Subrings of Quadratic Integer Ring

Solution 1:

Your ring $\mathbf Z[\omega]$ is the ring of integers ${\cal O}_{K}$ of the quadratic field $K=\mathbf Q (\omega)$. You can begin classifying the subrings (by definition, these will be supposed to contain $1$) $\cal O$ of ${\cal O}_{K}$ acording to their ranks as $\mathbf Z$-modules. Recall that since $\mathbf Z$ is a PID and $\cal O$ has no (additive) torsion, $\cal O$ is $\mathbf Z$-free, necessarily of rank $1$ or $2$. Neglecting $\mathbf Z$ (which corresponds to rank $1$), let us concentrate on the subrings $\cal O$ of rank $2$, usually called $orders$ (thus ${\cal O}_{K}$ is the maximal order of $K$). There is a well developped theory of orders, which you can find in any classical textbook on ANT such as Ireland-Rosen.

Back to your question. Let $D_K$ be the discriminant of the quadratic field. Putting $\delta_K=(d_K + \sqrt d_K)/2$, it is well known and easily checked that ${\cal O}_{K}$ admits a $\mathbf Z$-basis {$1,\delta_K$} (this is more convenient than the formulas with $\omega$). Since $\cal O$ and ${\cal O}_{K}$ have both $\mathbf Z$-rank $2$, the index $g=[{\cal O}_{K}:\cal O]$ is finite, and $g{\cal O}_{K}\subset \cal O$. Let us show that $\cal O$ has basis {$1, g\delta_K$}. Since ${\cal O}_{K}$ has basis {$1,\delta_K$}, clearly $\mathbf Z +g{\cal O}_{K}$ has basis {$1, g\delta_K$}, so we need only to show that $\mathbf Z + g\delta_K \mathbf Z$ has index $g$ in ${\cal O}_{K}=\mathbf Z + \delta_K \mathbf Z$. But this is obvious. Summarizing, we have obtained ${\cal O} =\mathbf Z + g\delta_K \mathbf Z= \mathbf Z + g{\cal O}_{K}$. This can straightforwardly translated back to your question in terms of $\omega$.

NB: The above computations can be skipped by using the theorem showing the existence of aligned bases for a submodule $N$ of a module $M$ of finite type over a PID, see e.g. www.math.uconn.edu/~kconrad/blurbs/.../modulesoverPID.pdf ./.

Solution 2:

I'll look at the special case $\omega=i=\sqrt{-1}$; the general case is really no harder.

If $R$ is a subring of $\Bbb Z[i]$ properly containing $\Bbb Z$, then it's an additive subgroup of $\Bbb Z[i]$. It corresponds to a subgroup of the quotient group $\Bbb Z[i]/\Bbb Z$ which is cyclic and generated by the class of $i+\Bbb Z$. This means that $$R=\Bbb Z+gi\Bbb Z $$ for some positive integer $g$. But $\Bbb Z+gi\Bbb Z$ is a ring, basically since $(gi)^2\in\Bbb Z+gi\Bbb Z$, etc.

Solution 3:

I am assuming that you want to characterize the unital subrings of $\mathbb{Z}[\omega]$ otherwise the statement is not true. For example $5 \mathbb{Z}[\omega]$ is a subring of $\mathbb{Z}[\omega]$ not containing an identity.

Assume $R$ is a unital subring of $\mathbb{Z}[\omega]$. Ofcourse $\mathbb{Z} \subseteq R$. If there is no non-integer element in $R$, then $R = \mathbb{Z}$ corresponding to the case $g = 0$.

Assume there exists at least one non-integer element in $R$. For simplicity, I will define the imaginary part of $a + b\omega \in \mathbb{Z}[\omega]$ as $b$. By our assumption there exists at least one element in $R$ with a non-zero imaginary part. Because $R$ is a group, it is closed under additive inverses so there exists at least one element in $R$ with a positive imaginary part. Then define $g \in \mathbb{Z}^+$ to be the minimum positive imaginary parts of elements of $R$. We claim that $R = \mathbb{Z}[g\omega]$.

Let $a + b\omega \in \mathbb{Z}[\omega]$. $b \omega \in \mathbb{Z}[\omega]$ because $\mathbb{Z} \subseteq \mathbb{Z}[\omega]$. Divide $b$ to $g$ and get $b = xg + r$ where $0 \leq r < g$. Notice $r\omega = b\omega - x \cdot g\omega \in \mathbb{Z}[\omega]$. But then, by definition of $g$, $r = 0$ implying $g \mid b$. This shows that $R \subseteq \mathbb{Z}[g\omega]$. The other direction is obvious so $R = \mathbb{Z}[g \omega]$ as claimed.