How do we know that reducing $E/K$ commutes with the addition law for $K$ local field I have an elliptic curve $E/K$ where $K$ is a local field. 
We know we can reduce this modulo the unique maximal ideal of the ring of integers of $K$. Suppose $E$ has good reduction so that $\tilde{E}/k$ is also an elliptic curve. 
Take $P,Q \in E(K)$. How do we know that $\widetilde{(P+Q)} = \tilde{P} + \tilde{Q}$? I wasn't sure how to see this and I would greatly appreciate any explanations. Here $\tilde{}$ means reduction. Thank you. 
 A: Addition on $E/K$ is defined in terms of solutions of polynomials in the base field $K$ (such as the intersection of the elliptic curve and the two points $P$ and $Q$), and reducing these polynomials gives addition on $\tilde E/k$. Reduction being a field quotient, reducing solutions to these equations gives solutions to the reduced equations.
Explicitly, suppose we have points $P=(X_P,Y_P)$ and $Q=(X_Q,Y_Q)$ on an elliptic curve defined by some equation $F(x,y) = 0$ (e.g. a Weierstrass equation). The straight line through both points is given by the equation $L(x,y) = 0$ where $$L(x,y) = (X_P - X_Q) y - (Y_P - Y_Q) x + (Y_PX_Q+X_PY_Q)$$ and therefore the points where the line and the curve intersect are solutions to the equation $F(x,y) = L(x,y) = 0$. These points of intersection are (by definition of addition on $E$) the three points $P$, $Q$ and $-(P+Q)$, and hence solutions to these equations define $P+Q$.
Now if you reduce everything everything in the above paragraph to the residue field, it remains true. In particular, if $R=-(P+Q)$ is the third solution to the system $F(x,y)=L(x,y)=0$, then $\tilde R$ is a solution to the reduced system $\tilde F(x,y) = \tilde L(x,y) = 0$ and therefore $\tilde R = -(\tilde P + \tilde Q)$.
