Let $E/\mathbf{Q}$ be an elliptic curve of rank $>0$. It is easy to see that there is a positive-density set of primes $p$ such that the reduction map $\mathrm{red}\_p : E(\mathbf{Q}) \rightarrow \widetilde{E}(\mathbf{F}\_p)$ is *not* surjective. Namely, for an integer $n>1$, take any rational prime $p$ that splits completely in the field $K\_n = \mathbf{Q}([n]^{-1}(E(\mathbf{Q})))$, i.e. the field that results from adjoining to $\mathbf{Q}$ the coordinates of all preimages of points in $E(\mathbf{Q})$ under multiplication by $n$. Note that the $K\_n$ are finite field extensions of $\mathbf{Q}$ (this is equivalent to the weak Mordell-Weil theorem). To show that these $p$ work: take $P \in E(\mathbf{Q})$ and $Q \in E(K\_n)$ with $nQ=P$, then $\mathrm{red}\_p(P) = n ( \mathrm{red}\_p(Q))$, with $\mathrm{red}\_p(Q)$ lying in $\widetilde{E}(\mathbf{F}\_p)$ by the assumption on $p$, so $\mathrm{red}\_p(E(\mathbf{Q}))$ lies in $n \widetilde{E}(\mathbf{F}\_p)$, which is an index-$n^2$ subgroup of $\widetilde{E}(\mathbf{F}\_p)$.


More generally, for any isogeny $\phi : E' \rightarrow E$ of elliptic curves over $\mathbf{Q}$ with non-trivial kernel, take any prime $p$ that splits completely in the finite field extension $\mathbf{Q}(\phi^{-1}(E(\mathbf{Q})))$ of $\mathbf{Q}$. 


My questions are in the opposite direction:


1. Do there exist infinitely many $p$ such that $\mathrm{red}\_p$ is surjective?
2. Do there exist arbitrarily large sets of primes $\{ p\_1, p\_2, \ldots, p\_m \}$ such that the combined reduction map $E(\mathbf{Q}) \rightarrow \prod\_{i=1}^m \widetilde{E}(\mathbf{F}\_{p\_i})$ is surjective?
3. For a prime $p$ such that $\mathrm{red}\_p$ is not surjective, is the failure of surjectivity explained by some isogeny $\phi$, by the argument sketched in the first paragraph?


**Edit.** The answer to 1. and 2. is obviously "no" in general by Maarten's answer. By the Gupta-Murty paper mentioned by Felipe, the answer to 1. becomes "yes" once the rank of $E$ is sufficiently large. As for 2., I would like to ask: is there even a single elliptic curve $E$ over $\mathbf{Q}$ for which question 2. has a positive answer? 



Surjectivity of reduction maps of elliptic curves over Q
**Part 1** is not true for rank 1 curves. If $\mathbb{Q}(\phi^{-1}(E(\mathbb{Q})))=\mathbb{Q}$ then every prime trivially splits in $\mathbb{Q}(\phi^{-1}(E(\mathbb{Q})))$. I will now give an explicit example of an isogeny between rank 1 curves $\phi:E' \to E$ such that $\mathbb{Q}(\phi^{-1}(E(\mathbb{Q})))=\mathbb{Q}$.


Let $E'$ be the curve with Cremona label 189b2, it is given by $y^2 + y = x^3 - 54x - 88$. One has that $E'(\mathbb{Q})\cong\mathbb{Z} \times \mathbb{Z}/3\mathbb{Z}$. The free part is generated by $P':=(-6 : 4 : 1)$ and the torsion part is generated by $T':=(12 : -32 : 1)$. 
Now let $\phi:E' \to E$ be the isogeny whose kernel is generated by $T'$. Then $E$ is the elliptic curve with cremona label 189b3. Now $E(\mathbb{Q}) \cong \mathbb Z$ and with an explicit calculation one can show that $E(\mathbb{Q})$ is generated by $P:=\phi(P')$. So $\mathbb{Q}(\phi^{-1}(E(\mathbb{Q})))=\mathbb{Q}$ as requested.


A computer search of an isogeny between rank 1 curves $\phi:E' \to E$ such that $\mathbb{Q}(\phi^{-1}(E(\mathbb{Q})))=\mathbb{Q}$ of all elliptic curves up to conductor 1000 gave 225 counter examples. The example above is the one with smallest conductor. Code for performing this search can be found at <https://sage.mderickx.nl/home/pub/9>


**Update on part 1:**


I extended the search to rank > 1 curves and also found multiple examples of an isogeny between rank 2 curves $\phi:E' \to E$ such that $\mathbb{Q}(\phi^{-1}(E(\mathbb{Q})))=\mathbb{Q}$. An example is where $E'$ is the elliptic curve defined by $y^2 + xy + y = x^3 + x^2 - 71x - 196$ and the kernel of $\phi$ is generated by $(9 : -5 : 1)$. The Cremona label of $E'$ is '3315b2'.
I did not find any examples of rank 3 after searching trough all elliptic curves of conductor < 100000.


ps. Note that my counter examples to part one are not counter examples to Lang-Trotter. The reason is that the Lang-Trotter conjecture is a conjecture about the density of the of the primes such that the reduction is surjective. In my examples both the conjectured density and the actual density are both 0.


**Part 3:**
Let $E$ be a non CM rank 1 elliptic curve that is the only one in it's isogeny class. Then the only isogenies to $E$ are the multiplication by $n$ maps. For concreteness I let $E$ be the curve among all curves with these properties of smallest conductor. This curve $E$ is given by $y^2 + y = x^3 - x$ and $E(\mathbb Q)= \mathbb Z$ is generated by $P=(0 : -1 : 1)$. Now let $p=23$ then $\\#E(\mathbb F\_p)=22$ but the order of $P$ after reduction is $11$ so that the index is $2$. This means that the obstruction cannot come from an isogeny because this would mean that it comes from some multiplication by $n$ map and hence that the index should not be squarefree.


In the article of Lang and Trotter where they state their conjecture they give a criterion in terms of $\mathbb Q(l^{-1}E(\mathbb Q))$ that is equivalent to $l$ being a divisor of the index. If you read that obstruction carefully you will realize that its really easy to cook up counter examples to part $3$, in particular using their criterion one can show that the set of $p$ such that reduction mod $p$ is a counter example has positive density for the above $E$.


