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Suppose $a \in \mathbb{Z}$. Does $a = 0$ iff $a \equiv 0 \ (\operatorname{mod} p)$ for every prime $p$?

It's probably a silly question, but I don't know how to go about it. I'm motivated by the common contest math trick of considering a Diophantine equation $\operatorname{mod} p$. By naturality of $\mathrm{eval}_{x}: (-[t]) \to \mathrm{id}_{\mathbf{CRing}}$ the above question is at least as strong as the following:

Is an integer a root of a diophantine equation iff it is a root of the said equation mod every prime number?

It's also cool that this question is equivalent to the following:

Are functions on $\operatorname{Spec} \mathbb{Z}$ completely defined by their values on points of the underlying topological space?

I don't know how to approach this question, but I'd be grateful for an answer or a hint (if it's not much harder than a typical exercise in an abstract algebra textbook).

Aleksei Averchenko
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    If $a\in \mathbb Z \setminus {0}$, choose a prime $p$ with $0 < |a| < p$. Then $a \not\equiv 0 \pmod p$. – martini Oct 28 '12 at 08:34
  • @martini Ah, this reminds me of how integers are encoded in computers :) Thank you! – Aleksei Averchenko Oct 28 '12 at 08:37
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    I guess you really balcked out here - I wonder how one can know what Spec is without noticing martini's observation. :) – Hagen von Eitzen Oct 28 '12 at 08:57
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    @Hagen I'm weird :D – Aleksei Averchenko Oct 28 '12 at 09:07
  • Fun, related fact: http://en.wikipedia.org/wiki/Carmichael_number – Per Alexandersson Oct 28 '12 at 17:38
  • This reminds me of an amusing result I saw a number of years ago: The function$f(x) = (x^2-2)(x^2-3)(x^2-6)$ has a integral root for every prime $p$ (apply quadratic reciprocity and $2*3=6$) but not over the reals. (Don't know if this should be a comment or answer, so flog me appropriately if I am wrong.) – marty cohen Oct 28 '12 at 19:13
  • By martini's observation, we see that not every sequence of residues comes from an integer. So what is the nature of the ring $\mathbb{F}_2 \times \mathbb{F}_3 \times \mathbb{F}_5 \times \cdots$? Hmmm... – Zhen Lin Oct 28 '12 at 19:50

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Since you mentioned $\text{Spec}$, the appropriate generalization to arbitrary commutative rings is the following: an element of a commutative ring $R$ is completely determined by its value mod all prime ideals $P$ if and only if $R$ has trivial nilradical (hence is a reduced ring), and is completely determined by its value mod all maximal ideals $m$ if and only if $R$ has trivial Jacobson radical (hence is a semiprimitive or Jacobson semisimple ring).

An abstract form of the Nullstellensatz asserts that if $R$ is a Jacobson ring, then any finitely generated $R$-algebra is also a Jacobson ring. In a Jacobson ring, the nilradical and Jacobson radical coincide, so a Jacobson ring is reduced iff it is semiprimitive.

Qiaochu Yuan
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