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


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

  1. Krista says:

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  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.

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