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 fields.

All fields are integral domains. The converse is false, but we can construct a field of fractions from an integral domain.

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.