The Freshman's Dream in Characteristic p

Detailed Solution

Since $R$ is commutative, the binomial theorem applies: $$ (a+b)^p=\sum_{k=0}^{p}\binom{p}{k}a^{p-k}b^k=a^p+\sum_{k=1}^{p-1}\binom{p}{k}a^{p-k}b^k+b^p. $$ For each $k$ with $0<k<p$, the binomial coefficient $\binom{p}{k}=\dfrac{p!}{k!(p-k)!}$ has $p$ (a prime) dividing the numerator $p!$ but not dividing $k!$ or $(p-k)!$, since all factors in those are strictly smaller than $p$. Hence $p\mid\binom{p}{k}$ for every such $k$. Since $R$ has characteristic $p$, multiplying any element of $R$ by $p$ gives $0$. Therefore every middle term $\binom{p}{k}a^{p-k}b^k$ vanishes in $R$, leaving $$ (a+b)^p=a^p+b^p. $$ This identity requires only that $R$ be commutative (so the binomial theorem is valid) and have characteristic $p$; it does not require $R$ to be a field, an integral domain, or free of zero divisors. It also fails in general for noncommutative rings, since the binomial expansion itself is invalid there without commutativity. Thus the correct option is $\boxed{\mathrm{A}}$.