**Definition**. A ring is called **reduced** if it has no non-zero nilpotent element, i.e.

**Remark 1**. In a reduced ring if for some then As a result, the left and the right annihilators of an element of are equal.

*Proof.* For any we have and thus

**Remark 2**. A ring is a domain iff it is both prime and reduced. Moreover, every reduced ring is semiprime.

*Proof*. Let Suppose that is a domain. Then is reduced and if then and thus, since is a domain, either or So is a prime ring. Now suppose that is both prime and reduced and Then, by Remark 1, and so, since is prime, either or To prove that every reduced ring is semiprime, suppose that for some Then and so

**Remark 3**. If is reduced, then every idempotent element of is central.

*Proof*. Suppose that is an idempotent, i.e. and let See that

and thus

**Remark 4**. The intersection of all prime ideals of a reduced ring is zero.

*Proof*. Suppose that the intersection of all prime ideals of is non-zero and choose in that intersection. Since is reduced, for all Let and Note that and so Thus, by Zorn’s lemma, has a maximal element, say To get a contradiction, we only need to show that is a prime ideal: suppose that are some ideals of which properly contain and Since is a maximal element of we get Thus for some But then which is absurd.

**Remark 5.** In a ring every prime ideal contains a minimal prime ideal. So the intersection of all prime ideals of is equal to the intersection of all minimal prime ideals of If is reduced, then the intersection of all minimal prime ideals of is zero.

*Proof*. Let be a prime ideal of and the set of prime ideals of contained in Consider with . Clearly because Let be a totally ordered subset of Since Zorn’s lemma gives us a prime ideal which is maximal in i.e. is a minimal prime.

I have a blog as well, and I think I need to enhance the given information I have on the website.

Anyways, I simply wanted to compliment you on your blog.

Hi, I think there’s a small typo in the proof of Remark 2. It should read $x^3 \in xRx = \{ 0 \} \Rightarrow x = 0$. It should be x cubed instead of x square, since we don’t know if $R$ is unitary or not.

It’s OK because in this blog, unless specified otherwise, by “ring” I mean an associative ring with 1. Of course, many results are still true for rings without 1.