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Math Help - direct product of finite local rings

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    direct product of finite local rings

    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 .
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    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.
    Last edited by NonCommAlg; June 26th 2010 at 11:00 PM.
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