A Rational Root of a Monic Polynomial with Integer Coefficients is an Integer
Problem 489
Suppose that $\alpha$ is a rational root of a monic polynomial $f(x)$ in $\Z[x]$.
Prove that $\alpha$ is an integer.
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Proof.
Suppose that $\alpha=\frac{p}{q}$ is a rational number in lowest terms, that is, $p$ and $q$ are relatively prime integers.
Let
\[f(x)=x^n+a_{n-1}x^{n-1}+\cdots+ a_1x+a_0\]
be a monic polynomial in $\Z[x]$ and $\alpha$ is a root of $f(x)$.
Since $\alpha$ is a root of $f(x)$, we have
\begin{align*}
0&=f(\alpha)\\
&=\alpha^n+a_{n-1}\alpha^{n-1}+\cdots+ a_1\alpha+a_0\\[6pt]
&=\frac{p^n}{q^n}+a_{n-1}\frac{p^{n-1}}{q^{n-1}}+\cdots+ a_1 \frac{p}{q}+a_0.
\end{align*}
Multiplying by $q^n$, we obtain
\begin{align*}
0&=q^n f(\alpha)\\
&=p^n+a_{n-1}qp^{n-1}+\cdots+ a_1 q^{n-1}p+a_0q^n\\[6pt]
&=p^n+q\left(\, a_{n-1}p^{n-1}+\cdots+ a_1 q^{n-2}p+a_0 q^{n-1} \,\right).
\end{align*}
Hence we have
\[q\left(\, a_{n-1}p^{n-1}+\cdots+ a_1 q^{n-2}p+a_0 q^{n-1} \,\right)=-p^n,\]
and this implies that $q$ divides $p^n$, and so $q$ divides $p$.
Since $p$ and $q$ are relatively primes, it yields that $q=1$.
Therefore, $\alpha=p/1=p$ is an integer.
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