Endomorphism ring of $y^2=x^3-x$ over $\Bbb F_p$ Consider the elliptic curve $E$ defined by $y^2 = x^3-x$ over $\Bbb Q$. Let $p \equiv 3 \pmod 4$ be a prime, and $E_p$ be the reduction of $E$ modulo $p$.
By Silverman "Advanced topics...", prop. II.4.4, we have an injective ring morphism $\mathrm{End}(E) \to \mathrm{End}(E_p)$.
It can be shown (e.g. by counting points) that $\mathrm{Frob}_p = \sqrt{-p} \in \mathrm{End}(E_p)$, so that $\mathrm{End}(E_p) \subset \Bbb Q(\sqrt{-p})$. Moreover, $\mathrm{End}(E) \cong \Bbb Z[i]$.
Thus we should get an embedding $\Bbb Z[i] \subset \Bbb Q(\sqrt{-p})$, which doesn't exist ! So what is wrong about my reasoning ? 
 A: $\newcommand{\into}{\hookrightarrow}$
$\newcommand{\End}{\mathrm{End}}$
$\newcommand{\Frob}{\mathrm{Frob}}$
$\newcommand{\Q}{\mathbb{Q}}$
$\newcommand{\F}{\mathbb{F}}$
$\newcommand{\Z}{\mathbb{Z}}$
I think that you should keep track of the fields of definition.
Namely, according to Silverman's proposition 4.4, we have two embeddings 
$$j_1 : \End_{\Q}(E) \into \End_{\F_p}(E_p) $$
and
$$j_2 : \End_{\Q(i)}(E) \into \End_{\F_{p^2}}(E_p)$$
As for $j_1$, according to this question and that one, we have
$$j_1 : \End_{\Q}(E) \cong \Bbb Z \into \End_{\F_p}(E_p) \into \Q(\sqrt{-p}).$$
However, since $p \equiv 3 \pmod 4$, $E_p$ is supersingular and thus has quaternionic endomorphism ring, so it makes perfect sense to have the injective ring morphism
$$j_2 : \End_{\Q(i)}(E) \cong \Bbb Z[i] \into \End_{\F_{p^2}}(E_p) \into Q_{p, \infty},$$
where $Q_{p, \infty}$ is the unique rational (definite) quaternion algebra ramified exactly only at $\infty$ and $p$. Indeed, when $p \equiv 3 \pmod 4$, $Q_{p, \infty}$ is generated by $i,j$ satisfying $i^2=-1, j^2=-p,ij=-ji$ (see e.g. 16.5, bottom of p. 486 in Martinet's book "Perfect Lattices in Euclidean Spaces").

Let me be more explicit about the $\F_{p^2}$-endomorphism ring of $E_p$. 
First, one can show that the $\F_{p}$-endomorphism ring of $E$ is the maximal order in $\Q(\sqrt{-p})$ (which is bigger than $\Z[\Frob_p] \cong \Z[\sqrt{-p}]$, since $p \equiv 3 \pmod 4$).
Notice that the Frobenius automorphism $\Frob_{p^2} = \Frob_{p}^2 \in \End_{\F_{p^2}}(E_p)$ equals $[-p]$ if $p \equiv 3 \pmod 4$, so that's why we don't have a commutative endomorphism ring anymore.
There is an element $a \in \F_{p^2}$ such that $a^2 = -1$. Define the endomorphism of $E$ by
$\phi : (x,y) \mapsto (-x, ay)$, it is defined over $\F_{p^2}$ but not over $\F_p$ (notice that $a \not \in \F_p$ since $p \equiv 3 \pmod 4).
We have
$$\phi^2 = [-1] : (x,y) \mapsto (x,-y),  
\qquad
\phi \circ \Frob_p : (x,y) \mapsto (-x^p, ay^p),
\qquad
\Frob_p \circ \phi : (x,y) \mapsto ((-x)^p, (ay)^p) = (-x^p, -ay^p)$$
where $a^p = a^{4k+3} = a^3 = -a$ using $p = 4k+3 \equiv 3 \pmod 4$ once more. Thus, we get elements
$\phi, \Frob_p \in \End_{\F_{p^2}}(E_p)$ such that
$\phi^2 = -1,\Frob_p^2 = -p, \phi \Frob_p = - \Frob_p \phi$, which are exactly the conditions of a quaternion algebra, and we do have $\Z[\phi] \cong \Z[i]$ inside! (I guess that $\End_{\F_{p^2}}(E_p) \cong \Z[ i, (1+j\sqrt{p})/2 ]$, but I would have to think more to be sure of a proof - e.g. this is a maximal order in $Q_{p, \infty}$).
