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Let $K/\mathbb{Q}$ be a finite extension. If $\{a_{1}, a_{2}, \ldots, a_{n}\}$ is an integral basis for $K$, does it imply that the ring of integers of $K$ is $\mathbb{Z}[a_{1}, a_{2}, \ldots, a_{n}]$?

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    I suppose it's this reasoning? As each element of $\mathcal{O}_{K}$ can be written unique as a $\mathbb{Z}$-linear combination of the $a_{i}$'s, $\mathcal{O}_{K} \subset \mathbb{Z}[a_{1}, \ldots, a_{n}]$. But $\mathbb{Z}[a_{1}, \ldots, a_{n}] \subset \mathcal{O}_{K}$ since sums and products of algebraic integers are also algebraic integers. Therefore we have equality.2011-11-27

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By definition of integral basis, every element in $\mathcal{O}_{K}$ can be written uniquely as a $\mathbb{Z}$-linear combination of the set. This implies that $\mathcal{O}_{K} \subset \mathbb{Z}[a_{1}, \ldots, a_{n}]$. But $\mathbb{Z}[a_{1}, \ldots, a_{n}] \subset \mathcal{O}_{K}$ since sums and products of algebraic integers are also algebraic integers. Therefore we have equality of the two sets.