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For some polynomial $f(x)$ where deg($f(x)$) $\le 3$ to be reducible in $\mathbb{Z}_n [x]$, then it must have at least 1 root in $\mathbb{Z}_n [x]$, right? Since for a cubic or quadratic to be reduced, it needs a factor of a polynomial of degree 1 to "multiply" the other factors, and a polynomial of degree 1 would be a root? Am I correct, or am I missing something? Thanks.

Max
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    You are right, since the degree is $3$, the polynomial can only be reducible if it has a linear factor, hence it must have a rational root to be reducible. – Peter Nov 28 '16 at 22:57

2 Answers2

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${\rm mod}\ n^2\!:\ 1\!+\!knx^{\large 3} \equiv (1\!+\!nx^{\large 3})^{\large k}\,$ is reducible if $\,k>1,\,$ with no roots: $\ n^{\large 2}\mid 1\!+\!kn x^{\large 3}\,\Rightarrow\,n\mid 1$

Bill Dubuque
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This is not true if $n$ is not prime because the factors of a reducible polynomial may have degree equal to or greater than the polynomial.

For instance, $6x+1=(2x+1)(4x+1)$ is reducible in $\mathbb{Z}_{8}[x]$, but $6x+1$ has no zero in $\mathbb{Z}_{8}$.

lhf
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  • See also http://math.stackexchange.com/questions/645818/why-doesnt-the-polynomial-factor-theorem-hold-for-polynomials-in-a-non-field-ri. – lhf Nov 28 '16 at 23:36
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    Simpler: $,1!+!2nx \equiv (1!+!nx)^2 \pmod{!n^2},,$ e.g. $,1!+!6x \equiv (1!+!3x)^2 \pmod{!9},$ as in my answer. – Bill Dubuque Nov 17 '23 at 23:37