A number being both Gaussian integer and Eisenstein integer If a complex number is Gaussian integer and Eisenstei integer at the same time, is it true that the number can only be ordinary integer? Why?
 A: Such a number would lie in $\Bbb Q(i)\cap\Bbb Q(i\sqrt3)$, but that is $\Bbb Q$.
Otherwise, $\Bbb Q(i)$ would equal $\Bbb Q(i\sqrt3)$, since their intersection
must then be a $\Bbb Q$-vector space of dimension $\ge2$. Then $-3$ would be
a square in $\Bbb Q(i)$: $-3=(a+bi)^2=a^2-b^2+2abi$, $a$, $b\in\Bbb Q$.
Either $a=0$ or $b=0$ so that $-3=a^2$ or $-3=b^2$. Both are impossible.
The same argument shows that $\Bbb Q(\sqrt r)\cap\Bbb Q(\sqrt s)=\Bbb Q$
for non-square $r$, $s\in\Bbb Q$ unless $rs$ is a square in $\Bbb Q$.
A: The ring of Gaussian integers, $\mathbf{Z}[i]$, and the ring of Eisenstein integers, $\mathbf{Z}[\zeta]$ (where $\zeta=e^{2\pi i/3}$), are the the respective integral closures of the ring $\mathbf{Z}$ of integers in their fields of fractions, $\mathbf{Q}(i)$ and $\mathbf{Q}(\zeta)$. The intersection $\mathbf{Q}(i)\cap\mathbf{Q}(\zeta)$ is an intermediate field of the quadratic extension $\mathbf{Q}(i)/\mathbf{Q}$, and so must either be equal to $\mathbf{Q}(i)$ or $\mathbf{Q}$. Suppose the former is true. Then $\mathbf{Q}(i)\subseteq\mathbf{Q}(\zeta)$. Since both fields are quadratic over $\mathbf{Q}$, this inclusion must in fact be in equality. Therefore $\mathbf{Z}[i]=\mathbf{Z}[\zeta]$. But, for instance, $\mathbf{Z}[i]$ has exactly four units ($\pm 1$ and $\pm i$) while $\mathbf{Z}[\zeta]$ has six units ($\pm 1$, $\pm \zeta$, and $\pm\zeta^2$). This is a contradiction. 
Thus $\mathbf{Q}(i)\cap\mathbf{Q}(\zeta)=\mathbf{Q}$, which implies that $\mathbf{Z}[i]\cap\mathbf{Z}[\zeta]\subseteq\mathbf{Q}$. Because $\mathbf{Z}$ is integrally closed, this forces $\mathbf{Z}[i]\cap\mathbf{Z}[\zeta]\subseteq\mathbf{Z}$.
