Shared Roots of Polynomials Let $f,g\in \mathbb{Z}[x]$ be two irreducible polynomials over $\mathbb{Z}$ with $g$ having degree at least 2. Let $\alpha$ be their common root in $\mathbb{C}$. Prove that $f$ and $g$ share another common root in $\mathbb{C}$ different from $\alpha$.
If $\alpha$ is complex, I can guarantee another root in $\overline{\alpha}$, since $f(\overline{x})=\overline{f(x)}$ for polynomials.
I'm completely lost in the case that $\alpha$ isn't complex.
 A: Hint: Since $f$ and $g$ both have $\alpha$ as a root, then as complex polynomials we have $$(x-\alpha)\mid \gcd(f,g)$$ (This leads us to a much stronger conclusion than "they share another root".)
A: Let $S = \{p(x) \in \mathbb{Q}[x]\setminus\{0\}: p(\alpha) = 0\}$. Then $f,g \in S$, so $S$ is non-empty, and it follows that we may choose some $m(x) \in S$ of least degree. Moreover, we may assume that $m(x)$ is monic, since dividing by the leading coefficient does not change the degree.
Lemma: The polynomial $m(x)$ is a common divisor of $f(x)$ and $g(x)$ in $\mathbb{Q}[x]$.
Proof: By the division algorithm for $\mathbb{Q}[x]$, we have
$$
f(x) = q(x)m(x) + r(x)
$$
for some rational polynomials $q(x), r(x)$ with $\deg r(x) < \deg m(x)$. Since $f(\alpha) = m(\alpha) = 0$, it follows that $r(\alpha) = 0$, so $r(x) = 0$ by minimality of $\deg m(x)$. Therefore $m(x)\mid f(x)$ in $\mathbb{Q}[x]$. Similarly, $m(x) \mid g(x)$, and the result follows.
$$\tag*{$\blacksquare$}$$
Now, $m(\alpha) = 0$, so $(x - \alpha)\mid m(x)$ in $\mathbb{C}[x]$. Therefore $\deg m(x) > 1$, so $m(x)$ is not a unit in $\mathbb{Q}[x]$, which means that $f(x) = \lambda m(x)$ and $g(x) = \mu m(x)$ for some rational $\lambda, \mu$ (this is by definition of irreducibility). So in fact $f(x)$ and $g(x)$ share all their roots. The result then follows from the following lemma.
Lemma: Let $f(x) \in \mathbb{Q}[x]$ be irreducible. Then $f(x)$ has no repeated roots in $\mathbb{C}$.
Proof: Let $f'(x)$ be the formal derivative of $f(x)$. That is, if
$$
f(x) = a_nx^n + \ldots + a_0,
$$
we define
$$
f'(x) = na_nx^{n-1} + \ldots + a_1.
$$
Since $\deg f' < \deg f$, and $f$ is irreducible, the polynomials $f$ and $f'$ are coprime, so by Bezout's Lemma there exist polynomials $a(x),b(x)\in\mathbb{Q}[x]$ such that
$$
a(x)f(x) + b(x)f'(x) = 1.
$$
Suppose that $f(x)$ has a repeated root $\alpha$. Then $(x-\alpha)^2 \mid f(x)$ in $\mathbb{C}[x]$, so $f(x) = (x - \alpha)^2q(x)$ for some $q(x) \in \mathbb{C}[x]$. It follows that
$$
f'(x) = 2(x-\alpha)q(x) + (x-\alpha)^2q(x),
$$
so $f(\alpha) = f'(\alpha) = 0$. But then
$$
1 = a(\alpha)f(\alpha) + b(\alpha)f'(\alpha) = 0,
$$
which is a contradiction.
$$\tag*{$\blacksquare$}$$
