# $\mathbb Z[\frac{2+i}{5}]\cap \mathbb Q =\mathbb Z$

As it says in the title I want to show that $$\mathbb Z[\frac{2+i}{5}]\cap \mathbb Q =\mathbb Z$$.

Set $$\omega=\frac{2+i}{5}$$. $$\mathbb Z[\omega]$$ is the smallest ring that contains $$\mathbb Z$$ and $$\omega$$. Therefore,

$$$$\mathbb Z[\omega] = \{a_0+a_1 \omega+\cdots+a_k \omega^k\mid a_i\in \mathbb Z,k\in \mathbb N\}$$$$

Let, then, $$q\in\mathbb Z[\frac{2+i}{5}]\cap \mathbb Q$$. Since $$q\in\mathbb Q$$ we can write $$q=\frac{m}{n}$$, where $$m,n$$ are relatively prime integers and $$m\neq 0$$. Now, since $$q\in\mathbb Z[\frac{2+i}{5}]$$ we have that

$$$$q=a_0+a_1 \omega+\cdots+a_k \omega^k$$$$

for some $$a_i\in \mathbb Z$$. Then,

\begin{align} &\frac{m}{n}=a_0+a_1 \frac{2+i}{5}+\cdots+a_k \left(\frac{2+i}{5}\right)^k \\ \Longrightarrow \ &5^km=5^kna_0+5^{k-1}na_1(2+i)+\cdots+a_kn(2+i)^k \end{align}

Hence, $$n\mid 5^km$$ and since $$\gcd(n,m)=1$$ we have that $$n|5^k$$.

Thus, every rational number in $$\mathbb Z[\omega]$$ is of the form $$\frac{m}{5^l}$$ where $$m\in\mathbb Z$$, $$l$$ is a positive integer and when $$l\geq 1$$ it holds that $$5\nmid m$$. It suffice then to show that it always hold that $$l=0$$. I am stuck here. I would very much appreciate some help!

Hint: Look at the equation $$5^km = n(5^ka_0+5^{k-1}a_1(2+i)+ \cdots + a_k(2+i)^k)$$ as an equation between Gaussian integers. Note that $$2+i$$ is a prime in the ring of Gaussian integers arising from the factorization of the integer prime $$5$$, namely $$(2+i)(2-i)=5$$. You already noticed that $$n$$ is a power of $$5$$. In particular, if $$m/n$$ is not an integer, you may assume that $$m$$ is not divisible by $$5$$. Hence, viewing it as a Gaussian integer, it should not be divisible by $$2+i$$ or $$2-i$$. Can you see the contradiction from here?