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Consider the polynomial ring $\Bbb F_m[x]$, and two polynomials $f,g $ in $\Bbb F_m[x]$.

Is there any necessary and sufficient conditions for $f,g$ such that $\Bbb F_m[x]/(f) \cong \Bbb F_m[x]/(g)$?

Or if there are two particular $f,g$ in $\Bbb F_m[x]$, how to check that the quotient rings generated by $f,g$ are isomorphic? I have this question when I encounter the following question:

Are $\Bbb F_3[x]/(x^3+x^2+x+1), \Bbb F_3[x]/(x^3-x^2+x-1) $ isomorphic?

I could not see a suitable isomorphism so I assume they are not isomorphic, yet cannot come to a contradiction so far.

Any help for the two problems is appreciated, thanks.

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    $\begingroup$ In your specific example, $\Bbb F[x]\to \Bbb F[x]$, $x\mapsto -x$ induces an isomorphism $\endgroup$ – Hagen von Eitzen Apr 15 '17 at 16:39
  • $\begingroup$ Well an obvious necessary condition is that $deg(f) = deg(g)$. Then the comment from Hagen von Eitzen could give a lead towards a solution, because if $f$ and $g\circ h$ are associated, $h\in K[X]$ of degree $1$, then it seems it would work. I don't know of this is necessary. $\endgroup$ – Max Apr 15 '17 at 16:49
  • $\begingroup$ If $m=p^t$ with $p$ prime, then the given isomorphism of rings is an isomorphism of $\mathbb F_p$-vector spaces and counting the dimension leads to $\deg f=\deg g$. $\endgroup$ – user26857 Apr 21 '17 at 20:28
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If $f$ and $g$ are irreducible, then the two quotient rings are fields and so are isomorphic iff $f$ and $g$ have the same degree, because there is only one finite field of each possible cardinality.

This argument can be generalized for squarefree polynomials, in which case the quotient rings are products of finite fields.

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