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

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. $\Box$