Let $K$ be a number field and let $O$ be a subring of the ring of integers $O_K$ of $K$. Show that $O$ contains a $\mathbb{Q}$-basis of $K$ if and only if the field of fractions of $O$ is $K$.

I proved the "only if" part.

Now for the "if" part I have a hint: Show that it makes sense to talk about the largest subfield $K_0$ of $K$ such that $O$ contains a $\mathbb{Q}$-basis for $K_0$. Then assume for contradiction that $K_0 \neq K$.

Clearly $\mathbb{Q}$ is a subfield of $K$ and $O$ contains a $\mathbb{Q}$-basis of it (namely $1$). So it has sense to talk about the largest subfield of $K$ satisfying this condition. I have two main ideas:

$1$) Since $K_0 \neq K$ there exists $\alpha \in K-K_0$. I can write $\alpha=\frac{\beta}{\gamma}$ for some $\beta, \gamma \neq 0 \in O$ since $K$ is the field of fractions of $O$. Now I would like to write $\alpha$ as a $\mathbb{Q}$-linear combination of the basis but I don't know how to proceed.

$2$) Since $K$ is the minimal subfield in which $O$ can be embedded (by definition of field of fractions), then $O$ is not contained in $K_0$ and thus there exists $\alpha \in O-K_0$.

Any suggestions?


You know $\mathbb{Q}O_L=\bigcup_n \frac1nO_L=L$ for any number field $L$ (Recall the usual proof: for $\alpha\in L$ look at the minimal polynomial $f(X)=a_dX^d+a_{d-1}X^{d-1}+\dots+a_0\in\mathbb{Z}[X]$ of $\alpha$. Then $a_d\alpha\in O_L$ since it satisfies $(a_d\alpha)^d+a_{d-1}a_d(a_d\alpha)^{d-1}+\dots+a_0a_d^d=0$.). With the same notation we have $\alpha^{-1}\in a_0^{-1} O\subseteq\mathbb{Q}O$ (for $0\neq\alpha\in O$) since $a_0\alpha^{-1}=-a_1-a_2\alpha-\dots-a_d\alpha^{d-1}\in\mathbb{Z}[\alpha]\subseteq O$. Hence $O\subseteq K_0$ gives $K=\mathbb{Q}O\subseteq K_0$, which is only possible if $K_0=K$.

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  • $\begingroup$ The first equality is a restatement of what is asked to prove. $\endgroup$ – Rybin Dmitry May 15 '19 at 15:22
  • $\begingroup$ I assumed the OP has seen that before and just a problem of translating that to another description, and no it isn't a restatement of the problem because our $O$ need not be integrally closed. Anyway, a sketch of proof is now given. $\endgroup$ – user10354138 May 15 '19 at 15:29
  • $\begingroup$ But if $K_0 \neq K=\mathrm{Frac}(O)$ then $O \not \subseteq K_O$ by minimality of $K$ $\endgroup$ – user289143 May 15 '19 at 15:51
  • $\begingroup$ There is no need to assume $K_0\neq K$. $K_0\subseteq K$ exists, and we just need to prove it contains all elements of $K$, which is what we did here. Of course adding the artificial $K_0\neq K$ changes it from a direct proof to a proof by contradiction. $\endgroup$ – user10354138 May 15 '19 at 15:59
  • $\begingroup$ Still I don't get your proof. You say $\mathbb{Q}O_L=L$ for any number field $L$, but then you use $K=\mathbb{Q}O$ and $O \subseteq O_K$ $\endgroup$ – user289143 May 15 '19 at 21:58

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