# a finite integral domain is a field

A finite integral domain is a field.

Proof:

Let $R$ be a finite integral domain. Let $a$ be nonzero element of $R$.

Define a function^{} $\phi :R\to R$ by $\phi (r)=ar$.

Suppose $\phi (r)=\phi (s)$ for some $r,s\in R$. Then $ar=as$, which implies $a(r-s)=0$. Since $a\ne 0$ and $R$ is a cancellation ring, we have $r-s=0$. So $r=s$, and hence $\phi $ is injective^{}.

Since $R$ is finite and $\phi $ is injective, by the pigeonhole principle^{} we see that $\phi $ is also surjective^{}. Thus there exists some $b\in R$ such that $\phi (b)=ab={1}_{R}$, and thus $a$ is a unit.

Thus $R$ is a finite division ring. Since it is commutative, it is also a field.

Note:

A more general result is that an Artinian integral domain is a field.

Title | a finite integral domain is a field |
---|---|

Canonical name | AFiniteIntegralDomainIsAField |

Date of creation | 2013-03-22 12:50:02 |

Last modified on | 2013-03-22 12:50:02 |

Owner | yark (2760) |

Last modified by | yark (2760) |

Numerical id | 11 |

Author | yark (2760) |

Entry type | Theorem |

Classification | msc 13G05 |

Related topic | FiniteRingHasNoProperOverrings |