Freshman's dream

Statement

Let $p$ be a prime number and let $R$ be a commutative ring of characteristic $p$. Then $(x + y)^p = x^p + y^p$ for all $x, y \in R$.

Remarks

The proof uses the binomial theorem. For $0 < k < p$, the prime $p$ divides the binomial coefficient $\binom{p}{k}$, so these terms vanish in $R$. By induction, $(x + y)^{p^n} = x^{p^n} + y^{p^n}$ for every $n \ge 0$. The identity gives the Frobenius endomorphism. In $\mathbb{R}$ the identity is false: $(1 + 1)^2 = 4$ and $1^2 + 1^2 = 2$.

Explanations

L1 human

The name refers to a common error: $(x + y)^2 = x^2 + y^2$ is false for real numbers. In characteristic $2$, the middle term $2 x y$ is $0$, and the identity becomes true. The same thing happens for every prime $p$ in characteristic $p$.