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In a commutative ring, the nilpotent elements form an ideal called the nilradical. The proof that the nilradical is an ideal uses the binomial theorem, which doesn't hold in noncommutative rings.

Is there an example of a noncommutative rings where the nilpotent elements do not form an ideal?

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up vote 8 down vote accepted

In the $2\times 2$ matrices over $\mathbb{Z}$ (or over $\mathbb{Q}$, or over $\mathbb{R}$), the elements $$x = \left(\begin{array}{cc} 0 & 1\\ 0 & 0 \end{array}\right)\quad\text{and}\quad y = \left(\begin{array}{cc} 0 & 0\\ 1 & 0 \end{array}\right)$$ are nilpotent: $x^2=y^2 = 0$. But $$x+y = \left(\begin{array}{cc} 0 & 1\\ 1 & 0 \end{array}\right)$$ is not only not nilpotent, it's a unit: $(x+y)^2 = 1$. So the set of nilpotent elements is not a subgroup, hence cannot be an ideal.

In noncommutative rings, you want to consider instead the notion of "nilpotent (left/right/two-sided) ideal": an ideal $I$ for which there exists $k\gt 0$ such that $I^k=0$. The sum of any finite family of nilpotent (left/right/two-sided) ideals is again a nilpotent (left/right/two-sided) ideal. An alternative is to consider "nil ideals", which are ideals in which every element is nilpotent. This is generally the case: when going from the commutative setting to the non-commutative setting, one often switches from "element-wise" to "ideal-wise" conditions. Thus, an ideal $P$ in a non-commutative ring is a prime ideal if for any two ideals $I$ and $J$, if $IJ\subseteq P$ then either $I\subseteq P$ or $J\subseteq P$ (compare to the definition of prime ideal in a commutative setting, which is called a "totally prime" or "completely prime" ideal in the noncommutative setting).

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Thanks, will look more into this. – Nils Sep 24 '11 at 2:35

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