**Definition**. Let be rings. For every we let be the natural projection. Then is called a **subdirect product** of if the following conditions are satisfied:

1) There exists an injective ring homomorphism

2) For every the map is surjective.

**Note**. Suppose that are some ideals of and put Then we can define by Clearly the second condition in the above definition is satisfied. Thus is a subdirect product of if and only if is injective, i.e.

**Remark 6**. If is a minimal prime ideal of the ring then is multiplicatively closed iff , for all

*Proof.* Suppose that for any and Let be the set of all elements of which are a finite product of some elements of Clearly is multiplicatively closed, and is multiplicatively closed iff So we’ll be done if we show that . Let We have because Therefore, by Zorn’s lemma, has a maximal element and is a prime ideal of Since we have and thus Thus because is a minimal prime. So , which means Hence

**Remark 7**. If is reduced and is a minimal prime, then is a domain.

*Proof*. Clearly is a domain iff is multiplicatively closed. Let be as defined in Remark 6. By that remark, we only need to show that So suppose that for some , where the integer is assumed to be minimal. Then by, Remark 1, we have Now, since is prime, cannot be a subset of because otherwise we’d have either or which is clearly nonsense. Thus Let Then

Hence which contradicts the minimality of

**The Structure Theorem For Reduced Rings**. A ring is reduced iff is a subdirect product of domains.

*Proof*. If is reduced, then, by Remarks 5 and 7, is a subdirect product of the domains where is the set of all minimal prime ideals of Conversely, suppose that is a subdirect product of domains and is an injective ring homomorphism. Suppose that and Let Then Thus for all and so for all because every is a domain. Hence and so is reduced.