Reduced ring

In ring theory, a branch of mathematics, a ring 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 ${\mathcal {N}}_{R}$ be nilradical of any commutative ring $R$ . There is a natural functor $R\mapsto R/{\mathcal {N}}_{R}$ of category of commutative rings ${\text{Crng}}$ into category of reduced rings ${\text{Red}}$ and it is left adjoint to the inclusion functor $I$ of ${\text{Red}}$ into ${\text{Crng}}$ . The bijection ${\text{Hom}}_{\text{Red}}(R/{\mathcal {N}}_{R},S)\cong {\text{Hom}}_{\text{Crng}}(R,I(S))$ is induced from the universal property of quotient rings.

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

Over a Noetherian ring R, we say a finitely generated module M has locally constant rank if ${\mathfrak {p}}\mapsto \operatorname {dim} _{k({\mathfrak {p}})}(M\otimes k({\mathfrak {p}}))$ is a locally constant (or equivalently continuous) function on Spec R. Then R is reduced if and only if every finitely generated module of locally constant rank is projective.^[2]

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.

Notes

^ 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.)
^ Eisenbud 1995, Exercise 20.13.

References

N. Bourbaki, Commutative Algebra, Hermann Paris 1972, Chap. II, § 2.7
N. Bourbaki, Algebra, Springer 1990, Chap. V, § 6.7
Eisenbud, David (1995). Commutative Algebra with a View Toward Algebraic Geometry. Graduate Texts in Mathematics. Springer-Verlag. ISBN 0-387-94268-8.

[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.)

[2] Eisenbud 1995, Exercise 20.13.

[1]

[2]

Examples and non-examples

Generalizations

See also

Notes

References