Can I always choose $1$ in the integral basis of a number field?

Question

I am trying to show that given any number field $K$ of degree $n$, I can always find an integral basis $\{ w_1, ..., w_n \}$ where $w_1 = 1$. I am interested in how to proving this, but I am stuck at the moment.
I would greatly appreciate any suggestions. Thank you very much!

Answer 1

Yes.

Suppose one has an integer basis $\alpha_1,\ldots,\alpha_n$. Then $$1=\sum_{i=1}^n c_i\alpha_i$$ with $c_i\in\Bbb Z$. I claim that $g:=\gcd(c_1,\ldots,c_n)=1$. This is because $1/g\in\Bbb Q\cap\mathcal{O}_K=\Bbb Z$. Then there is an integer matrix $M$ of determinant $\pm1$ with first row $(c_1,\ldots,c_n)$. Then the entries of the column vector $$M\pmatrix{\alpha_1\\\vdots\\\alpha_n}$$ are also an integer basis, and the top entry is $1$.

Answer 2

Here’s another proof, or maybe the same proof as @LordSharttheUnknown’s but in different words:

Let our field $K$, of degree $n$ over $\Bbb Q$, have ring of integers $R$. We know that $R$ is a free $\Bbb Z$-module of rank $n$. I’m going to argue that $R/\langle1\rangle$ is free of rank $n-1$, thus with a $\Bbb Z$-basis $\{\beta_2,\cdots,\beta_n\}$. Lift these to elements of $R$, which, together with $1$, clearly will form an integal basis.

The question is whether $R/\langle1\rangle$ has torsion. If not, my claim above is verified. Let then $\xi\in R/\langle1\rangle$, torsion of order $m$, so that $m\xi=0$; lift to $R$ to get $x\in R$ with $mx=r\in\Bbb Z$. Now, certainly $g=\gcd(m,r)=1$, otherwise $\frac mg\xi=0$ and the order of $\xi$ would be less than $m$.

From $Am+Br=1$ we get $mBx=Br=(1-Am)\cdot1$, $m(Bx+A)=1$ so that $\frac1m\in R$, contradiction unless $m=\pm1$, so there’s no torsion in $R/\langle1\rangle$ after all. And that does it, when you grant that a finitely-generated torsion-free $\Bbb Z$-module is free.