About reduced rings (2)

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


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