Science Fair Project Encyclopedia
Talk:Chinese remainder theorem
I wonder how general this theorem is true? It works for principal ideal domains, but is it still true for unique factorization domains? --AxelBoldt
Just noticed that it is not true for UFD's: the map F[X,Y]/(XY) -> F[Y] x F[X] is not surjective (F some field). --AxelBoldt
I'd like to say that the result is true for three or more numbers in the place of a and b (pairwise coprime), but I'm having difficulty phrasing this without making it seem more complicated than it is. --Matthew Woodcraft
I learned it to be true for any Euclidean ring. In that case one is able to perform the Euclidean Algorithm. Is one always able to perform the Euclidean Algorithm on principal ideal domains? -- Georg Muntingh
- You can't always perform the Euclidean Algorithm in principal ideal domains, but you don't need this for the Chinese Remainder Theorem. In a PID, if x and y have gcd 1, then xyR = xR ∩ yR and xR + yR = R. So the Chinese Remainder Theorem for PIDs follows from the version of the theorem for arbitrary rings, which I've now added to the article. (We still need to add the Chinese Remainder Theorem for integers, and this should be dealt with first, without mentioning isomorphisms and factor rings.) --Zundark, 2002 Jan 12
I added the versions for more than two factors. The description of how to solve the congruencies and of the inverse isomorphisms are still missing. --AxelBoldt
- This is a lot better than my writings. --Georg Muntingh
What I want is a more practical (not theoretical) approach to this theorem. For instance, let's say I have some arcade tickets. I don't know how many I have, but when I count them out by 10's I have 8 left over; when I count them out by 14's I have 6 left over, and when I count them out by 18's I have 12 left over. What is the set of possible numbers of tickets I might have?
- Yup, pretty practical that one. Happens to me all the time. AxelBoldt 05:49 Feb 19, 2003 (UTC)
- Have you ever noticed, when counting out a large number of objects, that it helps to count modulo a small number? Besides, I might add, who uses, or even understands, the technical mumbo-jumbo in this article?
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