Let $F$ be a field. Let $p$ and $q$ be monic members $F[x]$. Let $I = \{p\cdot r : r\in F[x]\}$ and $J = \{q\cdot r : r\in F[x]\}$. I know that if $F[x]/I$ is isomorphic to $F[x]/J$ then ($\operatorname{deg}(p) = \operatorname{deg}(q)$ and (p is reducible if and only if q is reducible)). For cases where we do have $\operatorname{deg}(p) = \operatorname{deg}(q)$ and (p is reducible if and only if q is reducible), is there any faster way to check whether $F[x]/I$ is isomorphic to $F[x]/J$ than seeing if any element of $F[x]/I$ has q as its minimal polynomial?


(In the particular cases I'm interested in, $F$ is very finite.)



checking if F[x]/I is isomorphic to F[x]/J
If $F$ is finite, $p$ and $q$ are irreducible, and have the same dagree $d$, then $F[x]/I$ and $F[x]/J$ are isomorphic $F$-algebras, since there is (up to isomorphism) only one extension field of a finite field of any given degree (see for example Birkhoff & Mac Lane, A survey of Modern Algebra, Section 15.6).


Things get a little more interesting when $p$ and $q$ are not irreducible. Suppose that $p=p\_1^{n\_1}p\_2^{n\_2}\ldots$. Then $F[x]/I$ is isomorphic to $\bigoplus\_i F[x]/(p\_i^{n\_i})$. It turns out that, for any perfect field $F$, if $E\_i=F[x]/(p\_i)$, then $F[x]/(p\_i^{n\_i})$ is isomorphic to $E\_i[t]/(t^{n\_i})$ (this result is well-known, if not well-documented; a proof appears [here (see Theorem A.19)](http://arxiv.org/abs/0712.4051)). Thus for finite fields the general answer to you question is the following:



> 
> For finite fields
> -----------------
> 
> 
> $F[x]/I$ and $F[x]/J$ are isomorphic if and only if the numerical invariants of the polynomials $p$ and $q$ are the same, in other words, if $p=p\_1^{n\_1}p\_2^{n\_2}\ldots p\_r^{n\_r}$ and $q=q\_1^{m\_1}q\_2^{m\_2}\ldots q\_s^{m\_s}$ are the decompositions of $p$ and $q$ into irreducible factors, then there is a bijection $w:\{1,\ldots,r\}\to \{1,\ldots,s\}$ such that the degree of $q\_{w(i)}$ is the same as the degree of $p\_i$ and $m\_{w(i)}=n\_i$ for all $i\in \{1,\ldots,r\}$.
> 
> 
> 


For general perfect fields, this does not give a complete solution. What we get is 



> 
> For perfect fields
> ------------------
> 
> 
> $F[x]/I$ and $F[x]/J$ are isomorphic if and only if when$p=p\_1^{n\_1}p\_2^{n\_2}\ldots p\_r^{n\_r}$ and $q=q\_1^{m\_1}q\_2^{m\_2}\ldots q\_s^{m\_s}$ are the decompositions of $p$ and $q$ into irreducible factors, then there is a bijection $w:\{1,\ldots,r\}\to \{1,\ldots,s\}$ such that $F[x]/(q\_{w(i)})$ is isomorphic to $F[x]/(p\_i)$ and $m\_{w(i)}=n\_i$ for each $i\in \{1,\ldots,r\}$.
> 
> 
> 


Thus the problem is reduced to determining whether or not two irreducible polynomials give rise to isomorphic field extensions.


Edit: Ricky Demer pointed out that the "only if" parts of the above assertions are not obvious, and so here is a way out: We can recover the subalgebras $F[x]/(p\_i^{m\_i})$ from $F[x]/I$ as the indecomposable ideals.
It remains to show that if $p\_i$ is irreducible then the field $E\_i=F[x]/p\_i$ and the integer $m\_i$ can be recovered from $F[x]/(p\_i^m)$. This is not hard:


1. The field $F[x]/p\_i$ can be recovered from $F[x]/(p\_i^n)$ as the quotient of the algebra by its radical.
2. The integer $m\_i$ can be recovered as the length of the ring $F[x]/p\_i^{m\_i}$.


