# Reduced ring

In ring theory, a ring R is called a reduced ring if it has no non-zero nilpotent elements. Equivalently, a ring is reduced if it has no non-zero elements with square zero, that is, x2 = 0 implies x = 0. A commutative algebra over a commutative ring is called a reduced algebra if its underlying ring is reduced.

The nilpotent elements of a commutative ring R form an ideal of R, called the nilradical of R; therefore a commutative ring is reduced if and only if its nilradical is zero. Moreover, a commutative ring is reduced if and only if the only element contained in all prime ideals is zero.

A quotient ring R/I is reduced if and only if I is a radical ideal.

Let D be the set of all zerodivisors in a reduced ring R. Then D is the union of all minimal prime ideals.[1]

## Examples and non-examples

• Subrings, products, and localizations of reduced rings are again reduced rings.
• The ring of integers Z is a reduced ring. Every field and every polynomial ring over a field (in arbitrarily many variables) is a reduced ring.
• More generally, every integral domain is a reduced ring since a nilpotent element is a fortiori a zero divisor. On the other hand, not every reduced ring is an integral domain. For example, the ring Z[x, y]/(xy) contains x + (xy) and y + (xy) as zero divisors, but no non-zero nilpotent elements. As another example, the ring Z×Z contains (1,0) and (0,1) as zero divisors, but contains no non-zero nilpotent elements.
• The ring Z/6Z is reduced, however Z/4Z is not reduced: The class 2 + 4Z is nilpotent. In general, Z/nZ is reduced if and only if n = 0 or n is a square-free integer.
• If R is a commutative ring and N is the nilradical of R, then the quotient ring R/N is reduced.
• A commutative ring R of characteristic p for some prime number p is reduced if and only if its Frobenius endomorphism is injective. (cf. perfect field.)

## Generalizations

Reduced rings play an elementary role in algebraic geometry, where this concept is generalized to the concept of a reduced scheme.

1. ^ Proof: let $\mathfrak{p}_i$ be all the (possibly zero) minimal prime ideals.
$D \subset \cup \mathfrak{p}_i:$ Let x be in D. Then xy = 0 for some nonzero y. Since R is reduced, (0) is the intersection of all $\mathfrak{p}_i$ and thus y is not in some $\mathfrak{p}_i$. Since xy is in all $\mathfrak{p}_j$; in particular, in $\mathfrak{p}_i$, x is in $\mathfrak{p}_i$.
$D \supset \mathfrak{p}_i:$ (stolen from Kaplansky, commutative rings, Theorem 84). We drop the subscript i. Let $S = \{ xy | x \in R - D, y \in R - \mathfrak{p} \}$. S is multiplicatively closed and so we can consider the localization $R \to R[S^{-1}]$. Let $\mathfrak{q}$ be the pre-image of a maximal ideal. Then $\mathfrak{q}$ is contained in both D and $\mathfrak{p}$ and by minimality $\mathfrak{q} = \mathfrak{p}$. (This direction is immediate if R is Noetherian by the theory of associated primes.)