Let \(R\) be a ring with the property that \(a^2=a\) for all \(a\in R\).
Compute the Jacobson radical of \(R\).
What is the characteristic of \(R\)?
Prove that \(R\) is commutative.
Prove that if \(R\) is finite, then \(R\) is isomorphic (as a ring) to \((\mathbb Z/2\mathbb Z)^d\) for some \(d\).
Solution
1. The Jacobson radical of \(R\) is zero.
Let \(x\in J(R)\). Since every element is idempotent, \[ x^2=x, \] so \[ x(1-x)=0. \] Because \(x\in J(R)\), the element \(1-x\) is a unit. Multiplying by its inverse gives \(x=0\). Hence \[ J(R)=0. \]
2. The characteristic of \(R\) is \(2\) (unless \(R=0\), in which case the usual convention gives characteristic \(1\)).
Applying the identity \(a^2=a\) to \(a=1+1\) gives \[ (1+1)^2=1+1. \] Thus \[ 4\cdot1=2\cdot1, \] so \(2\cdot1=0\). For a nonzero unital ring this means \(\operatorname{char}R=2\).
3. The ring \(R\) is commutative.
For arbitrary \(a,b\in R\), \[ (a+b)^2=a+b. \] Expanding and using \(a^2=a\), \(b^2=b\), we obtain \[ ab+ba=0. \] By 2, \(-ba=ba\), so \(ab=ba\). Hence \(R\) is commutative.
4. If \(R\) is finite, then\[ R\cong (\mathbb Z/2\mathbb Z)^d \]for some \(d\ge0\).
Because \(R\) is finite, it has only finitely many maximal ideals, say \[ \mathfrak m_1,\dots,\mathfrak m_d. \] In any commutative ring, \[ J(R)=\bigcap_{i=1}^d \mathfrak m_i. \] By 1 this intersection is zero. Distinct maximal ideals are comaximal, so the Chinese remainder theorem gives an isomorphism \[ R\cong \prod_{i=1}^d R/\mathfrak m_i. \] Each quotient \(R/\mathfrak m_i\) is a field and still satisfies \(x^2=x\) for every element \(x\). If \(0\ne x\) in such a field, then \[ x(x-1)=0 \] forces \(x=1\). Thus each residue field has exactly two elements, so \[ R/\mathfrak m_i\cong\mathbb F_2. \] Therefore \[ R\cong \mathbb F_2^d\cong(\mathbb Z/2\mathbb Z)^d. \]