# Nilpotent elements in a quotient ring.

Let R be a ring. Let I be an ideal of R. If R/I doesn't have nonzero nilpotent element, every nilpotent element in R is contained in I. Then, if I contains every nilpotent element in R, there is no nilpotent element in R/I?

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The desired conclusion holds if and only if $I$ is a radical ideal: an ideal $I$ is radical if and only if whenever $a^n\in I$, it follows that $a\in I$. – Arturo Magidin Apr 19 '12 at 15:28

No.

Consider $\mathbb{Q}[x]$ (rational polynomials) and the ideal $I=(x^2)$ (the principal ideal generated by $x^2$).

Notice that $\mathbb{Q}[x]$ is an integral domain (so no zero divisors so no nilpotent elements). Thus $I$ contains all of the nilpotent elements (since there are none).

On the other hand, $\mathbb{Q}[x]/I$ does have nilpotent elements. In particular $(x+I)^2=x^2+I=0+I$ and $x+I \not= 0+I$.

So quotienting can create new nilpotent elements.

Another example (along the same lines), consider $\mathbb{Z}$ (integral domain so no nilpotents). Then $8\mathbb{Z}$ contains all nilpotents (again since there aren't any). But $\mathbb{Z}/8\mathbb{Z} = \mathbb{Z}_8$ does have nilpotent elements. For example, $2 \not=0$ in $\mathbb{Z}_8$ but $2^3=8=0$.

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