Given a number field $K$, showing that $\mathbb{Q}O_K=K$

Question

Let $K$ be a number field. I want to show that $\mathbb{Q}O_K=K$.

$\mathbb{Q}O_K \subseteq K$ is clear. Now let $\alpha \in K$, then there exists the minimal polynomial $f(X)=X^n +\cfrac{a_{n-1}}{b_{n-1}}X^{n-1}+\dots+\cfrac{a_1}{b_1}X + \cfrac{a_0}{b_0} \in \mathbb{Q}[X]$. We need to find some $g(X)\in \mathbb{Z}[X]$ such that $g(\alpha)=0$ so that $\alpha \in O_K$.

(I do not know what $\mathbb{Q}O_K $ is actually, I am guessing that it consists of elements $q\beta$, i.e., $q\in\mathbb{Q},\beta \in O_K$. ) So, let $D=lcm(b_0,b_1,\dots,b_{n-1})$, then $Df(X)= f(X)=DX^n +D\cfrac{a_{n-1}}{b_{n-1}}X^{n-1}+\dots+D\cfrac{a_1}{b_1}X + D\cfrac{a_0}{b_0} \in \mathbb{Z}[X]$ and $Df(\alpha)=0$ however it is not monic so I can not conclude that $\alpha \in O_K$. How can I solve this problematic part and what is $\mathbb{Q}O_K$ exactly?

Answer 1

This is a classical result on $\mathcal{O}_K$, which is in every text on algebraic number theory. A detailed proof is, say, given here, Corollary $1.5$. It uses the minimal polynomial $f(x)$ of an element $\alpha\in K$ to show that there exists $d\in \mathbb{Z}$ such that $\alpha d\in \mathcal{O}_K$ (that is, $\alpha d=\beta\in \mathcal{O}_K$, or equivalently, $\alpha=\beta/d$). And indeed, the difficulty is, to construct a monic $g(x)\in \mathbb{Z}[x]$ with $\alpha d$ as root. The solution is given by choosing $d$ to be the least common multiple of the denominators of the coefficients of the monic polynomial $f(x)$, and to put $$ g(x)=d^{\deg(f)}f(x/d). $$

Answer 2

For those who are interested in and having problem understanding the proof like me, I will write down my work to show that $K \subseteq\mathbb{Q}O_K$. Note that instead of $\mathbb{Q}O_K$, we can roughly use $=\cfrac{O_K}{\mathbb{Z}}$.

So let $\alpha \in K$, there exists $f(X) \in \mathbb{Q}[X]$, the minimal polynomial of $\alpha$. $$f(X)=X^n+\cfrac{a_{n-1}}{b_{n-1}}X^{n-1}+\dots+\cfrac{a_1}{b_1}X+\cfrac{a_0}{b_0}$$ and we have $f(\alpha)=\alpha^n+\cfrac{a_{n-1}}{b_{n-1}}\alpha^{n-1}+\dots+\cfrac{a_1}{b_1}\alpha+\cfrac{a_0}{b_0}=0$. Our aim is to find a polynomial $g(X)\in \mathbb{Z}[X]$ such that $g(\beta)=0$ where $\beta=z.\alpha$ for some $z \in \mathbb{Z}$, so $\beta \in O_K$.

Our polynomial $f$ has rational coefficients, we have to get rid of these rational coefficients and get integer coefficients. (Of course if $f$is already in $\mathbb{Z}[X]$, there is nothing to show.)

The most natural idea is to multiply these coefficients with least common multiples of denominators of them, set $D=lcm(b_0,b_1,b_{n-1})$. Then, $Df \in \mathbb{Z}[X]$. Simply saying that $Df(X)=g(X)$ does not work since we have to plug some $\cfrac{\alpha}{z}$ in $g$ and the terms $D\cfrac{a_i}{b_i}X^i$ must be "consistent", I mean we should get $0$ in the end. Check for example that what happens if we plug $\alpha$ in $Df(X)=g(X)$:

$Df(\alpha)=g(\alpha)=D\alpha^n+D\cfrac{a_{n-1}}{b_{n-1}}\alpha^{n-1}+\dots+D\cfrac{a_1}{b_1}\alpha+D\cfrac{a_0}{b_0}$.

Due to this inconsistency, let us manipulate $g(X)$ a little. Notice that $f(\alpha)=\alpha^n+\cfrac{a_{n-1}}{b_{n-1}}\alpha^{n-1}+\dots+\cfrac{a_1}{b_1}\alpha+\cfrac{a_0}{b_0}=0$$\implies D^nf(\alpha)=D^n\alpha^n+D^n\cfrac{a_{n-1}}{b_{n-1}}\alpha^{n-1}+\dots+D^n\cfrac{a_1}{b_1}\alpha+D^n\cfrac{a_0}{b_0}=0 $ $\implies (D\alpha)^n+D\cfrac{a_{n-1}}{b_{n-1}}(D\alpha)^{n-1}+\dots+D^{n-1}\cfrac{a_1}{b_1}(D\alpha)+D^n\cfrac{a_0}{b_0}=0$

Now say $g(X)= X^n+D\cfrac{a_{n-1}}{b_{n-1}}X^{n-1}+\dots+D^{n-1}\cfrac{a_1}{b_1}X+D^n\cfrac{a_0}{b_0} \in \mathbb{Z}[X].$

Finally, remember we are looking for some integer $z$ and $\beta=z.\alpha$ so that $\beta$ is an algebraic integer, thus, set $z=D$ and we have $\beta= z.\alpha=D.\alpha$. $$\implies g(\beta)=g(D.\alpha)= (D\alpha)^n+D\cfrac{a_{n-1}}{b_{n-1}}(D\alpha)^{n-1}+\dots+D^{n-1}\cfrac{a_1}{b_1}(D\alpha)+D^n\cfrac{a_0}{b_0}=0$$; $K \subseteq \mathbb{Q}O_K.$