alpha \in \mathbb{C}$ is such that $f(\alpha) = 0$ and $g(\alpha) = 0$ is it true that $f$ divides $g$?
Thanks
 A: Lemma: Let $\alpha$ be a root of irredcucible $f$. Then $f$ is the minimal polynomial of $\alpha$.
Proof. Let $p$ be the minimal polynomial of $\alpha$. Suppose $f=rp+s$, $\deg s<\deg p$. Then $s(\alpha)=0$, so $s=0$ because $p$ is minimal. Hence $f=rp$, then $r$ is of degree zero because $f$ is irreducible. This proves that $f$ is minimal. 
Now we back to the problem. Suppose $g=qf+t$ with $\deg t<\deg f$. Since $f$ is minimal and $g(\alpha)=0=t(\alpha)$, we can deduce that $t=0$ as above. Therefore $g=qf$, establishing the assertion.
A: This is a possible value-add to Ming's answer from a ring-theoretic point of view.
Now, since we've established that $f$ is the minimal polynomial since it is irreducible, we proceed as follows.
Note that since $\mathbb{Q}$ is a field, we have that $\mathbb{Q}[x]$ is a Euclidean domain where we define $N(p(x)) = \deg(p(x))$. So we can write $g = q\cdot f + r$ where $\deg(r(x)) < \deg(f(x))$. Now since $g(\alpha) = f(\alpha) = 0$, we have that $r(\alpha) = 0$. By the minimality of $f(t)$, we have that $r(t) = 0$, whence $g = q\cdot f$.
