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

Proof:  Choose any basis $\alpha_{1},\ldots,\alpha_{n}$ 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 $\alpha_{1},\ldots,\alpha_{n}$ with each $\alpha_{i}\in\mathcal{O}_{K}$.
Consider the group homomorphism  $\varphi:K\rightarrow\mathbb{Q}^{n}:\gamma\mapsto(\operatorname{Tr}_{\mathbb{Q}% }^{K}(\gamma\alpha_{1}),\ldots,\operatorname{Tr}_{\mathbb{Q}}^{K}(\gamma\alpha% _{n}))$

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

 $n=\operatorname{Tr}_{\mathbb{Q}}^{K}(1)=\operatorname{Tr}_{\mathbb{Q}}^{K}(% \gamma\gamma^{-1})=\operatorname{Tr}_{\mathbb{Q}}^{K}(\gamma\sum r_{i}\alpha_{% i})=\sum r_{i}\operatorname{Tr}_{\mathbb{Q}}^{K}(\gamma\alpha_{i})=0$

where the last equality holds since $\gamma\in\ker\varphi$.

Hence $\varphi:\mathcal{O}_{K}\hookrightarrow\mathbb{Z}^{n}$, so $\mathcal{O}_{K}$ is finitely generated   and torsion-free. It has rank $\geq n$ since the $\alpha_{i}$ are linearly independent  , and rank $\leq n$ 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 $\mathbb{Z}$ ProofOfTheRingOfIntegersOfANumberFieldIsFinitelyGeneratedOvermathbbZ 2013-03-22 16:03:07 2013-03-22 16:03:07 rm50 (10146) rm50 (10146) 9 rm50 (10146) Proof msc 13B22