Involution on false elliptic curve

Question

Let B be a indefinite quaternion algebra with discriminant $d>1$, maximal order $\mathcal{O}$ and standard involution $'$, then there exists $t\in{B}$ such that $t^2=-d$ and a new involution on B given by $a^*=t^{-1}a^\prime{t}$. A fundamental result says that for scheme S on which d is invertible, and any abelian surface A together with an embedding $\mathcal{O}\to\mathrm{End}(A)$, i.e. a false elliptic curve over S, there exists a unique principal polarization on A such that for any geometric point $s\in{S}$, the Rosati involution on $A_s$ induces $*$ on $\mathcal{O}$. For any prime $p\nmid{d}$, we know $\mathcal{O}_p$ is isomorphic to $M_2(\mathbb{Z}_p)$, does there exists a non-trivial idempotent element $e\in{\mathcal{O}_p}$ such that $e^*=e$? This is claimed in the paper "non-optimal levels of mod l modular representations" before lemma 7, but they do not give any reason. Thanks.

Answer 1

I think that you can construct an idempotente $e=e^*$ in some quadratic base change $B\otimes F$ where $F$ is a subfied of $B$ fixed by the involution.

Now, if you choose $F$ so the prime $p$ splits in it, after localizing $e\in({\cal O}\otimes{\cal O}_F)_p={\rm M}_2({\Bbb Z}_p)$

Answer 2

I find it helpful to think over a field first, to get my bearings. For notational reasons that will hopefully become evident, let $i := t$. Then (Exercise 2.5 in http://quatalg.org, which basically boils down to Gram-Schmidt orthogonalization) there exists $j \in B$ such that $B=(d,b\,|\,\mathbb{Q})$, where $j^2=b \in \mathbb{Q}^\times$. In particular, $B$ is spanned by $1,i,j,ij$ and $ji=-ij$. We then compute that $1^*=1$ and $i^*=-i$ and then $$ j^*=i^{-1}\overline{j}i = i^{-1}(-j)i=i^{-1}(ij)=j. $$ Thus $(ij)^* = ij$ (e.g. $(ij)^*=j^* i^* = -ji = ij$). So $V := \{\alpha \in B : \alpha^*=\alpha\}$ is a $\mathbb{Q}$-vector space of dimension $3$, spanned by $1,j,ij$. (The standard involution negates the pure quaternions; conjugation by $i$ negates $j,ij$, we just put these together.) This argument applies over any base field $F$ not of characteristic 2.

We want an element $e=t+yj+zij$ of reduced norm $0$ and reduced trace $1$, so then $e^2-e=0$ and $e$ is idempotent. We compute $\mathrm{trd}(e)=2t=1$ so $t=1/2$, and $\mathrm{nrd}(e)=t^2-by^2+bdz^2=0$ so $y^2-dz^2=1/(4b)$. We now pass to $\mathbb{Q}_p$ with $p$ odd, $p \nmid d$. We recall (Main Theorem 5.4.4) that $(d,b\,|\,\mathbb{Q}_p) \simeq \mathrm{M}_2(\mathbb{Q}_p)$ if and only if $b \in \mathrm{Nm}_{K|\mathbb{Q}_p}(K^\times)$, where $K=\mathbb{Q}_p[i]$. Since we are taking $p$ splitting $B$, we conclude $b$ is a norm and hence so is $(4b)^{-1}$, so we find an idempotent. It is nontrivial because it is nonscalar, else $y^2-dz^2=0=1/(4b)$, a contradiction.

This construction did not respect integrality, but maybe now we are ready for analyzing denominators! I will ask that $i \in \mathcal{O}$ to start (without loss of generality, see 43.6.6) and that $p$ is odd (but still with $p \nmid d$). Then $i \in \mathcal{O}_p^\times$ (since $i^2=d \in \mathbb{Z}_p^\times$), and $\mathbb{Z}_p[i]$ is the valuation ring in the $\mathbb{Q}_p$-algebra $\mathbb{Q}_p[i]$ (which is a product of fields when $d$ is a quadratic residue modulo $p$). So $\mathcal{O}_p$ is a free left $\mathbb{Z}_p[i]$-module of rank $2$ (indeed, it is torsion free and rank $2$ tensoring with $\mathbb{Q}_p$), so has a basis $1,j$. Applying the Gram-Schmidt orthogonalization process (finding the orthogonal complement of $1,i$ in $\mathcal{O}_p$) gives such a $j$ with $j^2=b \in \mathbb{Z}_p$, so $1,i,j,ij$ is a $\mathbb{Z}_p$-basis for $\mathcal{O}_p$. The reduced discriminant computed in this basis is $4db$; since $\mathcal{O}_p$ is maximal this discriminant is a unit, and we conclude $p \nmid b$. Then the argument above proceeds as before, noting we can solve the norm equation integrally (e.g., lifting a solution over a finite field using Hensel's lemma).

I suppose we can make this work if $p=2$, how important is that?