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Thread: direct product of finite local rings

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    direct product of finite local rings

    Let $\displaystyle R$ be a commutative ring with identity. Let $\displaystyle S$ be the set of all non-unit elements of $\displaystyle R$ . If $\displaystyle S-J(R)$ ( $\displaystyle J(R)$ is the jacobson radical of $\displaystyle R$ ) is finite , then prove that $\displaystyle R \cong R_1\times R_2\times ...\times R_k$ for some finite local rings $\displaystyle R_i$ and every non-unit is a zero-divisor .
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    we also need the condition $\displaystyle S \setminus J(R) \neq \emptyset.$ fix an element $\displaystyle y \in S \setminus J(R).$

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

    Proof. clearly $\displaystyle x+y \notin J(R).$ proving $\displaystyle x+y \in S$ is by contradiction: suppose that $\displaystyle x+y \notin S.$ so $\displaystyle (x+y)u=1,$ for some $\displaystyle u \in R.$ but then $\displaystyle yu=1-xu \notin S,$ because $\displaystyle x \in J(R).$ that means $\displaystyle yu$, and hence $\displaystyle y$, is a unit and that's a contradiction. Q.E.D.

    claim 2. $\displaystyle S$ is finite.

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

    claim 3. $\displaystyle R \cong R_1 \times R_2 \cdots \times R_k,$ where each $\displaystyle R_i$ is a local ring.

    Proof. since every proper ideal of $\displaystyle R$ is contained in $\displaystyle S,$ our ring has only a finite number of ideals and so $\displaystyle R$ is artinian. hence $\displaystyle J(R)^n=\{0\},$ for some positive integer $\displaystyle n.$ let $\displaystyle M_i, \ 1 \leq i \leq k,$ be the maximal ideals of $\displaystyle R$ and put $\displaystyle R_i=R/M_i^n.$ it is clear that each $\displaystyle R_i$ is a local ring and by the chinese remainder theorem for rings we have $\displaystyle R \cong R_1 \times R_2 \times \cdots \cdot R_k.$ Q.E.D.

    claim 4. each $\displaystyle R_i$ is finite.

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

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

    Proof. $\displaystyle \{s^m : \ m \geq 1 \} \subseteq S$ and so, since $\displaystyle S$ is finite, there exist $\displaystyle p \neq q \in \mathbb{N}$ such that $\displaystyle s^p = s^q$ and $\displaystyle p+q$ is minimal. so $\displaystyle s(s^{p-1}-s^{q-1})=0.$ if we prove that $\displaystyle s^{p-1} - s^{q-1} \neq 0,$ then we've proved that $\displaystyle s$ is a zero divisor. well, if $\displaystyle s^{p-1}=s^{q-1}$ and $\displaystyle p=1$ or $\displaystyle q=1,$ then $\displaystyle s$ would be a unit, which is absurd. if $\displaystyle p,q > 1,$ then, since $\displaystyle p-1+q-1 < p+q,$ we must have $\displaystyle s^{p-1} \neq s^{q-1},$ by the minimality of $\displaystyle p+q.$ Q.E.D.
    Last edited by NonCommAlg; Jun 26th 2010 at 10:00 PM.
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