Nilpotent Elements

Let $R$ be a ring. We say $x\in R$ is a zero divisor if for some nonzero $y\in R$ we have $x y = 0$.

Example: 2 is a zero divisor in $\mathbb{Z}_4$. 5,7 are zero divisors in $\mathbb{Z}_{35}$.

A nonzero ring in which there are no nonzero zero divisors is called an integral domain.

Example: $\mathbb{Z},\mathbb{Z}[i],\mathbb{Q},\mathbb{R},\mathbb{C}$ are integral domains.

The following is easily verified:

Proposition: If $R$ is an integral domain then so is $R[x]$.

Corollary: If $R$ is an integral domain then so is $R[x_1,...,x_n]$.

An element $x \in R$ is nilpotent if $x^n = 0$ for some $n \ge 0$. Note all nilpotent elements are zero divisors, but the converse is not always true, for example, $2$ is a zero divisor in $\mathbb{Z}_6$ but not nilpotent.

We say $x \in R$ is a unit if $x y = 1$ for some $y \in R$. Note if such a $y$ exists, it must be unique so we write $y = x^{-1}$. Also note that the units form an abelian group under multiplication. If $x = y u$ for some unit $u$ then we say $x$ is an associate of $y$.

Example: The units in $\mathbb{Z}$ are $\pm 1$. The units of $\mathbb{Z}[i]$ are $\pm 1, \pm i$. We have $x \in \mathbb{Z}_n$ a unit if and only if $x,n$ are coprime.

A field is a nonzero ring in which all nonzero elements are units.

Example: $\mathbb{Q}, \mathbb{R}, \mathbb{C}, \mathbb{Z}_p$ for prime $p$ are all fields.

Note that all fields are integral domains. The converse is not true, but integral domains are closely related to fields as will be seen when constructing fields of fractions.

A principal ideal $P$ of $R$ is an ideal generated by a single element, in other words, for some $x\in R$,

\[ P = A x =\{a x | a \in R\} \]

For example, $A 1 = A, A 0 = \{0\}$. Clearly $x$ is a unit if and only if $A x = A$.

Proposition: Let $R$ be a nonzero ring. The following are equivalent: 1. $R$ is a field 2. The only ideals of $R$ are $\{0\}$ and $R$. 3. Every homomorphism of $R$ onto a nonzero ring is injective.

Proof: $(1)\implies(2)$ since any ideal containing a unit must contain all of $R$. $(2)\implies(3)$: suppose the only ideals of $R$ are $\{0\}, R$ and let $f:R\rightarrow S$ be a ring homomorphism onto a nonzero ring $S$. We have $ker f\ne A$ since $f$ has at least one nonzero value. But $ker f\triangleleft R$, so $ker f = \{0\}$. Hence $f$ is injective. $(3)\implies(1)$: let $x\in R$ be a nonunit. Then $x R \ne R$ so $R/x R$ is a nonzero ring. By assumption, the natural map $A\rightarrow A/x A$ is injective, thus its kernel is injective and hence $x R = \{0\}$. In particular $x = x \cdot 1 = 0$ hence all nonzero elements of $R$ are units.