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.

Solution 1:

$(1,0,0,0,0,....)$ 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 $2$.

Solution 2:

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.