# ring-finite integral extensions are module-finite

Theorem If $B$ is a subring of $A$ and ${u}_{1},\mathrm{\dots},{u}_{s}\in A$ are integral over $B$, then $B[{u}_{1},\mathrm{\dots},{u}_{s}]$ is module-finite over $B$.

Proof. If $s=1$ then ${u}^{n}+{b}_{1}{u}^{n-1}+\mathrm{\cdots}+{b}_{n}=0$, so $\{1,u,\mathrm{\dots},{u}^{n-1}\}$ spans $B[u]$ over $B$.

If $s>1$, use induction on $B\subset B[{u}_{1}]\subset B[{u}_{1},{u}_{2}]\subset \mathrm{\dots}\subset B[{u}_{1},\mathrm{\dots},{u}_{s}]$ and multiply the spanning sets together.

Title | ring-finite integral extensions are module-finite |
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Canonical name | RingfiniteIntegralExtensionsAreModulefinite |

Date of creation | 2013-03-22 17:01:28 |

Last modified on | 2013-03-22 17:01:28 |

Owner | rm50 (10146) |

Last modified by | rm50 (10146) |

Numerical id | 6 |

Author | rm50 (10146) |

Entry type | Theorem |

Classification | msc 16D10 |

Classification | msc 13C05 |

Classification | msc 13B02 |

Related topic | ModuleFiniteExtensionsAreIntegral |