nLab reduced ring

Definition

Given a commutative ring $R$, $R$ is reduced or has a trivial nilradical if $x \cdot x = 0$ implies that $x = 0$ for all $x \in R$.

Theorem

For every natural number $n$, $x^{n + 1} = 0$ implies that $x = 0$ for all $x \in R$.

Proof

Let the function $f:\mathbb{N} \to \mathbb{N}$ be defined as the ceiling of half of $n$, $f(n) \coloneqq \lceil n/2 \rceil$. Then $x^{n + 1} = 0$ implies that $x^{f(n + 1)} = 0$, and for every natural number $n$, the $(n + 1)$-th iteration of the function $f$ evaluated at $n + 1$ is always equal to $1$, $f^{n + 1}(n + 1) = 1$, thus resulting in $x^{f^{n + 1}(n + 1)} = x = 0$. Thus, the nilradical of $R$ is trivial.

As a result, the theory of a reduced ring is a coherent theory.

Properties

• Every integral domain is a reduced ring. Thus, every field is a reduced ring.

• An example of a reduced ring which is not an integral domain is the quotient ring $R[x, y]/(x \cdot y)$.

• Given a square-free integer $n$, the integers modulo n $\mathbb{Z}/n\mathbb{Z}$ is a reduced ring. Since for every integer $n$, $\mathbb{Z}/n\mathbb{Z}$ is a prefield ring, $\mathbb{Z}/n\mathbb{Z}$ is an integral domain and thus a field if and only if $n$ is a prime number.

• Given a discrete field $F$, let $\overline{F}$ denote its algebraic closure. Given a square-free polynomial $q \in \overline{F}[x]$, the quotient ring $\overline{F}[x]/q\overline{F}[x]$ is a reduced ring. Since for every polynomial $q \in \overline{F}[x]$, $\overline{F}[x]/q\overline{F}[x]$ is a prefield ring, $\overline{F}[x]/q\overline{F}[x]$ is an integral domain and thus a field if and only if $q$ is a prime polynomial in $\overline{F}[x]$, a monic polynomial of degree one; the resulting quotient ring is equivalent to $\overline{F}$.

See also

• radical

• reduction modality

commutative ring	reduced ring	integral domain
local ring	reduced local ring	local integral domain
Artinian ring	semisimple ring	field
Weil ring	field	field

References

• Wikipedia, Reduced ring