Nilpotent Elements in Rings

# Nilpotent Elements in Rings

Definition: Let $(R, +, *)$ be a commutative ring with additive identity $0$. An element $a \in R$ is said to be Nilpotent if there exists an $n \in \mathbb{N}$ such that $a^n = 0$. |

*By definition, the additive identity $0$ is always a nilpotent element in a ring $(R, +, *)$.*

For example, consider the commutative ring $(\mathbb{Z}_4, +, *)$. Then the element $2 \in \mathbb{Z}_4$ is nilpotent since:

(1)\begin{align} \quad 2^2 = 4 \equiv 0 \pmod 4 \end{align}

Theorem 1: Let $(R, +, *)$ be a commutative ring and let $a, b \in R$ be nilpotent. Then $a + b$ is nilpotent. |

**Proof:**Let $a, b \in R$ be nilpotent. Then there exists $n, m \in \mathbb{N}$ such that $a^n = 0$ and $b^m = 0$. Let $t = n * m + 1$. Then $t$ is a positive integer and by the binomial theorem:

\begin{align} \quad (a + b)^t = \binom{t}{0} a^tb^0 + \binom{t}{1} a^{t-1}b^1 + ... + \binom{t}{t} a^0b^{t} \end{align}

- Each term in the expression is of the form $\binom{t}{k} a^{t-k}b^k$. Since $t = n^m + 1$, we must have at least $t - k > n$ or $k > m$ (since if $t - k < n$ AND $k < m$ then $t = (t-k) + k < n + m \leq n * m < t$). So each term in the expression above is equal to zero. So $(a + b)$ is nilpotent. $\blacksquare$

Theorem 2: Let $(R, +, *)$ be a commutative ring and let $a \in R$ be nilpotent. Then for all $r \in R$, $r * a$ and $a * r$ are nilpotent. |

**Proof:**Let $a \in R$ be nilpotent. Then there exists a positive integer $n \in \mathbb{N}$ such that $a^n = 0$. Let $r \in R$. Then since $R$ is a commutative ring:

\begin{align} \quad (r * a) = r^n * a^n = r^n * 0 = 0 \quad \mathrm{and} \quad (a * r)^n = a^n * r^n = 0 * r^n = 0 \end{align}

- So $r * a$ and $a * r$ are nilpotent.

Theorem 3: Let $(R, +, *)$ be a commutative ring and let $u, a \in R$. If $u$ is a unit and $a$ is nilpotent, then $u - a$ is a unit. |

**Proof:**Let $u \in R$ be a unit. Then there exists an element $w \in R$ such that $u * w = 1$ and $w * u = 1$. Let $a \in R$ be nilpotent. Then there exists an $n \in \mathbb{N}$ such that $a^n = 0$.

- First consider the case when $u = 1$. Then $w = 1$. So $a^n + 1 = 1$ and hence:

\begin{align} \quad 1 = 1 - a^n = (1 - a)(a^{n-1} + a^{n-2} + ... + a + 1) \end{align}

- So $1 - a = u - a$ is a unit with multiplicative inverse $a^{n-1} + a^{n-2} + ... + a + 1$.

- Now in general, if $u$ is a unit with $u * w = 1$ and $a$ is nilpotent then:

\begin{align} \quad (u - a) = u(1 - w * a) \end{align}

- Since $w$ is a unit, $1 - w * a$ is a unit from the remarks made above. So $u - a$ is a product of two units, namely $u$ and $(1 - w * a)$. So $u - a$ is a unit. $\blacksquare$