## A finiteness condition for commutative rings with unity

Posted: June 26, 2010 in Elementary Algebra; Problems & Solutions, Rings and Modules
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Let$R$ 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 $p+q.$