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30 votes
1 answer
2k views

Do the algebraic integers form a free abelian group?

It is a well-known fact, proved in every introductory textbook on algebraic number theory, that if $K$ is an algebraic number field, i.e. a finite extension of $\mathbb{Q}$, then its ring $\mathcal{O}...
Robert Kucharczyk's user avatar
14 votes
1 answer
696 views

$\mathbb{Z}$-module structure of the subring generated by an algebraic number

Let $a$ and $b$ be algebraic numbers which are not necessarily algebraic integers. Is there some invariant that allows us to determine whether $\mathbb Z[a]$ and $\mathbb Z[b]$ are isomorphic as $\...
user108921's user avatar