A finiteness condition for commutative rings with unity

Posted: June 26, 2010 in Elementary Algebra; Problems & Solutions, Rings and Modules
Tags: , , ,

LetR be a commutative ring with 1 and J(R) its Jacobson radical. Let S be the set of non-unit elements of R and let T:=S \setminus J(R).

Problem. Suppose that T is non-empty and finite. Then R is finite and R \cong R_1 \times R_2 \times \cdots \times R_k, for some finite local rings R_i and k \geq 2.

Solution.

\boxed{1} x+y \in T for all y \in T and x \in J(R).

Proof. Clearly x+y \notin J(R). Proving x+y \in S is by contradiction: suppose that x+y \notin S, i.e. (x+y)u=1, for some u \in R. But then yu=1-xu \notin S, because x \in J(R). That means yu, and hence y, is a unit. Contradiction!

\boxed{2} S is finite.

Proof. Let y \in T and define f: J(R) \longrightarrow T by f(x)=x+y. Clearly f is one-to-one. Thus J(R), and therefore S=T \cup J(R), is finite.

\boxed{3} R \cong R_1 \times R_2 \cdots \times R_k, where each R_i is a local ring.

Proof. Since every proper ideal of R is contained in S, our ring has only a finite number of ideals and so it is artinian. Hence J(R)^n=\{0\}, for some positive integer n. Let M_i, \ 1 \leq i \leq k, be the maximal ideals of R and R_i=R/M_i^n.  It is clear that each R_i is a local ring and, by the Chinese remainder theorem, R \cong R_1 \times R_2 \times \cdots \cdot \times R_k.

\boxed{4} k \geq 2 and R, and so each R_i, is finite.

Proof. Since T \neq \emptyset, R is not local and thus k \geq 2. Now suppose, to the contrary, that R is infinite. Then at least one of R_i is infinite and so the set

A=\{ (x_1, x_2, \cdots , x_k) \in R: \ x_i = 0, \ \text{for some} \ i \}

is infinite. But A \subseteq S, which is a contradiction. \Box

Remark. Every 0 \neq s \in S is a zero divisor.

Proof. We have \{s^m : \ m \geq 1 \} \subseteq S and so, since S is finite, there exist some p \neq q \in \mathbb{N} such that s^p = s^q and p+q is minimal. So s(s^{p-1}-s^{q-1})=0 and thus we need to prove that s^{p-1} - s^{q-1} \neq 0. Well, if s^{p-1}=s^{q-1} and p=1 or q=1, then s would be a unit, which is absurd. Otherwise, we’re done by the minimality of p+q.

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