If $x_2=a x_1$ for $a \in \mathbb{Q}$ different from -1 ,where $x_1,x_2$ are roots of $f(x) \in \mathbb{Q}[x]$, prove that $f(x)$ is reducible.

Question

Suppose that $f(x)$ is a polynomial in $\mathbb{Q}[x]$ of degree $d>1$ with $d$ roots $x_1, \ldots, x_d$ in $\mathbb{C}$. If $x_2=a x_1$ for $a \in \mathbb{Q}$ different from -1 , prove that $f(x)$ is reducible.

This question comes from UCLA Spring 2004 Algebra Qualifying exam.

After trying induction on $n$ and failed, I tried Galois theory but cannot hit the point. Any hint or solution will be appreciated.

Answer 1

(Edited to be more clear about the shape of the argument as suggested below)

Assume that $a\neq \pm 1$ and handle the trivial $a=0$ case separately. WLOG assume that $f(x)$ is monic.

The monic polynomials $f(x)$ and $a^{-d}f(ax)$ share the common root $x_{1}$, so their gcd (in $\mathbb{Q}[x]$) is of positive degree.

On the other hand, $a^{-d}\neq 1$, so $f(x)$ and $a^{-d}f(ax)$ have distinct constant term and thus are distinct monic polynomials of equal degree.

So the gcd of $f(x)$ and $a^{-d}f(ax)$ is a positive degree proper divisor of $f(x)$, a contradiction to $f$’s irreducibility.

Answer 2

Consider the polynomials $f(x)$ and $f(ax)$. If $x_1,\cdots,x_d$ are the roots and $x_2 = a x_1$, let us split cases.

• If $a=0$, then $x$ divides $f(x)$, so $f$ is reducible. We are also in this case if $x_1 = 0$, so we can exclude that case moving forward.
• If $a=1$, factor $f(x)$ into irreducibles and let $m_{x_1}(x)$ be the minimal polynomial of $x_1$. Since $x_2 = x_1$, the root $x_1$ has multiplicity two. The field $\mathbb Q$ is separable, so $m_{x_1}(x)^2$ divides $f(x)$, showing $f(x)$ is reducible.
• If $a \neq -1$, then the polynomials $f(x)$ and $f(ax)$ do not have the same leading coefficient. Their GCD is not $1$ since they have the root $x_1$ in common because $f(x_1) = 0$ and $f(ax_1) = f(x_2) = 0$. If their GCD is equal to $f(x)$, this means that $f(ax) = Cf(x)$ for some non-zero constant $C$. Assume $f(x)$ is irreducible and that $f(ax) = Cf(x)$. This leads to $f(0) = f(a \cdot 0) = C f(0)$, thus $C=1$ or $f(0) = 0$. In the first case (where $C=1$), by comparing the leading terms of $f(x)$ and $f(ax)$, we get $a^d = 1$, which leads to $a= \pm 1$, and those are cases we already discussed. If $f(0) = 0$, then $x$ divides $f(x)$, contradicting the irreducibility of $f(x)$ satisfying $\deg f > 1$. Since we reached a contradiction, we rollback and notice that this implies that the GCD of $f(x)$ and $f(ax)$ is not $f(x)$. It wasn't constant either, so it's some divisor of $f(x)$, showing it is reducible.

Hope that helps,

Answer 3

Suppose $f(x)$ irreducible and let $K$ be its splitting field. Let $\alpha$ and $\beta\in K$ two roots of $f(X)$. Then $\mathbb{Q}(\alpha)$ and $\mathbb{Q}(\beta)$ are isomorphic subextension of $K$ and if $$\varphi:\mathbb{Q}(\alpha)\longrightarrow\mathbb{Q}(\beta) $$ is the isomorphism we must have $\varphi(\alpha)=\beta$

The isomorphism $\varphi$ extends to an embedding $$ \phi:K\rightarrow\overline{\mathbb{Q}} $$ and since $K$ is normal in fact $\phi\in{\rm Aut}(K)$. But the latter set is a finite group so that if $\beta=a\alpha$ with $\pm1\neq a\in\mathbb{Q}$ we get a contradiction because $\phi$ must have finite order.