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 ${\mathcal {N}}_{R}$ denote nilradical of a commutative ring $R$ . There is a functor $R\mapsto R/{\mathcal {N}}_{R}$ of the category of commutative rings ${\text{Crng}}$ into the category of reduced rings ${\text{Red}}$ and it is left adjoint to the inclusion functor $I$ of ${\text{Red}}$ into ${\text{Crng}}$ . The natural 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]

^ 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.

[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.

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