4
$\begingroup$

Given a monic polynomial $f(x)$ in $\mathbb{Z}[x]$ such that $\alpha$ and $3 \alpha$ are complex roots of $f(x),$ prove that the constant term of $f(x)$ is divisible by 3.

I have attempted this problem several times and have not found a satisfactory proof. We will denote the minimal polynomial of $\alpha$ over $\mathbb{Q}$ by $p(x)$ and the minimal polynomial of $3 \alpha$ over $\mathbb{Q}$ by $q(x).$ Clearly, we have that $\mathbb{Q}(\alpha) = \mathbb{Q}(3 \alpha),$ from which it follows that $\deg(p) = \deg(q).$ Consider the polynomial $r(x) = p(\frac{x}{3})$ in $\mathbb{Q}[x].$ We note that $r(3 \alpha) = p(\alpha) = 0$ and $\deg(r) = \deg(p) = \deg(q),$ from which I would like to conclude that $r(x) = 3^{-k} q(x)$ for $k = \deg(p).$ But I don't believe this does anything to solve the problem, and this is where I am now stuck. Thanks for your time.

$\endgroup$
0

2 Answers 2

Reset to default
3
$\begingroup$

Consider the splitting field $F$ of the polynomial, and the ring of integers $\mathcal{O}_F$ of this field (i.e. the integral closure of $\mathbb{Z}$ in $F$). Then the constant coefficient $f_0$ of $f$ is equal to the product of the roots of $f$ in $F$, which are all integral over $\mathbb{Z}$; therefore, if $\beta$ is the product of the roots other than $\alpha$ and $3\alpha$, then $f_0 = 3 \alpha^2 \beta$ and $\beta \in \mathcal{O}_F$. It follows that $\frac{1}{3} f_0 \in \mathcal{O}_F \cap \mathbb{Q} = \mathbb{Z}$.

$\endgroup$
2
  • $\begingroup$ Unfortunately, I am not familiar with the notion of integral closures of a ring in a field, so I would not have thought of this approach on my own; however, with this additional information, the problem does have a very elegant solution. I appreciate your contribution. $\endgroup$ Commented Aug 8, 2018 at 23:20
  • $\begingroup$ I did come up with an expression $3 \alpha^2 c$ for $f(0),$ but I could not establish that $\alpha^2 c$ was an integer. I appreciate that your solution makes use of the fact (of which I was heretofore unaware) that the integral closure of $\mathbb{Z}$ in $\mathbb{Q}$ is $\mathbb{Z}.$ $\endgroup$ Commented Aug 8, 2018 at 23:23
2
$\begingroup$

You can replace $3$ by any integer $m\neq 0$, and $\alpha$ is not necessarily a non-real complex number. I do not assume either that $f(x)\in\mathbb{Z}[x]$ is monic except that the leading coefficient of $f(x)$ is coprime to $m$.

Let $f(x)\in\mathbb{Z}[x]$ is a polynomial such that, for some algebraic number $\alpha$, both $\alpha$ and $m\alpha$ are roots of $f(x)$. Suppose further that the leading coefficient of $f(x)$ is relatively prime to $m$. Let $p(x)\in\mathbb{Z}[x]$ be a primitive minimal polynomial of an arbitrary algebraic number $\alpha$. Let $n$ be the degree of $p(x)$.

Show that $q(x):=m^n\,p\left(\frac{x}{m}\right)\in\mathbb{Z}[x]$ is a primitive minimal polynomial of $m\alpha$. Verify that, if $\alpha\neq 0$ and $|m|>1$, then $p(x)$ and $q(x)$ are relatively prime over $\mathbb{Q}$. Prove also that both $p(x)$ and $q(x)$ divide $f(x)$. In the case that $\alpha$ is not a rational number, we know that $n\geq 2$. Since $m^n$ divides the constant term of $f(x)$, we conclude that at least $m^2$ divides the constant term of $f(x)$.

$\endgroup$
2
  • $\begingroup$ I appreciate that your solution attempts to vastly generalize the problem, but I think it also unnecessarily complicates things. I appreciated your original proof because it was clear and elegant. We did not differ too much in our lines of thinking in your original post. Please allow me to clarify, therefore. $\endgroup$ Commented Aug 8, 2018 at 23:28
  • $\begingroup$ Everything that I have written should be correct up to the introduction of the polynomial $r(x)$; instead, assuming that $\deg(p) = n,$ we can verify that $q(x) = 3^n p(\frac{x}{3})$ since the polynomial on the right-hand side is a monic polynomial of the same degree as $q(x)$ with a root in common with $q(x).$ We have therefore that $f(x) = p(x) q(x) r(x)$ for some polynomial $r(x)$ in $\mathbb{Z}[x]$ so that $f(0) = 3^n [p(0)]^2 r(0)$ is an integer that is divisible by 3. $\endgroup$ Commented Aug 8, 2018 at 23:32

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.