Proof of the weak Mordell-Weil theorem and showing that torsion part of $A(k)$ is finite Reading a proof of the weak Mordell-Weil theorem, I'm stuck somewhere. We have the following theorem :

Let $A$ be an abelian variety defined over a number field $k$ and $v$ a finite place of $k$ at which $A$ has good reduction. Let $\tilde{k}$ be the residue field of $v$ and let $p$ be the characteristic of $\tilde{k}$. Then, the map :
$$ A_m(k) \rightarrow \tilde{A}(\tilde{k})$$
is injective for any $m \geq 1$, $p \nmid m$, where $A_m$ design the kernel of the elements of $m$-torsion on $A$.

From there, it is written that we can directly show that $A_{\text{tors}}$, i.e the torsion part of $A$, is finite. Actually, we choose two places $v$ and $w$ at which $A$ has good reductions, and $v$ and $w$ of different characteristics $p$ and $q$. It is then written that we obtain an injection :
$$ i : A(k)_{\text{tors}} \hookrightarrow \tilde{A}_v(\tilde{k_v}) \times  \tilde{A}_w(\tilde{k_w})$$
But I can't justify that this map is an injection. By composing with the projection over $\tilde{A}_v(\tilde{k_v})$ or $\tilde{A}_w(\tilde{k_w})$, if we take some $x$ in the kernel of $i$, $x$ of $m$-torsion, with $p \nmid m$ or $q \nmid m$, we can then deduce that $x= 0$ using the theorem below. But if $x$ is of $pq$-torsion, this argument doesn't work anymore, right ?
I really don't succeed to justify the fact that $i$ is injective...
Thank you for the help !
 A: It follows from the first fact that anything in the kernel of the reduction
$$A_{tors}(k) \rightarrow \tilde{A}(\tilde{k})$$
must have order a power of $p$. Actually, that fact is how one usually proves the first statement (the kernel is associated to a formal group, and the torsion of formal groups can only have elements of prime power order, for the prime in question) so your trouble here might just be from picking a slightly less than optimal statement to apply.
Take some $x$ on the LHS of order $p^r m$ with $p\nmid m$. That means we can decompose it into a sum $y+z$ where $y$ has order $p^r$ and $z$ has order $m$.
Reducing, $\tilde y$ has order dividing $p^r$ and possibly strictly smaller than $p^r$, and $\tilde z$ has order exactly $m$, because the above map is injective on prime-to-$p$ torsion, and hence preserves orders of elements prime to $p$. In an abelian group, when you add two elements of coprime order, the order of the sum is the product of their orders, and so the order of $\tilde y + \tilde z$ is $m$ times whatever power of $p$ is associated to $\tilde y$. In particular it is at least $m$ and so it can only have order 1 (i.e. be the identity) when $m=1$, which is to say that $x$ has order a power of $p$.
Thus when you put together two such maps, the kernel is the intersection of the two reduction maps. By the above, that means everything in the kernel has order dividing $p$ and order dividing $q$, but the primes $p,q$ are distinct, so that requires everything in the kernel has order $1$, i.e. it is just the identity.
In the case of your particular example, when you reduce such an $x$ mod two such primes, in the first component, only the ``$p$'' part of its order can be killed, and so it has order $p$ or $pq$ in the first component. Likewise in the second component it has order $q$ or $pq$. In any of those cases the whole thing must have order at least $pq$, hence exactly $pq$.
