## 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 $x\in R$,

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.