Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
3  
It's true. Your "can be written as a integral linear combination" line is exactly why. –  Sanchez 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
2  
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
2  
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
show 3 more comments

Know someone who can answer? Share a link to this question via email, Google+, Twitter, or Facebook.

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.