1

Is there an analogue of Chinese remainder theorem over the rationals? If I knew the residues modulo a few primes of the numerator and denominator (say $r = x/y$) of a rational number, and if I also know the range in which the rational number lies (say $[a,b]$), can I reconstruct the rational number from this information?

mathstudent42
  • 321
  • 1
  • 9
  • 2
    CRT is for all commutative rings with $1$, see here. – Dietrich Burde Mar 20 '18 at 20:27
  • 2
    While the CRT is true for all commutative rings, it's of no utility for fields. I think the posters is actually asking about an analogous task rather than the actual theorem. – rschwieb Mar 20 '18 at 20:39
  • @rschwieb Yes, my application is rather algorithmic. I have a large expression involving rational numbers, which for my purpose has to be evaulated modulo a bunch of small primes and then to test some simple properties of the rational number represented by this expression. And unfortunately, I cannot simplify the whole expression to division of two integers (a rather technical reason) and implement CRT. Is there any other way out like using p-adics etc? Does the fact that I know the range help at all? – mathstudent42 Mar 20 '18 at 20:42
  • @rschwieb Can you elaborate on why you say it is useless? – mathstudent42 Mar 20 '18 at 20:53
  • @mathstudent42 the CRT is a statement about collections of comaximal ideals, but a field does not have any proper ideals which can be comaximal. – rschwieb Mar 20 '18 at 21:35
  • 1
    As stated, the answer to the question would be No. Let $N$ be the product of the supposedly coprime moduli. You cannot distinguish $x$ from $x+aN$ with $a\in\mathbb{Z}$, likewise you cannot distinguish $y$ from $y+bN$ where $b\in\mathbb{Z}$. With large enough $a,b$, we can bring $(x+aN)/(y+bN)$ arbitrarily close to $a/b$, which itself can be chosen arbitrarily. – ccorn Mar 20 '18 at 22:23
  • 1

0 Answers0