Algebraic number field basis

• Nov 23rd 2008, 03:25 AM
Banach
Algebraic number field basis
Hi! I tried to show the following but i failed: We have a finite galois extension $\displaystyle L/\mathbb{Q}$ with commutative galois group and $\displaystyle O_L$ the ring of algebraic integers in L. Suppose there exists $\displaystyle x \in O_L$ so that the set $\displaystyle \{\sigma(x): \sigma \in$ Gal($\displaystyle L/\mathbb{Q})\}$ forms a $\displaystyle \mathbb{Z}$-Basis of $\displaystyle O_L$.
Then for every field $\displaystyle K \subset L$ there also exists $\displaystyle y \in O_K$ so that $\displaystyle \{\phi(y): \phi \in$ Gal($\displaystyle K/\mathbb{Q})\}$ forms a $\displaystyle \mathbb{Z}$-Basis of $\displaystyle O_K$.

I could show that $\displaystyle K/\mathbb{Q}$ is normal and hence each $\displaystyle \phi \in$ Gal$\displaystyle (K/\mathbb{Q})$ is just a restriction of $\displaystyle \sigma \in$ Gal $\displaystyle (L/\mathbb{Q})$. But i dont know how to continue the argument. Can anybody please help me?

Greetings
Banach
• Nov 23rd 2008, 10:37 AM
ThePerfectHacker
Quote:

Originally Posted by Banach
forms a $\displaystyle \mathbb{Z}$-Basis of $\displaystyle O_L$.

I want to help but I do not what a Z-basis is.
Can you tell me?
• Nov 23rd 2008, 10:57 AM
Banach
Hi! It is very nice that you want to help me. Here it just means that every $\displaystyle a \in O_L$ can be written as $\displaystyle a=\sum \limits_{i=1}^n \lambda_i \sigma_i(x)$ with $\displaystyle \lambda_i \in \mathbb{Z}$ and $\displaystyle \sigma_i$ the elements of the galois group.
• Nov 25th 2008, 10:03 PM
Banach
Another idea: It also follows that K is galois, so maybe the primitive element theorem helps?