## Posts Tagged ‘subdirect product’

Definition 1. A ring $R$ is called Dedekind-finite if $\forall a,b \in R: \ ab=1 \Longrightarrow ba=1.$

Remark 1. Some trivial examples of Dedekind-finite rings: commutative rings, any direct product of Dedekind-finite rings, any subring of a Dedekind-finite ring.

Definition 2. A ring $R$ is called reversible if $\forall a,b \in R : \ ab = 0 \Longrightarrow ba = 0.$

Example 1. Every reversible ring $R$ is Dedekind-finite. In particular, reduced rings are Dedekind-finite.

Proof. Suppose that $ab=1$ for some $a,b \in R.$ Then $(ba-1)b=b(ab)-b=0$ and thus $b(ba-1)=0.$ So $b^2a=b$ and hence $ab^2a=ab=1.$ It follows that $ba=(ab^2a)ba=(ab^2)(ab)a=ab^2a=1.$ So $R$ is Dedekind-finite. Finally, note that every reduced ring is reversible because if $ab=0,$ for some $a,b \in R,$ then $(ba)^2=b(ab)a=0$ and thus $ba=0. \Box$

Example 2. Every (left or right) Noetherian ring $R$ is Dedekind-finite.

Proof. We will assume that $R$ is left Noetherian. Suppose that $ab=1$ for some $a,b \in R.$ Define the map $f: R \longrightarrow R$ by $f(r)=rb.$ Clearly $f$ is an $R$-module homomorphism and $f$ is onto because $f(ra)=(ra)b=r(ab)=r,$ for all $r \in R.$ Now we have an ascending chain of left ideals of $R$

$\ker f \subseteq \ker f^2 \subseteq \cdots.$

Since $R$ is left Noetherian, this chain stabilizes at some point, i.e. there exists some $n$ such that $\ker f^n = \ker f^{n+1}.$ Clearly $f^n$ is onto because $f$ is onto. Thus $f^n(c)=ba-1$ for some $c \in R.$ Then

$f^{n+1}(c)=f(ba-1)=(ba-1)b=b(ab)-b=0.$

Hence $c \in \ker f^{n+1}=\ker f^n$ and therefore $ba-1=f^n (c) = 0. \Box$

Example 3. Finite rings are obviously Noetherian and so Dedekind-finite by Example 2. More generally:

Example 4. If the number of nilpotent elements of a ring is finite, then the ring is Dedekind-finite. See here.

Note that Example 4 implies that every reduced ring is Dedekind-finite; a fact that we proved in Example 1.

Example 5. Let $k$ be a field and let $R$ be a finite dimensional $k$-algebra. Then $R$ is Dedekind-finite.

Proof. Every left ideal of $R$ is clearly a $k$-vector subspace of $R$ and thus, since $\dim_k R < \infty,$ any ascending chain of left ideals of $R$ will stop at some point. So $R$ is left Noetherian and thus, by Example 2, $R$ is Dedekind-finite. $\Box$

Remark 2. Two important cases of Example 5 are $M_n(R),$ the ring of $n \times n$ matrices over a field, and, in general, semisimple rings. As a trivial result, $M_n(R)$ is Dedekind-finite for any commutative domain $R$ because $M_n(R)$ is a subring of $M_n(Q(R))$, where $Q(R)$ is the quotient field of $R$.
So the ring of $n \times n$ matrices, where $n \geq 2,$ over a field is an example of a Dedekind-finite ring which is not reversible, i.e. the converse of Example 1 is not true. Now let $R_i = \mathbb{Z}, \ i \geq 1.$ Then $R= \prod_{i=1}^{\infty} R_i$ is obviously Dedekind-finite but not Noetherian. So the converse of Example 2 is not true.

Example 6 and Example 7 are two generalizations of Example 5.

Example 6. Every algebraic algebra $R$ over a field $k$ is Dedekind-finite.

Proof. Suppose that $ab=1$ for some $a,b \in R.$ Since $R$ is algebraic over $k,$ there exist integers $n \geq m \geq 0$ and some $\alpha_i \in k$ with $\alpha_n \alpha_m \neq 0$ such that $\sum_{i=m}^n \alpha_i b^i = 0.$ We will assume that $n$ is as small as possible. Suppose that $m \geq 1.$ Then, since $ab=1,$ we have

$\sum_{i=m}^n \alpha_i b^{i-1}=a \sum_{i=m}^n \alpha_i b^i = 0,$

which contradicts the minimality of $n.$ So $m = 0.$ Let $c = -\alpha_0^{-1}\sum_{i=1}^n \alpha_i b^{i-1}$ and see that $bc=cb=1.$ But then $a=a(bc)=(ab)c=c$ and therefore $ba=bc=1. \ \Box$

Remark 3. Regarding Examples 5 and 6, note that although any finite dimensional $k$-algebra $R$ is algebraic over $k,$ but $R$ being algebraic over $k$ does not necessarily imply that $R$ is finite dimensional over $k.$ For example, if $\overline{\mathbb{Q}}$ is the algebraic closure of $\mathbb{Q}$ in $\mathbb{C},$ then it is easily seen that $\dim_{\mathbb{Q}} \overline{\mathbb{Q}}=\infty.$ Thus the matrix ring $R = M_n(\overline{\mathbb{Q}})$ is an algebraic $\mathbb{Q}$-algebra which is not finite dimensional over $\mathbb{Q}.$ So, as a $\mathbb{Q}$-algebra, $R$ is Dedekind-finite by  Example 6 not Example 5.

Example 7. Every PI-algebra $R$ is Dedekind-finite.

Proof. Let $J(R)$ be the Jacobson radical of $R.$ If $J(R)=\{0\},$ then $R$ is a subdirect product of primitive algebras $R/P_i,$ where $P_i$ are the primitive ideals of $R.$ Since $R$ is PI, each $R/P_i$ is PI too and thus, by Kaplansky’s theorem, $R/P_i$ is a matrix ring over some division algebra and thus Dedekind-finite by Example 2. Thus $\prod R/P_i$ is Dedekind-finite and so $R,$ which is a subalgebra of $\prod R/P_i,$ is also Dedekind-finite. For the general case, let $S=R/J(R).$ Now, $S$ is PI, because $R$ is PI, and $J(S)=\{0\}.$ Therefore, by what we just proved, $S$ is Dedekind-finite. Suppose that $ab = 1$ for some $a,b \in R$ and let $c,d$ be the image of $a,b$ in $S$ respectively. Clearly $cd=1_S$ and so $dc=1_S.$ Thus $1-ba \in J(R)$ and so $ba=1-(1-ba)$ is invertible. Hence there exists $e \in R$ such that $e(ba)=1.$ But then $eb=(eb)ab=e(ba)b=b$ and hence $ba=(eb)a=e(ba)=1. \Box$

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. $\Box$