Discussion
Let R be a Boolean ring with identity having \(n\) elements, then which of the following values of \(n\) is possible ?
Answer: A
The possible value is \(n = 32\).
In a Boolean ring every element satisfies \(x^{2} = x\). Expanding \((x + x)^{2} = x + x\) gives \(4x = 2x\), so \(2x = 0\) and the ring has characteristic 2 and its additive group is a vector space over \(\mathbf{Z}_{2}\). A finite one therefore has \(2^{k}\) elements; in fact a finite Boolean ring with identity is isomorphic to \(\mathbf{Z}_{2}^{k}\).
\(32 = 2^{5}\) is a power of 2. 23 is prime, 27 is \(3^{3}\) and 15 is \(3 \times 5\), none of them a power of 2.
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