Subrings of Quadratic Integer Ring Define:
$$
\omega = \begin{cases} \sqrt{D }& D \equiv 2,3 \mod 4
\\ 
\frac{1 + \sqrt{D}}{2} & D \equiv 1 \mod 4
\end{cases}
$$
for (fixed) $D$ squarefree.
Let $\mathbb{Z}[\omega]$ denote the ring:
$$
\mathbb{Z}[\omega] = \{a+ b \omega | a,b \in \mathbb{Z}\}
$$
Show that the subrings of $\mathbb{Z}[\omega]$ are given by $\mathbb{Z}[g\omega]$ where $ g\in \mathbb{N}$ (or is just plain old $\mathbb{Z}$.
Does anyone have any ideas? I am not sure where to begin. I have proved some results about the norm and conjugate on this ring, as well as the units, but do not know how to prove this. In general, it seems like "characterizing" subrings, is quite hard.
 A: 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  ./.
A: 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.
A: 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.
