I am reading A. Knapp's book on elliptic curves right now. In Proposition 5.6 the author wants to prove that the reduction map (modulo $p$, where $p$ does not divide the discriminant) of an elliptic curve over $\Bbb Q$ preserves the group structure.
All you need to show is that the chord-tangent construction is preserved, that is, if $E$ is the curve, $r$ is the reduction map and $P \cdot Q$ denotes the third intersection of the line through $P$ and $Q$ with $E$, then \begin{align*} r(P) \cdot r(Q) = r(P \cdot Q) \end{align*} To prove this, the author uses the following proposition: Given a curve $F$ and a line $L$ in $P_2(\Bbb Q)$ (the two-dimensional prjective space over $\Bbb Q$) and a point $P$, we have $i(P, L, F) \le i(r(P), r(L), r(F))$ where $i$ is the intersection multiplicity.
His proof for the above identity is now as follows: "We apply Proposition 5.5. Since the sum of intersection multiplicities over a line is $\le 3$, the proposition gives $r(P \cdot Q) = r(P) \cdot r(Q)$.
Now here is my problem: What if $P$, $Q$ and $P \cdot Q$ all reduce to the same point modulo $p$? Then the assertion is wrong unless this point is a flex. Clearly, if $L$ is the line through $P$ and $Q$, then its reduction modulo $p$ has intersection multiplicity $1$ with $r(P) = r(Q) = r(P \cdot Q)$. But this does not mean that there is no tangent (over $\Bbb Z_p$) to $r(P)$ so that $r(P) \cdot r(Q) \ne r(P \cdot Q)$, does it?
Ok, I found a full and very detailed solution to this problem on A. Knapp's homepage, see here. Apparently, the above problem is indeed a gap in the proof (and a quite big one, as it seems).