nil is a radical propertyNilIsARadicalProperty2004-02-28 13:55:432004-02-28 13:57:47Proofradical theory0% this is the default PlanetMath preamble. as your knowledge
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\newcommand{\nilrad}{\mathcal{N}}We must show that the nil property, $\nilrad$, is a radical property, that is that it satisfies the following conditions:
\begin{enumerate}
\item The class of $\nilrad$-rings is closed under homomorphic images.
\item Every ring $R$ has a largest $\nilrad$-ideal, which contains all other $\nilrad$-ideals of $R$. This ideal is written $\nilrad(R)$.
\item $\nilrad(R/\nilrad(R)) = 0$.
\end{enumerate}
It is easy to see that the homomorphic image of a nil ring is nil, for if $f \colon R \to S$ is a homomorphism and $x^n = 0$, then $f(x)^n = f(x^n) = 0$.
The sum of all nil ideals is nil (see proof \PMlinkid{here}{5650}), so this sum is the largest nil ideal in the ring.
Finally, if $N$ is the largest nil ideal in $R$, and $I$ is an ideal of $R$ containing $N$ such that $I/N$ is nil, then $I$ is also nil (see proof \PMlinkid{here}{5650}). So $I \subseteq N$ by definition of $N$. Thus $R/N$ contains no nil ideals.