# alternative proof that a finite integral domain is a field

###### Proof.

Let $R$ be a finite integral domain^{} and $a\in R$ with $a\ne 0$. Since $R$ is finite, there exist positive integers $j$ and $k$ with $$ such that ${a}^{j}={a}^{k}$. Thus, ${a}^{k}-{a}^{j}=0$. Since $$ and $j$ and $k$ are positive integers, $k-j$ is a positive integer. Therefore, ${a}^{j}({a}^{k-j}-1)=0$. Since $a\ne 0$ and $R$ is an integral domain, ${a}^{j}\ne 0$. Thus, ${a}^{k-j}-1=0$. Hence, ${a}^{k-j}=1$. Since $k-j$ is a positive integer, $k-j-1$ is a nonnegative integer. Thus, ${a}^{k-j-1}\in R$. Note that $a\cdot {a}^{k-j-1}={a}^{k-j}=1$. Hence, $a$ has a multiplicative inverse^{} in $R$. It follows that $R$ is a field.
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Title | alternative proof that a finite integral domain is a field |
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Canonical name | AlternativeProofThatAFiniteIntegralDomainIsAField |

Date of creation | 2013-03-22 16:21:54 |

Last modified on | 2013-03-22 16:21:54 |

Owner | Wkbj79 (1863) |

Last modified by | Wkbj79 (1863) |

Numerical id | 6 |

Author | Wkbj79 (1863) |

Entry type | Proof |

Classification | msc 13G05 |