# Sufficient and necessary conditions for $f,g$ such that $\Bbb F_m[x]/(f) \cong \Bbb F_m[x]/(g)$

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.

• In your specific example, $\Bbb F[x]\to \Bbb F[x]$, $x\mapsto -x$ induces an isomorphism – Hagen von Eitzen Apr 15 '17 at 16:39
• 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. – Max Apr 15 '17 at 16:49
• 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 $\deg f=\deg g$. – user26857 Apr 21 '17 at 20:28

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.