CLAIM. The polynomial $h(x)=x^5+x^3+x+1$ is irreducible over the integers modulo 3.
因为５＜３＋３，假设它可约，则它有线性或二次因子。容易检查不存在 $\mathbf{Z}_3$ 上的根，所以我们要做的就是证明它没有二次因子。
There is a field $\mathbf{F}_9$ with exactly 9 elements which contains $\mathbf{Z}_3$ and also contains a square root $i$ of $-1$. This is true because $x^2+1$ has no roots in $\mathbf{Z}_3$. If there are irreducible quadratic factors, the theory of finite fields implies that there must be a root for $h(x) \bmod 3$ in the field $\mathbf{F}_9$. Every element in $\mathbf{F}_9$ is uniquely expressible in the form $a+b i$ where $a, b \in \mathbf{Z}_3$. One can check directly that none of these elements can be a root of the polynomial $h(x)$ reduced $\text{mod } 3$; an argument of this sort is definitely not elegant, but it works and does not require additional digressions.
