Is $\mathbb{Z}$ isomorphic to $\bigoplus_{p\in \mathbb{P}}\mathbb{Z}_p$? Is $\mathbb{Z}$ isomorphic to $\mathbb{Z}_2\oplus\mathbb{Z}_3\oplus\mathbb{Z}_5\oplus\mathbb{Z}_7\oplus \cdots$?
This seems like a natural conceptual extension of the Chinese remainder theorem but I'm not sure how it would work with an infinite sum.
 A: And, after the good points in the comments and other answers, there is something going on here... the projective limit of the quotients $\mathbb Z/n$ is the product of all p-adic integers as p varies over primes. :)
A: $(1,0,0,0,0,....)$ has order $2$ in $\mathbb{Z}_2\oplus\mathbb{Z}_3\oplus\mathbb{Z}_5\oplus\mathbb{Z}_7\oplus...$     
No element in $\mathbb{Z}$ has order $2$.
A: There is a natural map $\mathbb{Z} \to \prod_{p^k} \mathbb{Z}/p^k\mathbb{Z}$, where the product runs over all prime powers, given by taking remainders. The Chinese Remainder Theorem shows that this map is injective. Since the LHS is countable and the RHS is uncountable, it cannot be surjective. 
However, the following natural question presents itself: "suppose I describe a number by describing its residues modulo all prime powers in a consistent way, e.g. if the number is $1 \bmod 4$ then it must also be $1 \bmod 2$. What kind of object do I get if I don't get an integer back?" The answer is that you get a profinite integer. The profinite integers $\hat{\mathbb{Z}}$ are a direct product of the $p$-adic integers $\mathbb{Z}_p$ over all primes $p$, which is in some sense the correct salvage of your conjecture. 
A: Every element in the direct sum has a finite order. To see this, note that an element in the direct sum is a sequence of finitely many non-zero elements, and in the coordinate $n$ the element is less than $p_n$ (the $n$-th prime).
Let $\bar a= \langle a_i\mid i\in\Bbb N\rangle$ be an element, and let $a_{k_1},\ldots,a_{k_n}$ be its non-zero coordinates. Take $m$ to be the $\operatorname{lcm}(k_1,\ldots,k_n)$ then we have that $m\cdot\bar a$ has to be $\bar 0$, the zero sequence, since for every $a_{k_i}$ we have some $t_i$ such that $m\cdot a_{k_i}=t_i\cdot k_i\cdot a_{k_i}\equiv 0\pmod{k_i}$.
So not only the direct sum is not isomorphic, there is no embedding from $\Bbb Z$ into the direct sum or vice versa either, since no element of $\Bbb Z$ has a finite order and no element of the direct sum has an infinite order.
