A Rational Root of a Monic Polynomial with Integer Coefficients is an Integer

Field theory problems and solution in abstract algebra

Problem 489

Suppose that $\alpha$ is a rational root of a monic polynomial $f(x)$ in $\Z[x]$.
Prove that $\alpha$ is an integer.

 
LoadingAdd to solve later
Sponsored Links

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.


LoadingAdd to solve later

Sponsored Links

More from my site

You may also like...

Leave a Reply

Your email address will not be published. Required fields are marked *

More in Field Theory
Field theory problems and solution in abstract algebra
Cubic Polynomial $x^3-2$ is Irreducible Over the Field $\Q(i)$

Prove that the cubic polynomial $x^3-2$ is irreducible over the field $\Q(i)$.  

Close