# Algebraic Number Theory - Integral Basis

Let $K$ be a number field with $[K:Q] =n$. Let $O_k$ be its ring of algebraic integers.

I understand how there is an integral basis for $Q$, i.e. $\exists$ a $Q$-basis of $K$ consisting of elements of $O_k$. Let this integral basis be denoted by $\omega_1, \omega_2, \dots, \omega_n \in O_k$.

However, I do not understand how this leads to the fact that

$$\bigoplus_{i=1}^n Z\omega_i \subseteq O_k$$

Could someone elaborate please? Thank you.

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Gerry Myerson's answer is correct, but if his "of course" isn't obvious to you, then you should look in a textbook for the theorem that the sum and the product of any two algebraic integers is again an algebraic integer. (You implicitly used this fact in referring to "its ring of algebraic integers", but you might not have been aware of it.) –  Andreas Blass Jan 12 '13 at 19:43
Thanks Andreas, appreciate it. I was confused because the text that I looked at stated this fact as a consequence of integral basis. But from Gerry's answer, I now understand that it applies for any $\omega_1,\omega_2,\dots,\omega_n \in O_k$, and actually could be stated before the exposition of integral basis in the text. –  Conan Wong Jan 12 '13 at 19:46

That sum is (isomorphic to) the set of all numbers $\sum a_i\omega_i$ where the $a_i$ are integers. But if the $\omega_i$ are in $O_k$ then of course any integer linear combination of them is also in $O_k$.

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Gerry, thank you. But then this would be true for any $\omega_1,\omega_2,\dots,\omega_n \in O_k$? i.e. they do not have to be an integral basis of $K$ for the fact to be true. –  Conan Wong Jan 12 '13 at 19:35
@YACP I was confused because a text that I looked at stated this fact as a consequence of integral basis. Please do not assume that people who post questions that are trivial for you have not thought about them or looked in books beforehand. –  Conan Wong Jan 12 '13 at 19:41
@YACP, on this site, where peoploe regularly ask for $2+2$, no question about an integral basis for the ring of integers in an algebraic number field is trivial. –  Gerry Myerson Jan 12 '13 at 19:45

The text may be trying to make the following point:

• if $\omega_1,\ldots,\omega_r$ are any elements of $O_k$ then their $\mathbb Z$-span, which one might denote by $$\sum_i \mathbb Z \omega_i,$$ is contained in $O_k$ (since $O_k$ is closed under addition).

• However, unless that $\omega_i$ are linearly independent over $\mathbb Z$, their span in $O_k$ won't be isomorphic to the direct sum of the $\mathbb Z \omega_i$.

There is always a natural surjection $$\bigoplus_i \mathbb Z_i \to \sum_i \mathbb Z \omega_i$$ (the source being the direct sum and the target being the span in $O_k$), but in general it has a kernel; indeed, the kernel is the collection of all linear dependence relations between the $\omega_i$.

Now if the $\omega_i$ are an integral basis, then they are linearly independent over $\mathbb Z$, and so the direct sum is embedded into $O_k$.

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Thank you Matt. This is very useful too. –  Conan Wong Jan 13 '13 at 2:26