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How can I show that the polynomial $x-1$ is irreducible in $\mathbb{Z}_8[x]$?

$\mathbb{Z}_8[x]$ is not an integral domain so we cannot use degree considerations. I have tried reducing the problem mod 2 to conclude a factorization of $x-1$ must be of form $(1+a_1x+a_2x^2+...+a_mx^m)(1+b_1x+b_2x^2+...+b_nx^n)$ where $a_i,b_i$ are even, except $b_1$ is odd. I also tried using induction on the degrees $m,n$ but to no avail.

Any help is appreciated!

EDIT: Upon trial and error I have discovered $(2x^2-x-1)(4x^2-2x+1)=x-1$ in $\mathbb{Z}_8[x]$, however this does not disprove anything because we do not know if the 2 polynomials on the left are units or not...

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    $\begingroup$ Indeed, $4x^2-2x+1$ is a unit in $\Bbb Z_8[x]$: its reciprocal is $2x+1$. (Note that $\frac1{1+2x}=1-2x+4x^2-8x^3+16x^4\cdots=1-2x+4x^2$ in $\Bbb Z_8[x]$.) $\endgroup$ Mar 19, 2016 at 6:34
  • $\begingroup$ It had occurred to me that I have never showed x-1 to be non-unit, but that is easy: let f be its inverse, the leading term of f times the leading term of x-1 must be a term with coefficient 0, but that means the leading term of f has zero coefficient, contradiction. I'm putting this down for other's reference. $\endgroup$ Mar 19, 2016 at 9:52

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Aha, together we have the necessary pieces! You've remarked (by looking modulo $2$) that any factorization of $x-1$ in $\Bbb Z_8[x]$ must look like $$ x-1 = (1-x+2f(x))(1+2g(x)). $$ However, $1+2g(x)$ is a unit, since its reciprocal is $1-2g(x)+4g(x)^2$. Therefore $x-1$ is indeed irreducible in $\Bbb Z_8[x]$.

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    $\begingroup$ Nice answer, +1. Indeed, in general, the units of $R[X]$ are polynomials $a_{0} + a_{1}X+\cdots a_{n}X^{n}$ such that $a_{0}$ is a unit, and $a_{1}, \ldots, a_{n}$ are nilpotent. (Here, $a_{0} = 1$, and every coefficient of $2g(x)$ is nilpotent, since $2$ is nilpotent.) $\endgroup$ Mar 19, 2016 at 6:47
  • $\begingroup$ That's a very neat trick! Thanks a lot! $\endgroup$ Mar 19, 2016 at 8:16
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First of all I have to admit that I don't know what's an irreducible polynomial over a non-integral domain.
But if I understand well the OP wants to show that $X-1$ can't be written as a product of two non-invertible polynomials (of degree at least one).
Suppose the contrary, and write $X-1=fg$ with $\deg f\ge1$, $\deg g\ge1$.
If we work over a ring $R$ having only one prime ideal $\mathfrak p$ (as it is $\mathbb Z/8\mathbb Z$), then we arrive to a contradiction immediately: in $R/\mathfrak p$ we have $X-\overline 1=\overline f\overline g$ hence $\deg\overline f=0$ and $\deg\overline g=1$ (or vice versa). Then all coefficients of $f$, excepting $f_0$, belong to $\mathfrak p$, so they are nilpotent. But obviously $f_0$ is invertible, so $f$ is invertible.

Remark. We showed that $X-1$ is "irreducible" over any commutative ring having only one prime ideal (in particular for $\mathbb Z/p^n\mathbb Z$ with $p$ prime, $n\ge1$). One can't generalize this to commutative rings with at least two prime ideals: in $(\mathbb Z/6\mathbb Z)[X]$ we can write $X-1=(3X-1)(2X+1)$.

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  • $\begingroup$ Note that this is essentially the same proof as in my answer together with Alex Wertheim's comment. This is not a derogatory statement!—just a way of helping the OP assimilate the various viewpoints on this topic. $\endgroup$ Mar 19, 2016 at 17:23
  • $\begingroup$ @GregMartin I've thought I did a little more. $\endgroup$
    – user26857
    Mar 19, 2016 at 20:28

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