Polynomials in $\Bbb Q[x]$ with same real root dont have common divisor with degree more than 1

Question

Let $f,g\in \Bbb Q[x]$ polynomyals with the same real root $\alpha \in \Bbb R$. I'm asked wether or not $f$ and $g$ must have a common divisor $h\in \Bbb Q[x]$ with $\deg(h) \geq 1$.

I believe that the answer is yes, I just cant justify that.

Edit: I think is solved that. Let $I = \{p\in \Bbb Q[x] \mid p(\alpha)=0 \}$. Obviously $I$ is an Ideal of $\Bbb Q[x]$, and since $\Bbb Q$ is a field $\Bbb Q[x]$ is euclid Ring and therefore principal domain, so i'll just take the creator of I as a common divisor.

Answer 1

If a real (or complex) number $\alpha$ is the root of some non-zero polynomial in $\mathbb{Q}[x]$, then we say $\alpha$ is algebraic. Any algebraic number has a minimal polynomial $m_\alpha\in\mathbb{Q}[x]$ (Wikipedia link), which can be characterized in many ways:

• Out of the monic polynomials in $\mathbb{Q}[x]$ with $\alpha$ as a root, it is the one of least degree.

• Out of the monic polynomials in $\mathbb{Q}[x]$ with $\alpha$ as a root, it is the only one that is irreducible.

• Out of the polynomials that generate the kernel of the evaluation homomorphism $$\mathrm{ev}_\alpha:\mathbb{Q}[x]\to\mathbb{C},\quad \mathrm{ev}_\alpha(f)=f(\alpha)$$ it is the monic one (note that $\mathbb{Q}[x]$ is a PID and the kernel is an ideal so the kernel can be generated by a single element.)

Note that since $m_\alpha$ is irreducible, it certainly has degree $\geq 1$.

Since we know that the given $\alpha$ is the root of some non-zero polynomial in $\mathbb{Q}[x]$, it is algebraic, and has its minimal polynomial $m_\alpha$. Use any of the characterizations above to prove that $f$ and $g$ must both be multiples of $m_\alpha$. It would be useful to keep in mind that $\mathbb{Q}[x]$ is a Euclidean domain, i.e., it has a division algorithm (statement EF1 on Wikipedia).

Answer 2

If $\alpha \in \mathbb{Q}$, then the problem is trivial since $(x-\alpha)$ will be a common divisor for all polynomials with $\alpha$ as a root.

Suppose $\alpha \not\in \mathbb{Q}$, then there is a minimal polynomial $h(x) \in \mathbb{Q}[x]$ with $\alpha$ as a root. What we want to show is that, if there exists a $f(x) \in \mathbb{Q}[x]$ with $\alpha$ as a root, then $f(x) \in \langle h(x) \rangle$.

Well, consider the evaluation homomorphism $ev_\alpha:\mathbb{Q}[x] \rightarrow \mathbb{Q}[\alpha]$ defined as follows: $g(x) \mapsto g(\alpha)$. If you haven't seen this before, you'll want to write up a quick proof that this is indeed a homomorphism.

The kernel of this homomorphism is an ideal in $\mathbb{Q}[x]$, which will be the set of all polynomials with $\alpha$ as a root. Since $\mathbb{Q}$ is a field, then $\mathbb{Q}[x]$ is a principal ideal domain. That is, every ideal in $\mathbb{Q}[x]$ is generated by a single element. For completeness, you'll want to prove this fact as well. Therefore, $\ker(ev_\alpha)$ must be generated by a single element.

Can you see why that element must be $h(x)$? And why $f(x) \in \langle h(x) \rangle$? From here, you can conclude that $h(x)$ will be a common divisor of all polynomials with $\alpha$ as a root.

Answer 3

Since $f(\alpha) = g(\alpha) = 0$, $\alpha \in \Bbb R$ is algebraic over $\Bbb Q$; thus there exists a polynomial $0 \ne m(x) \in \Bbb Q[x]$ with $\deg m(x)$ minimal among all polynomials $0 \ne p(x) \in \Bbb Q[x]$ such that $p(\alpha) = 0$; that is, if $p(\alpha) = 0$ then $\deg m(x) \le \deg p(x)$. We can apply the division algorithm to $f(x)$ and $m(x)$, yielding $q_f(x), r_f(x) \in \Bbb Q[x]$ with

$f(x) = m(x)q_f(x) + r_f(x) \tag{1}$

and either $r_f(x) = 0$ or $\deg r_f(x) < \deg m(x)$; evaluating this equation at $\alpha$ yields

$0 = f(\alpha) = m(\alpha)q_f(\alpha) + r_f(\alpha) = r_f(\alpha), \tag{2}$

since $m(\alpha) = 0$. But since $\deg r_f(x) < \deg m(x)$, if $r_f(x) \ne 0$ this contradicts the choice of $m(x)$ as having minimal degree. Thus $r_f(x) = 0$ and hence $f(x) = m(x)q_f(x)$, so $m(x) \mid f(x)$. Applying the same technique to $g(x), m(x)$ we have

$g(x) = m(x) q_g(x) + r_g(x) \tag{3}$

with either $r_g(x) = 0$ or $\deg r_g(x) < \deg m(x)$. And again we see that this forces $r_g(\alpha) = 0$ whence $r_g(x) = 0$ and $g(x) = m(x) q_g(x)$, so that $m(x) \mid g(x)$. Note that since $m(\alpha) = 0$ we must have $\deg m(x) \ge 1$.

The question in the post may thus be answered in the affirmative: $f(x)$ and $g(x)$ have common divisor $m(x)$ with $\deg m(x) \ge 1$.

It should be noted that Snufsan's argument, edited into his question, is perilously close in spirit to mine; the key is to use the division algorithm on $m(x)$ and $f(x), g(x)$.

Hope this helps. Cheerio,

and as always,

Fiat Lux!!!