# freshman’s dream

###### Theorem 1 (Freshman’s dream).

If $k$ if a field of characteristic (http://planetmath.org/Characteristic) $p\mathrm{>}\mathrm{0}$ (so $p$ is prime) then for all $x\mathrm{,}y\mathrm{\in}k$ we have

$${(x+y)}^{{p}^{i}}={x}^{{p}^{i}}+{y}^{{p}^{i}}.$$ |

Therefore $x\mathrm{\mapsto}{x}^{{p}^{i}}$ is a field monomorphism (called a Frobenius monomorphism^{}.)

When $k$ is finite then it is indeed an automorphism.
A field $k$ is called a perfect field^{} when the map is surjective^{}.

The theorem is so named because it is a common mistake for freshman math students to make over the real numbers. However, as the characteristic of the real numbers is 0, this does not apply in any interesting way to that setting.

It should also be noted that the result applies only to powers of the characteristic, and not all exponents.

###### Proof.

The proof is an application of the binomial theorem. We prove it for $p$ first.

$${(x+y)}^{p}=\sum _{i=0}^{p}\left(\genfrac{}{}{0pt}{}{p}{i}\right){x}^{i}{y}^{p-i}.$$ |

Now observe

$$\left(\genfrac{}{}{0pt}{}{p}{i}\right)=\frac{p!}{(p-i)!i!}=p\cdot \frac{(p-1)!}{(p-i)!i!}.$$ |

As $p$ is prime and $1\le i\le p-1$ it follows $i!$ and $(p-i)!$ do not divide $p$. As the field $k$ has characteristic $p$, $\frac{(p-1)!}{(p-i)!i!}$ is an integer $m$ where

$$\left(\genfrac{}{}{0pt}{}{p}{i}\right)=pm\equiv 0.$$ |

Thus ${(x+y)}^{p}={x}^{p}+{y}^{p}$.

Now for ${p}^{i}$ simply use induction^{}:

$${(x+y)}^{{p}^{i}}={({(x+y)}^{p})}^{{p}^{i-1}}={({x}^{p}+{y}^{p})}^{{p}^{i-1}}={x}^{{p}^{i}}+{y}^{{p}^{i}}.$$ |

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Title | freshman’s dream |
---|---|

Canonical name | FreshmansDream |

Date of creation | 2013-03-22 15:51:17 |

Last modified on | 2013-03-22 15:51:17 |

Owner | Algeboy (12884) |

Last modified by | Algeboy (12884) |

Numerical id | 18 |

Author | Algeboy (12884) |

Entry type | Theorem |

Classification | msc 11T23 |

Classification | msc 11T30 |

Synonym | Frobenius Automorphism^{} |

Related topic | PolynomialCongruence |