# Fraction field and ring of integers

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 $$K_0=K$$.
• 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. – user10354138 May 15 '19 at 15:29
• But if $K_0 \neq K=\mathrm{Frac}(O)$ then $O \not \subseteq K_O$ by minimality of $K$ – user289143 May 15 '19 at 15:51
• 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. – user10354138 May 15 '19 at 15:59
• 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 $O \subseteq O_K$ – user289143 May 15 '19 at 21:58