$\Bbb{Z}[X]$ embeds in $\Bbb{Z}_2 \times \Bbb{Z}_3 \times \Bbb{Z}_5 \times \cdots$? Let $p_n$ be the $n$th prime.  Define $\psi: \Bbb{Z}[X] \to \prod\limits_{n \geq 1} \Bbb{Z}/(p_n): f \mapsto \prod\limits_{n\geq 1} \overline{f(n)}$ is this a ring monomorphism or does there exist a non-zero polynomial $f$ such that $p_n | f(n), n \geq 1$ ?

Let $\psi$ be the hom.  If $\ker \psi \neq 0$, then there exists nonzero $g \in \ker \psi$.  But $g \in \ker \phi_k$ as well where $\phi_k : \Bbb{Z}[X] \to \Bbb{Z}/(2) \times \Bbb{Z}/(3) \times \cdots \times \Bbb{Z}/(p_k)$, for all $k \geq 1$.  Clearly $\ker \phi_{k+1} \subset \ker\phi_k$ so that $\ker \phi_1 \supset \ker \phi_2 \supset \dots$ is a descending chain of ideals.

$\ker \phi_1 = \{ f \in \Bbb{Z}[X] : 2 | f(1) \} = \{ f : \sum_{i=0}^{\deg f} c_i = 0 \pmod 2, \ c_i = $ coefficients of $f\}$.
 A: Your question can be reworded as: if I know how some polynomial in $\mathbb{Z}[x]$ acts on each $\mathbb{Z}/(p\mathbb{Z})$, do I know the coefficients of such polynomial? The answer is yes. Assuming two different polynomials $f(x),g(x)$ act in the exactly same way over each $\mathbb{Z}/(p\mathbb{Z})$, then their difference $h(x)$ is a non-zero polynomial which is $\equiv 0\pmod{p}$ for any prime $p$ and any $x\in\mathbb{Z}$. The Chinese remainder theorem then implies $h(x)=0$ for any $x\in\mathbb{Z}$, hence $h(x)$ has an infinite number of roots and $h(x)\equiv 0$.
In particular $\mathbb{Z}[X]$ embeds in $\mathbb{Z}_2\times\mathbb{Z}_3\times\ldots$ via $1\mapsto(1,1,1,\ldots)$ and $X\mapsto(1,2,3,\ldots)$.
A: A theorem of Jordan of 1872 (see On a theorem of Jordan, by Serre) states that if $f\in\mathbb{Z}[X]$ is irreducible over $\mathbb{Q}$ and of degree 2 or more, then the set of primes $p$ such that $f$ has no zeros in $\mathbb{Z}_p$ has positive density (with $\pi(X)$ as denominator), so in particular this set is infinite.
