## 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. $\Box$

## 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. $\Box$