direct product of finite local rings

**xixi** · June 25th 2010, 11:08 PM

Let $R$ be a commutative ring with identity. Let $S$ be the set of all non-unit elements of $R$ . If $S-J(R)$ ( $J(R)$ is the jacobson radical of $R$ ) is finite , then prove that $R \cong R_1\times R_2\times ...\times R_k$ for some finite local rings $R_i$ and every non-unit is a zero-divisor .

**NonCommAlg** · June 26th 2010, 09:42 PM

we also need the condition $S \setminus J(R) \neq \emptyset.$ fix an element $y \in S \setminus J(R).$

claim 1. $x+y \in S \setminus J(R)$ for all $x \in J(R).$

Proof. clearly $x+y \notin J(R).$ proving $x+y \in S$ is by contradiction: suppose that $x+y \notin S.$ so $(x+y)u=1,$ 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 and that's a contradiction. Q.E.D.

claim 2. $S$ is finite.

Proof. define $f: J(R) \longrightarrow S \setminus J(R)$ by $f(x)=x+y.$ clearly $f$ is one-to-one and thus $J(R)$ , and therefore $S,$ is finite.

claim 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 $R$ 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 put $R_i=R/M_i^n.$ it is clear that each $R_i$ is a local ring and by the chinese remainder theorem for rings we have $R \cong R_1 \times R_2 \times \cdots \cdot R_k.$ Q.E.D.

claim 4. each $R_i$ is finite.

Proof. first note that $k \geq 2$ because $S \setminus J(R) \neq \emptyset.$ now suppose, to the contrary, that $R_j$ is infinite for some $j.$ then $A=\{(x_1, x_2, \cdots , x_k) \in R: \ x_i=0, \ \text{for some} \ i \neq j \}$ is infinite and $A \subseteq S,$ which is impossible because $S$ is finite. so each $R_i$ , and hence $R$ itself, is finite.

claim 5. every $0 \neq s \in S$ is a zero divisor.

Proof. $\{s^m : \ m \geq 1 \} \subseteq S$ and so, since $S$ is finite, there exist $p \neq q \in \mathbb{N}$ such that $s^p = s^q$ and $p+q$ is minimal. so $s(s^{p-1}-s^{q-1})=0.$ if we prove that $s^{p-1} - s^{q-1} \neq 0,$ then we've proved that $s$ is a zero divisor. well, if $s^{p-1}=s^{q-1}$ and $p=1$ or $q=1,$ then $s$ would be a unit, which is absurd. if $p,q > 1,$ then, since $p-1+q-1 < p+q,$ we must have $s^{p-1} \neq s^{q-1},$ by the minimality of $p+q.$ Q.E.D.

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