Quadratic orders embedded in matrices

Question

Let $\tau$ be a CM point and let $\mathcal O$ be the quadratic order corresponding to the lattice $[\tau,1]$, that is $$\mathcal O =\lbrace \lambda \in \mathbb C: \lambda[\tau,1]\subset[\tau,1]\rbrace.$$

Embed $\mathcal O$ in $2$-by-$2$ matrices via its action on $[\tau,1]$: $$f\colon \mathcal O\longrightarrow \operatorname{M}_2(\mathbb Z).$$

Let $N$ be a positive integer. The embedding $f$ induces a homomorphism $$\mathcal O/N\mathcal O \longrightarrow \operatorname{M}_2(\mathbb Z/N\mathbb Z),$$ $$a+N\mathcal O\longmapsto f(a)+N\cdot\operatorname{M}_2(\mathbb Z).$$

Is this homomorphism injective?

Answer 1

Yes, it is injective. Write $\Lambda$ in place of $[\tau,1]$. Let $a$ be in the kernel. This means that the matrix of $a$ is $0$ modulo $N$ or, in other words, $a \Lambda \subseteq N \Lambda$. But then $(a/N) \Lambda \subseteq \Lambda$ so, by definition of $\mathcal{O}$, we have $a/N \in \mathcal{O}$. This shows $a \in N \mathcal{O}$ and $a$ is $0$ in $\mathcal{O}/N \mathcal{O}$, as desired.