---
tag: 0396VGFA78
kind: statement
role: theorem
name: Freshman's dream
msc: [13A35]
uses: [0GYGHKX834, 0P7CH1CND5]
sources:
  - url: https://en.wikipedia.org/wiki/Freshman%27s_dream
    locator: Prime characteristic
same_as:
  wikidata: Q4120528
  wikipedia: Freshman's dream
---

## Statement

Let $p$ be a prime number and let $R$ be a [[commutative ring|0GYGHKX834]] of [[characteristic|0P7CH1CND5]] $p$.
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|0GHDDVAH8P]].
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 $p$.