Question: Let $L = \mathbb{Q}(\sqrt{m})$ be a quadratic number field and $p \in \mathbb{Z}$ be any prime number so that $p \nmid m$ in $\mathbb{Z}$. How do we show that $$\mathcal{O}_L/p \mathcal{O}_L \cong \mathbb{Z}(\sqrt{m})/p\mathbb{Z}(\sqrt{m})$$ where $\mathcal{O}_L$ denotes the ring of integers in $L$ and $p\mathcal{O}_L$ denotes the ideal in $\mathcal{O}_L$ generated by $p$.
My attempt: Using the following identity for the number rings in quadratic field $$\mathcal{O}_L = \begin{cases} \mathbb{Z}(\sqrt{m}) &\ \text{if}\ m \equiv 2\ \text{or}\ 3 \pmod 4, \quad \\ \mathbb{Z}\left(\frac{1 + \sqrt{m}}{2}\right)&\ \text{if}\ m \equiv 1 \pmod 4\\ \end{cases}$$ The above result holds trivially for $m \equiv 2\ \text{or}\ 3 \pmod 4$, however how do we show for $m \equiv 1 \pmod 4$?
Source: Page 24 of Kato et al's Introduction to Class Field Theory.
The book references two facts: $\mathcal{O}_L/\Bbb{Z}[\sqrt{m}]$ has index $1$ or $2$ ($2$ in the case of $m\equiv 1$ mod $4$), and $p$ is odd. Consider the composition of the inclusion map and the quotient map $$\Bbb{Z}[\sqrt{m}]\to \mathcal{O}_L\to \mathcal{O}_L/p \mathcal{O}_L=\Bbb{Z}/p^k\Bbb{Z}.$$ Here $k=1,2$, depending on whether or not $p$ is inert in $\mathcal{O}_L$. The kernel of the composition above is $\Bbb{Z}[\sqrt{m}]\cap p\mathcal{O}_L=p\Bbb{Z}[\sqrt{m}]$ (here we use $2\nmid p$), and so you map $\Bbb{Z}[\sqrt{m}]/p\Bbb{Z}[\sqrt{m}]$ to a subgroup of $\Bbb{Z}/p^k\Bbb{Z}$. This subgroup has at most index $2$ by our first fact* (see proof below), and since $p$ is odd, no subgroup of order $2$ exists. Thus the subgroup is the whole group, i.e. $\Bbb{Z}[\sqrt{m}]/p\Bbb{Z}[\sqrt{m}] = \mathcal{O}_L/p\mathcal{O}_L$.
Let me fully prove the above mentioned fact in general. Since our argument basically boils down to a cardinality argument, we can forget the ring structure and just talk about the underlying additive groups. Let $G$ be an abelian group with $H\subset G$ index $n$ subgroup and $K\subset G$ any subgroup. $H\cap K=:M$ is again a subgroup.
By the third isomorphism theorem, $(G/M)(H/M)\cong G/H$, so $H/M$ has index $n$ in $G/M$. Since $G/M$ surjects to $G/K$, and this quotient map preserves $H/M$, this subgroup must then have at most index $n$ in $G/K$.