Show that if \(p(x)\in\mathbb Z[x]\) is monic and \(r\in\mathbb Q\) is a root of \(p\), then \(r\in\mathbb Z\).
Solution
Write \[ p(x)=x^n+a_{n-1}x^{n-1}+\cdots+a_0 \] and suppose \(r=a/b\) in lowest terms with \(b>0\). From \(p(a/b)=0\), multiplying by \(b^n\) gives \[ a^n+b(a_{n-1}a^{n-1}+a_{n-2}a^{n-2}b+\cdots+a_0b^{n-1})=0. \] Hence \(b\mid a^n\). Since \(\gcd(a,b)=1\), also \(\gcd(a^n,b)=1\), so \(b=1\). Therefore \(r=a\in\mathbb Z\).