# Isomorphic quotient rings of polynomial rings over field

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?

• 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? – freakish Feb 1 '17 at 18:29
• In general there is no criterion: it might be quite hard to show that $F[x]/I$ is (or is not) isomorphic to $F[x]/J$. – Crostul Feb 1 '17 at 18:33
• – Watson Feb 1 '17 at 19:27
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.
• Is there a shortcut to showing that $F[x]/(x-1) \simeq F$ besides directly constructing isomorphism or using first isom thm? – shmth Feb 2 '17 at 8:26
• I think the first isomorphism theorem is the way to go. Just map $x$ to $1$. – carmichael561 Feb 2 '17 at 16:58