# proof of the ring of integers of a number field is finitely generated over $\mathbb{Z}$

Proof:
Choose any basis ${\alpha}_{1},\mathrm{\dots},{\alpha}_{n}$ of $K$ over $\mathbb{Q}$. Using the theorem in the entry multiples^{} of an algebraic number^{}, we can multiply each element of the basis by an integer to get a new basis ${\alpha}_{1},\mathrm{\dots},{\alpha}_{n}$ with each ${\alpha}_{i}\in {\mathcal{O}}_{K}$.

Consider the group homomorphism^{}

$$\phi :K\to {\mathbb{Q}}^{n}:\gamma \mapsto ({\mathrm{Tr}}_{\mathbb{Q}}^{K}(\gamma {\alpha}_{1}),\mathrm{\dots},{\mathrm{Tr}}_{\mathbb{Q}}^{K}(\gamma {\alpha}_{n}))$$ |

where ${\mathrm{Tr}}_{\mathbb{Q}}^{K}$ is the trace (http://planetmath.org/trace2) from $K$ to $\mathbb{Q}$. Note that $\phi $ is $1-1$, since if $\gamma \ne 0$ and $\phi (\gamma )=0$, then

$$n={\mathrm{Tr}}_{\mathbb{Q}}^{K}(1)={\mathrm{Tr}}_{\mathbb{Q}}^{K}(\gamma {\gamma}^{-1})={\mathrm{Tr}}_{\mathbb{Q}}^{K}(\gamma \sum {r}_{i}{\alpha}_{i})=\sum {r}_{i}{\mathrm{Tr}}_{\mathbb{Q}}^{K}(\gamma {\alpha}_{i})=0$$ |

where the last equality holds since $\gamma \in \mathrm{ker}\phi $.

Hence $\phi :{\mathcal{O}}_{K}\hookrightarrow {\mathbb{Z}}^{n}$, so ${\mathcal{O}}_{K}$ is finitely generated^{} and torsion-free. It has rank $\ge n$ since the ${\alpha}_{i}$ are linearly independent^{}, and rank $\le n$ since it injects into ${\mathbb{Z}}^{n}$, so it has rank $n$.

Title | proof of the ring of integers of a number field is finitely generated over $\mathbb{Z}$ |
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Canonical name | ProofOfTheRingOfIntegersOfANumberFieldIsFinitelyGeneratedOvermathbbZ |

Date of creation | 2013-03-22 16:03:07 |

Last modified on | 2013-03-22 16:03:07 |

Owner | rm50 (10146) |

Last modified by | rm50 (10146) |

Numerical id | 9 |

Author | rm50 (10146) |

Entry type | Proof |

Classification | msc 13B22 |