$\Bbb{Z}[X]$ 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 $f\}$.

• @JackD'Aurizio : how do I get $(1,0,0,...)$ as an integer multiple of $(1,1,...)$ in $\prod_n \Bbb Z/p_n \Bbb Z$? This product is uncountable. – Watson Feb 4 '18 at 17:36
• @JackD'Aurizio But the set of ring homomorphisms from $\mathbb{Z}[X]$ to any commutative ring is isomorphic to that ring. So there are plenty of choice for where to map it. – Tobias Kildetoft Feb 4 '18 at 17:52
• @JackD'Aurizio You can send $X$ to $(1,2,3,\ldots)$, i.e., to $(\overline X)_n$. – Jose Brox Feb 4 '18 at 18:07
• Uuuuu wow, 1 vote to close and 3 favorites. This is getting interesting! :D – Shine On You Crazy Diamond Feb 4 '18 at 18:12
• I realized I was on the wrong side of it. For any prime $p$, let $g_p$ a generator of $\mathbb{Z}/(p\mathbb{Z})^*$. We may simple send $X$ into $(g_2,g_3,g_5,\ldots)$ and $1$ into $(1,1,1,\ldots)$. – Jack D'Aurizio Feb 4 '18 at 18:16

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 $X\mapsto(1,2,3,\ldots)$.

• Does the chinese remainder theorem apply even for an infinite cartesian product? – Shine On You Crazy Diamond Feb 4 '18 at 23:28
• @EnjoysMath You could probably restrict to a number of ideals big enough, since the polynomial has a finite degree, but what I do not see is that $h(n)$ is 0 modulo $p$ for every $p$ and every $n\in\mathbb{Z}$. What I think we have is only that $h(n)$ is 0 modulo $p_n$ for every $n$ (so we have only restated the problem for the kernel of a linear map, from $f=g$ to $f=0$). – Jose Brox Feb 5 '18 at 10:15
• @EnjoysMath Btw, what can be said for an infinite number of ideals $I_i$ is that if they are pairwise coprime then there is a monomorphism from $R/\bigcap I_i$ to $R/\prod I_i$ via $f(x+\bigcap I_i):=(x+I_i)_i$, but it does not need to be surjective anymore – Jose Brox Feb 5 '18 at 10:19

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.

• Do you mean $\Bbb{Z}[X]$? – Shine On You Crazy Diamond Feb 4 '18 at 18:20
• @EnjoysMath Yeah, of course :D – Jose Brox Feb 4 '18 at 18:21
• I'm not quite seeing if you've said that it doesn't embed that way, or it does... please explaine more. Thanks – Shine On You Crazy Diamond Feb 4 '18 at 18:22
• Sorry, thought that it would arise from this result in several manners, but none of them actually seem to work in a obvious way. I'm still thinking about it... – Jose Brox Feb 4 '18 at 19:29