Polynomial divisibility Given $p(x) \in \mathbb Q[x] $ an irreducible polynomial, and $\alpha \in\mathbb C $ root of $p(x)$.
Prove that if $q(x) \in \mathbb Q[x]$ it's a polynomial, such $q(\alpha) = 0$ then $p(x) \mid q(x)$ suggestion: considerate $(p:q)$
OK, for this exercise, I thought that given that $\alpha$ is a root and is in $\mathbb C $ then $\overline \alpha $ its also a root of $p(x)$
Then, $p(x) = (x-\overline \alpha)(x-\alpha) h(x)$ where $h(x) \in \mathbb C[x]$
And given that $q(\alpha) = 0$ it also can be written as $q(x) = (x-\overline \alpha)(x-\alpha) j(x)$ where $j(x) \in \mathbb C[x]$
Then, using the suggestion $(p:q) = (x-\overline \alpha)(x-\alpha) (h:j)$
And then I get stuck, its $h(x) = 1$?
How can I justify that?
 A: Let me introduce a lemma for this.

Lemma: If $p(x) \in \mathbb{Q}[x]$ is irreducible over $\mathbb{Q}$ and has a root $\alpha \in \mathbb{C}$, then $p(x)$ is a minimal degree polynomial over $\mathbb{Q}$ which has root $\alpha$.

Proof of the lemma:
Let $m(x)$ be a minimal degree polynomial which has $\alpha$ as root. From division algorithm, there exist unique polynomials $s(x), r(x) \in \mathbb{Q}[x]$ such that
$$ p(x) = s(x)m(x) + r(x), \qquad \deg(r) < \deg(m). $$
Since $m(x)$ and $p(x)$ have the common root $\alpha$,
$$ r(\alpha) = p(\alpha) - s(\alpha)m(\alpha) = 0. $$
As the degree of $r(x)$ is less than that of $m(x)$, and $m(x)$ is of minimal degree, the remainder $r(x)$ must be the zero polynomial.
Hence $p(x) = s(x)m(x)$. The irreducibility of $p(x)$ forces $s(x)$ to be a nonzero constant polynomial, and $p(x)$ and $m(x)$ have the same degree. $\square$
Now for the main result:
Proof:
From division algorithm, there exist unique polynomials $s(x), r(x) \in \mathbb{Q}[x]$ such that
$$ q(x) = s(x)p(x) + r(x), \qquad \deg(r) < \deg(p). $$
Since $p(x)$ and $q(x)$ have a common root $\alpha$,
$$ r(\alpha) = q(\alpha) - s(\alpha)p(\alpha) = 0. $$
From the above lemma, the remainder $r(x)$ must be the zero polynomial. Therefore $p(x) \mid q(x)$ as required. $\square$
A: Using the complex conjugate root may be useful for real polynomials, but not for polynomials with rational coefficients (though it gives another root, there may be other roots of $p$ still).
Substitution for $x$ of an element$~a$ of any ring$~R$ containing$\def\Q{\Bbb Q}~\Q$ always defines a ring morphism $\Q\to R: p[x]\mapsto p[a]$.* You can apply this for $R=\Bbb C$ and $a=\alpha$, so that $p[x]\mapsto p[a]$ is a ring morphism $f:\Q\to\Bbb C$, and since $p[\alpha]=0$ is given one has $p[x]\in\ker(f)$. Now $\ker(f)$ is an ideal of $\Q[x]$ which is a PID, so the ideal is generated by a single element $d[x]\in\Q[x]$, which visibly is a divisor of$~p[x]$. Now $p[x]$ is irreducible and clearly $1\notin\ker(f)$ (substitution does not change constant polynomials), so the $d[x]$ must be associated to $p[x]$ (a nonzero scalar multiple of it); then $p[x]$ itself also generates $\ker(f)$. This means that all $q[x]\in\ker(f)$, that is, those with $q[\alpha]=0$, are (polynomial) multiples of $p[x]$, which is precisely what was asked to prove.
*I always use square brackets to denote substitution, as this avoids ambiguities that using parentheses can cause. I also will follow the habit in he question of denoting plain polynomials by substituting $x$ into them (for $x$, a rather pointless operation), though as you see this habit quickly gets extremely tiresome.
