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

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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