A question about rings of algebraic integers Let $R$ be a subring of the field of algebraic numbers. If $R\cap \mathbb{Q}= \mathbb{Z}$, does it follow that all of the elements of $R$ are algebraic integers?
 A: No, this is not true in general. For example, let $R=\mathbb{Z}[\alpha]$ with an algebraic number $\alpha$ which is not an algebraic integer, and such that $R\cap \mathbb{Q}=\mathbb{Z}$, e.g., 
$$
\alpha=\frac{3-\sqrt{2}}{7}.
$$
Then indeed $R\cap \mathbb{Q}=\mathbb{Z}$, but not all elements are algebraic integers.
In fact, the minimal polynomial of $\alpha$ is given by $f(x)=x^2 - \frac{6}{7}x + \frac{1}{7}$, which does not have integer coefficients. 
A: Let $\alpha=(3-\sqrt{2})/7$.
Here is a proof-sketch of the statement $$\mathbb{Z}[\alpha]\cap \mathbb{Q}= \mathbb{Z},$$ based on the facts that the ring $\mathbb{Z}[\sqrt{2}]$ is a unique factorization domain, and $3-\sqrt{2}$ is a prime element of this ring.
Let $L$ be a $\mathbb{Z}$-linear combination of powers of $\alpha$. Plainly $L$ has the form $\beta/7^k$ with $\beta\in \mathbb{Z}[\sqrt{2}]$ and $k\ge0$. Suppose  that $\beta\in \mathbb{Z}$. We can assume that the fraction $\beta/7^k$ is in lowest terms. Then $k$ must be 0: Otherwise the prime $3-\sqrt{2}$
appears to a negative power in $\beta/7^k$.  But $3-\sqrt{2}$ appears to the power 0 in $\alpha$, hence to a non-negative power in $L$.
