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I read such a judgement,it says:

If $p$ is a prime integer,then the non-constant polynomials in $\mathbb Z/(p^k)[x]$ have a unique factorization, for positive integer $k$.

So how can I prove it? since $\mathbb Z(p^k)$ is not a field.

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  • $\begingroup$ You have to be careful with the definitions. For example, when $p=k=2$ we have factorizations like $$(x+2)^2=x^2+4x+4\equiv x^2\pmod4.$$ So uniqueness is violated unless the factorizations $(x+2)^2$ and $x^2$ are somehow the same. $\endgroup$ Aug 30, 2018 at 15:28
  • $\begingroup$ See Bill Dubuque's old answer for more information and links. I guess this is related, too. $\endgroup$ Aug 30, 2018 at 15:40
  • $\begingroup$ You are right,my judgement should be fake. $\endgroup$
    – Daniel Xu
    Aug 31, 2018 at 5:38

1 Answer 1

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$F_{p^k}$ is not the same thing as $Z/(p^k)$ because the former is a field with $p^k$ elements, whereas the latter is just a ring with $p^k$ elements (and not a field when $k>1$).

For instance, if $p^k = 2^2 = 4$ then $F_4$ is a field with 4 elements {0,1,a,b} in which $x+x = 0$ for all $x \in F_4$. However, $Z/(4)$ is a ring with 4 elements {0,1,2,3}, and in this ring, 1+1 is not 0.

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  • $\begingroup$ oh,sorry.i believe it is the Z/(pk) case $\endgroup$
    – Daniel Xu
    Aug 30, 2018 at 2:18
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    $\begingroup$ But now that you replaced $F_{p^k}$ by $\mathbb{Z}/(p^k)$, this $\mathbb{Z}/(p^k)[x]$ can not be a UFD for $k>1$ because it fails the "D" part of UFD. For $k>1$ this is not an integral Domain. $\endgroup$
    – Mark
    Aug 30, 2018 at 12:55
  • $\begingroup$ alright,thank you! $\endgroup$
    – Daniel Xu
    Aug 30, 2018 at 14:38

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