In Dorman's paper, Global orders in definite quaternion algebras as endomorphism rings for reduced CM elliptic curves, he considers the following situation:
$K = \mathbb{Q}(\sqrt{d})$ where $d$ is a fundamental, negative, odd discriminant, with ring of integers $\mathcal{O}_K;$
Let $D^{-1}$ be the inverse different of $\mathcal O_K;$
$l$ is a fixed prime;
We take a prime $q$ to be such that for every prime divisor $p\mid d$, we can find a solution $\lambda_p^2 = -lq\;(\bmod\; p)$. (We can take $q$ prime by Dirichlet's theorem on prime progressions and just use CRT.).
Now, by design, $(q) = \mathfrak q \mathfrak{\bar q}.$
We define the quaternion algebra $\mathbb B_{l,\infty} = \left\{[\alpha,\beta] = \begin{bmatrix}
\alpha & \beta \\
-lq\bar\beta & \bar\alpha
\end{bmatrix}: \alpha,\beta\in K\right\}.$
For any $\mathfrak a\subset \mathcal O_K,$ let
$\lambda'_p = (-1)^{v_{\mathfrak p}(\mathfrak a)}\lambda_p$ where $(p) = \mathfrak p^2.$ Then, you CRT these to get a $\lambda'\in \mathbb Z/d\mathbb Z$, and here comes the doozy of a definition that is giving me problems.
Now, define
$$R(\mathfrak a) = \left\{[\alpha,\beta]\in \mathbb{B}_{l,\infty}: \alpha\in D^{-1}, \beta\in \mathfrak{q}^{-1} D^{-1}\mathfrak a^{-1}\mathfrak {\bar a}, \alpha\equiv \lambda'\beta \;(\bmod\; \mathcal O)\right\}.$$
The claim is that $R(\mathfrak a)$ is a maximal order in the quaternion algebra, but I can't even show that it's a ring! I keep having a negative or a conjugate in the wrong place.
Any guidance is thoroughly appreciated! Thanks in advance! 🙂
What I've done so far:
Surely sums of such matrices are still of this form; so, my problem is with products. I am going to let $\pi = \sqrt{d}$ and take a product of two such matrices $A = 1/\pi[\alpha,\beta]$ and $B = 1/\pi[\gamma,\delta]$. (I do this to make the final condition in $R(\mathfrak a)$ easier to confirm.)
$AB = 1/d[\alpha\gamma-lq\beta\bar\delta, \alpha\delta + \beta\bar\gamma].$ Then, the final condition says we would like $$\alpha\gamma-lq\beta\bar\delta\equiv\lambda'( \alpha\delta + \beta\bar\gamma )\;(\bmod\; d)$$
given that $\alpha \equiv \lambda' \beta \;(\bmod\; \pi)$ and $\gamma\equiv \lambda'\delta \;(\bmod\; \pi).$
Well, taking the products of the given relations, we have
$$(\alpha – \lambda'\beta)(\gamma-\lambda'\delta) \equiv \alpha\gamma -lq\beta\delta -\lambda'(\alpha\delta + \beta\gamma)\equiv 0 \;(\bmod\; d).$$
So, this is a recurrent theme with showing this to be a ring– the conjugates are missing. By fiddling around, I realized that $\beta\lambda'((-\gamma+\lambda'\delta) +(\bar\gamma -\lambda'\bar\delta))\equiv 0\;(\bmod\; d)$ since each of the terms inside is divisible by $\pi$ and the difference of conjugates adds another factor of $\pi$. So, we can add this onto the previously computed product to get the desired relation.
As for showing that either of the first two relations still holds for the product of $A$ and $B$, I'm at a loss.
Best Answer
See Remark 6.1 in Singular Moduli for Arbitrary Discriminants. I think that what you're observing is this Remark 6.1, that this is a mistake in Dorman's work and $R(\mathfrak{a},\lambda)$ is not closed under multiplication necessarily. It is corrected in the linked article (``On Singular Moduli for Arbitrary Discriminants'' by Lauter and Viray, where they generalize the results of Dorman and Gross-Zagier). The upshot is that one needs an extra hypothesis $\lambda \mathfrak{q}^{-1}\overline{\mathfrak{a}}\mathfrak{a}^{-1}\subset \mathcal{O}$ so that $R(\mathfrak{a},\lambda)$ is closed under multiplication. See Lemma 6.3 for a proof that $R(\mathfrak{a},\lambda)$ is an order given this extra assumption.