Why is every element of a ring not a nilpotent?

Question

An element $a$ of a ring $R$ is nilpotent if there is an integer $n$ such that $a^n = 0$. Every element in a ring must have some $n$ that makes this true (the order of the element). So why is every element in a ring not a nilpotent?

Answer 1

It's not true that every element has a finite order, e.g. $(\mathbb{Z},+,\times)$.

Even if it's the case for a ring $R$ (e.g. $R$ is finite), your definition of a (multiplicative) order is wrong - it is the least $n$ such that $a^n = 1$, not $0$. In particular, every non-zero element in a finite field (e.g. $\mathbb{F}_p$) can't be nilpotent, as fields do not have zero divisors.

Answer 2

Note that every non-zero nilpotent element provides zero divisors. So, think of a ring with no zero divisors. They are known as integral domains, and the best examples are fields.

Another interesting class of rings with no non-zero nilpotents is Boolean rings. Think of the power set of any set $X$, with the addition being the symmetric difference, and the multiplication being the intersection. There every element satisfies $x^2=x$. Note that Boolean rings can have zero divisors.

There is a way to turn any commutative ring $R$ with nilpotents into a nilpotent-free ring. First, consider the set of all nilpotent elements $N$. This is called the nilradical. Then consider the quotient ring (if you have not studied them yet, think of quotient groups) $R/N.$ The resulting quotient ring has no nilpotents.

Answer 3

Some clarity is needed here. Elements in a ring may have both multiplicative and additive orders; however the definition of nilpotent uses neither. An element is nilpotent if a positive power of it (using * operation) is 0 (additive identity). Simple examples without nilpotent elements are $\mathbb{Z}$ and $\mathbb{Z/p}$ where $p$ is prime (the latter are fields) and examples with nilpotent elements are $\mathbb{Z/p^2}$ e.g. $5^2=0$ in $\mathbb{Z/25}$ (these are not integral domains).