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If the Unit of a Ring is Idempotent, It Can Be Expressed as a Direct Sum 📂Abstract Algebra

If the Unit of a Ring is Idempotent, It Can Be Expressed as a Direct Sum

Theorem

Given a ring RR with a unity element 11, if its zero divisors aa satisfy a2=aa^2 = a, that is, if it’s an idempotent, then RR is uniquely represented as the direct product of aRaR and (1−a)R(1-a)R. R=aR×(1−a)R R = a R \times (1-a)R

Explanation

This theorem is charmingly mathematical enough that one can understand it without defining the direct sum of rings separately.

Proof

Part (i). Existence

For r∈Rr \in R ar∈aR(1−a)r∈(1−a)R ar \in aR \\ (1-a) r \in (1-a) R and since r=ar+(1−a)rr = ar + (1-a)r, then R=aR×(1−a)RR = a R \times (1-a)R.

Part (ii). Uniqueness

For r1,r2,r3,r4∈Rr_{1}, r_{2}, r_{3} , r_{4} \in R, ar1=x1∈aRar3=x2∈aR(1−a)r2=y1∈(1−a)R(1−a)r4=y2∈(1−a)R \begin{align*} a r_{1} =& x_{1} \in aR \\ a r_{3} =& x_{2} \in aR \\ (1-a) r_{2} =& y_{1} \in (1-a) R \\ (1-a) r_{4} =& y_{2} \in (1-a) R \end{align*} Assuming x1=x2x_{1} = x_{2} and y1=y2y_{1} = y_{2} suffices. x1+y1=x2+y2  ⟹  ar1+(1−a)r2=ar3+(1−a)r4  ⟹  a2r1+a(1−a)r2=a2r3+a(1−a)r4 \begin{align*} & x_{1} + y_{1} = x_{2} + y_{2} \\ \implies& a r_{1} + (1-a) r_{2} = a r_{3} + (1-a) r_{4} \\ \implies& a^2 r_{1} + a(1-a) r_{2} = a^2 r_{3} + a(1-a) r_{4} \end{align*} Since a2=aa^2 = a and a(1−a)=a−a2=a−a=0a(1-a) = a - a^2 = a-a = 0, x1=ar1=a2r1=a2r3=ar3=x2 x_{1} = a r_{1} = a^2 r_{1} = a^2 r_{3} = a r_{3} = x_{2} Similarly, ar1+(1−a)r2=ar3+(1−a)r4 a r_{1} + (1-a) r_{2} = a r_{3} + (1-a) r_{4} and since ar1=ar3a r_{1} = a r_{3}, (1−a)r2=(1−a)r4 (1-a) r_{2} = (1-a) r_{4} Hence, the following is obtained. y1=(1−a)r2=(1−a)r4=y2 y_{1} = (1-a) r_{2} = (1-a) r_{4} = y_{2}

Part (iii). Exclusivity

Suppose x∈aR∩(1−a)Rx \in aR \cap (1-a)R, for some r1,r2∈Rr_{1} , r_{2} \in R x=ar1=(1−a)r2 x = a r_{1} = (1-a) r_{2} If both sides are multiplied by aa, ax=a2r1=a(1−a)r2=(a−a2)r2=0⋅r2=0 ax = a^2 r_{1} = a(1-a) r_{2} = (a - a^2) r_{2} = 0 \cdot r_{2} = 0 Since x=ar1=a2r1=0x=a r_{1} = a^2 r_{1} = 0, the following is obtained. aR∩(1−a)R={0} aR \cap (1-a)R = \left\{ 0 \right\}

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