Last edited by hbghlyj 2024-5-8 08:15$x^5+x^3+x+1$在整数mod p的分解:
GF(2)
$(x + 1) (x^4 + x^3 + 1)$
GF(3)
$x^5 + x^3 + x + 1$
GF(5)
$(x^2 + 3 x + 4) (x^3 + 2 x^2 + x + 4)$
GF(7)
$x^5 + x^3 + x + 1$
GF(11)
$(x + 6)^3 (x^2 + 4 x + 8)$
GF(13)
$(x^2 + 12 x + 3) (x^3 + x^2 + 12 x + 9)$
GF(17)
$(x + 3) (x^4 + 14 x^3 + 10 x^2 + 4 x + 6)$
GF(19)
$(x + 13) (x^4 + 6 x^3 + 18 x^2 + 13 x + 3)$
GF(23)
$(x + 16) (x^4 + 7 x^3 + 4 x^2 + 5 x + 13)$
GF(29)
$(x^2 + 22 x + 19) (x^3 + 7 x^2 + 2 x + 26)$
GF(31)
$(x + 19) (x^4 + 12 x^3 + 21 x^2 + 4 x + 18)$
GF(37)
$(x + 16)^2 (x + 32) (x^2 + 10 x + 5)$
GF(41)
$(x + 2) (x + 15) (x^3 + 24 x^2 + 14 x + 26)$
GF(43)
$(x + 41) (x^4 + 2 x^3 + 5 x^2 + 10 x + 21)$
GF(47)
$x^5 + x^3 + x + 1$
GF(53)
$(x + 11) (x + 26) (x^3 + 16 x^2 + 24 x + 48)$
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.
