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This is something I was curious about from my algebraic number theory class. Given any non-zero ideal $I$ of $A \cap K$ (algebraic integers in $K$ where $[K:\mathbb{Q}] < \infty$), we know it has an integral basis $\{\alpha_1,..., \alpha_n \}$ such that every element in $I$ can be uniquely written as a integral linear combination of $\{ \alpha_1,..., \alpha_n \}$.

Does it follow that $I$ is in fact $Span_{\mathbb{Z}} (\alpha_1,..., \alpha_n)$? (If not, when is this true?)

Here $\mathbb{Q}$ is rationals and $\mathbb{Z}$ is integers... Thanks!

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It's true. Your "can be written as a integral linear combination" line is exactly why. – user27126 Mar 31 '13 at 17:03
I'm a little confused by your definition $A\cap K$, if $[K:\mathbb{Q}] < \infty$ then $K$ is an algebraic number field, and therefore it's elements are already algebraic! Maybe you meant algebraic integers? – Alex J Best Mar 31 '13 at 17:07
@AlexJBest Yes, let me fix that. – Tom Mosher Mar 31 '13 at 17:09
The fact that everything in $I$ can be written as linear combinations of $\alpha_i$s tells you $I\subseteq{\rm Span}_{\bf Z}\{\alpha_i\}$. The fact that each $\alpha_i$ is in $I$ tells us that $I\supseteq{\rm Span}_{\bf Z}\{\alpha_i\}$. Since we have inclusion in both directions, we have equality $I={\rm Span}_{\bf Z}\{\alpha_i\}$. – anon Mar 31 '13 at 17:52
Amusing side fact: every ideal in a ring of integers can be generated (over the ring, not over $\mathbb{Z}$) by at most two elements. This is actually true in any Dedekind domain. – Paul VanKoughnett Mar 31 '13 at 18:11

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