## About reduced rings (1)

Posted: June 4, 2010 in Elementary Algebra; Problems & Solutions, Rings and Modules
Tags: , ,

Definition. A ring $R$ is called reduced if it has no non-zero nilpotent element, i.e. $x \in R, \ x^n=0 \Longrightarrow x=0.$

Remark 1. In a reduced ring $R$ if $xy=0,$ for some $x,y \in R,$ then $yRx=\{0\}.$ As a result, the left and the right annihilators of an element of $R$ are equal.

Proof. For any $z \in R$ we have $(yzx)^2=0$ and thus $yzx=0. \ \Box$

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

Proof. Let $x,y \in R.$ Suppose that $R$ is a domain. Then $R$ is reduced and if $xRy=\{0\},$ then $xy=0$ and thus, since $R$ is a domain, either $x=0$ or $y=0.$ So $R$ is a prime ring. Now suppose that $R$ is both prime and reduced and $xy=0.$ Then, by Remark 1, $yRx=\{0\}$ and so, since $R$ is prime, either $y=0$ or $x=0.$ To prove that every reduced ring is semiprime, suppose that $xRx=\{0\},$ for some $x \in R.$ Then $x^2 \in xRx = \{0\}$ and so $x=0. \ \Box$

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

Proof. Suppose that $x \in R$ is an idempotent, i.e. $x^2=x$ and let $y \in R.$ See that $(xy-xyx)^2=(yx-xyx)^2=0$

and thus $xy=yx=xyx. \ \Box$

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

Proof.  Suppose that the intersection of all prime ideals of $R$ is non-zero and choose $x \neq 0$ in that intersection. Since $R$ is reduced, $x^n \neq 0$ for all $n \in \mathbb{N}.$ Let $S = \{x^n: \ n \in \mathbb{N} \}$ and $\mathcal{A}=\{I \lhd R : \ I \cap S = \emptyset \}.$ Note that $(0) \in \mathcal{A}$ and so $\mathcal{A} \neq \emptyset.$ Thus, by Zorn’s lemma, $\mathcal{A}$ has a maximal element, say $P.$ To get a contradiction, we only need to show that $P$ is a prime ideal:  suppose that $I,J$ are some ideals of $R$ which properly contain $P$ and $IJ \subseteq P.$ Since $P$ is a maximal element of $\mathcal{A},$ we get $I \notin \mathcal{A}, \ J \notin \mathcal{A}.$ Thus $x^i \in I, \ x^j \in J,$ for some $i,j \in \mathbb{N}.$ But then $x^{i+j} \in IJ \subseteq P,$ which is absurd. $\Box$

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

Proof. Let $P$ be a prime ideal of $R$ and $\mathcal{B}$ the set of prime ideals of $R$ contained in $P.$ Consider $\mathcal{B}$ with $\supseteq$. Clearly $\mathcal{B} \neq \emptyset$ because $P \in \mathcal{B}.$ Let $\{Q_i \}_{i \in I}$ be a totally ordered subset of $\mathcal{B}.$ Since $\bigcap_{i \in I} Q_i \in \mathcal{B},$ Zorn’s lemma gives us a prime ideal $Q \subseteq P$ which is maximal in $\mathcal{B},$ i.e. $Q$ is a minimal prime. $\Box$

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Comments
1. Krista says:

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

2. yYukataYy says:

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.

• Yaghoub Sharifi says:

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.