Posts Tagged ‘reduced ring’

Definition. Let R, \ R_i, \ i \in I, be rings. For every j \in I we let \pi_j : \prod_{i \in I} R_i \longrightarrow R_j be the natural projection. Then R is called a subdirect product of R_i, \ i \in I, if the following conditions are satisfied:

1) There exists an injective ring homomorphism f: R \longrightarrow \prod_{i \in I} R_i,

2) For every j \in I the map \pi_j f: R \longrightarrow R_j is surjective.

Note. Suppose that A_i, \ i \in I, are some ideals of R and put R_i = R/A_i. Then we can define f: R \longrightarrow \prod_{i \in I} R/A_i by f(r)=(r+ A_i)_{i \in I}. Clearly the second condition in the above definition is satisfied. Thus R is a subdirect product of R/A_i, \ i \in I, if and only if f is injective, i.e. \bigcap_{i \in I} A_i = \{0\}.

Remark 6. If P is a minimal prime ideal of the ring R, then S=R \setminus P is multiplicatively closed iff s_1s_2 \cdots s_k \neq 0, for all s_i \in S, \ k \in \mathbb{N}.

Proof. Suppose that s_1s_2 \cdots s_k \neq 0, for any s_1,s_2, \cdots, s_k \in S and k \in \mathbb{N}. Let T be the set of all elements of R which are a finite product of some elements of S. Clearly T is multiplicatively closed, S \subseteq T and S is multiplicatively closed iff S=T. So we’ll be done if we show that S=T. Let \mathcal{C}=\{A \lhd R: \ A \cap T=\emptyset \}. We have \mathcal{C} \neq \emptyset because (0) \in \mathcal{C}. Therefore, by Zorn’s lemma, (\mathcal{C}, \subseteq) has a maximal element Q and Q is a prime ideal of R. Since Q \cap T = \emptyset, we have Q \cap S = \emptyset and thus Q \subseteq P. Thus Q=P because P is a minimal prime. So P \cap T= \emptyset, which means T \subseteq S. Hence T=S. \ \Box

 Remark 7. If R is reduced and P \lhd R is a minimal prime, then R/P is a domain.

Proof. Clearly R/P is a domain iff S = R \setminus P is multiplicatively closed. Let T be as defined in Remark 6. By that remark, we only need to show that 0 \notin T. So suppose that s_1s_2 \cdots s_k = 0, for some s_1, s_2, \cdots , s_k \in S, where the integer k \geq 2 is assumed to be minimal. Then by, Remark 1, we have s_k R s_1s_2 \cdots s_{k-1} = \{0\}. Now, since P is prime, s_k R s_1 cannot be a subset of P because otherwise we’d have either s_k \in P or s_1 \in P, which is clearly nonsense. Thus s_k Rs_1 \cap S \neq \emptyset. Let s \in s_kRs_1 \cap S. Then

ss_2 \cdots s_{k-1} \in s_kRs_1s_2 \cdots s_{k-1} = \{0\}.

Hence ss_2 \cdots s_{k-1}=0, which contradicts the minimality of k. \ \Box

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

Proof. If R is reduced, then, by Remarks 5 and 7, R is a subdirect product of the domains R/P_i, \ i \in I, where \{P_i \}_{i \in I} is the set of all minimal prime ideals of R. Conversely, suppose that R is a subdirect product of domains R_i, \ i \in I and f: R \longrightarrow \prod_{i \in I} R_i is an injective ring homomorphism. Suppose that x \in R and x^2=0. Let f(x)=(x_i)_{\in I}. Then (0_{R_i})_{i \in I} = f(x^2)=(f(x))^2=(x_i^2)_{i \in I}. Thus x_i^2=0, for all i \in I, and so x_i = 0, for all i \in I, because every R_i is a domain. Hence x=0 and so R is reduced. \Box


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