How can I determine when two quotient rings of a polynomial ring are isomorphic? For example, is $F[x]/(x^2) \cong F[x] / (x^2 - x)$? I know (or at least I think) that they are isomorphic as additive groups, but I don't think they are as rings. How can I show this? In general, is there some criterion for determining when two quotient rings by ideals generated by polynomials of the same degree are isomorphic?

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    $\begingroup$ That highly depends on the underlying field $F$. For example $\mathbb{R}[x]/(x^2+1)\simeq \mathbb{C}$ is a field but $\mathbb{C}[x]/(x^2+1)$ is not a field. In a simple case of $x^2$ or $x^2-x$ when the polynomial is a product of linear components then this is simplier. So are you interested in these concrete examples or in general? $\endgroup$
    – freakish
    Feb 1, 2017 at 18:29
  • $\begingroup$ In general there is no criterion: it might be quite hard to show that $F[x]/I$ is (or is not) isomorphic to $F[x]/J$. $\endgroup$
    – Crostul
    Feb 1, 2017 at 18:33
  • $\begingroup$ Related: math.stackexchange.com/questions/869335, math.stackexchange.com/questions/1831305 $\endgroup$
    – Watson
    Feb 1, 2017 at 19:27
  • $\begingroup$ See also mathoverflow.net/questions/65109 $\endgroup$
    – Watson
    Feb 1, 2017 at 21:55

1 Answer 1


I doubt that a general criterion exists, but the Chinese remainder theorem can sometimes be helpful in simplifying quotient rings. In your example, the ideals $(x)$ and $(x-1)$ are coprime since $1=x-(x-1)$, hence $$ F[x]/(x^2-x)\simeq F[x]/(x)\times F[x]/(x-1)\simeq F\times F $$

If $F$ is a field then $F\times F$ has no nilpotent elements. On the other hand, $F[x]/(x^2)$ does have a nilpotent element, namely the image of $x$. So $F[x]/(x^2-x)$ and $F[x]/(x^2)$ are not isomorphic.

  • $\begingroup$ Is there a shortcut to showing that $F[x]/(x-1) \simeq F$ besides directly constructing isomorphism or using first isom thm? $\endgroup$
    – shmth
    Feb 2, 2017 at 8:26
  • $\begingroup$ I think the first isomorphism theorem is the way to go. Just map $x$ to $1$. $\endgroup$ Feb 2, 2017 at 16:58

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