# integrality is transitive

Let $C\subset B\subset A$ be rings. If $B$ is integral over $C$ and $A$ is integral over $B$, then $A$ is integral over $C$.

Proof. Choose $u\in A$. Then ${u}^{n}+{b}_{1}{u}^{n-1}+\mathrm{\cdots}+{b}_{n}=0,{b}_{i}\in B$. Thus $C[{b}_{1},\mathrm{\dots},{b}_{n},u]$ is integral and thus module-finite over $C[{b}_{1},\mathrm{\dots},{b}_{n}]$. Each ${b}_{i}$ is integral over $C$, so $C[{b}_{1},\mathrm{\dots},{b}_{n}]$ is integral hence module-finite over $C$. Thus $C[{b}_{1},\mathrm{\dots},{b}_{n},u]$ is module-finite, hence integral, over $C$, so $u$ is integral over $C$.

Title | integrality is transitive |
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Canonical name | IntegralityIsTransitive |

Date of creation | 2013-03-22 17:01:25 |

Last modified on | 2013-03-22 17:01:25 |

Owner | rm50 (10146) |

Last modified by | rm50 (10146) |

Numerical id | 6 |

Author | rm50 (10146) |

Entry type | Theorem |

Classification | msc 13B21 |