Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In Surprising Generalizations, it is mentioned that Chinese remainder theorem and Lagrange interpolation are specific instances of the same thing, my question is what is their common generalisation/abstraction ?

Thank you

PS : Should there be a Generalisation tag ? to be used when one knows a specific concept and is looking for it's generalisation/more-abstract forms?

share|cite|improve this question
Did you read the wikipedia article…? I think it answers your question (at any rate, it's the answer I would give). Namely, both the classical CRT and Lagrange interpolation are special cases of a CRT in more general ring $R$. The most direct generalization is when $R$ is a PID: in particular then one still has an "explicit solution". A more general generalization is to any finite set of comaximal ideals in any commutative ring. – Pete L. Clark Jul 28 '11 at 10:31
You know, you could have clicked on the names in the question you link to... You would have landed here (Harry's name) and here (Qiaochu's name) - the latter because Q. linked to that thread in a comment to Harry's answer, in case you wonder... – t.b. Jul 28 '11 at 11:03
@Theo : I assumed the names link to profiles and not the content that was being referred to, That's why I didn't click them, I did wonder why someone would give refrences to people for helping with an intresting result but not the result itself. Funy enogh I looked at other posts/replies to see if anyone would mention the original post. – Arjang Jul 28 '11 at 12:16
Yeah, it is well-hidden :) It took me a moment to figure that out. – t.b. Jul 28 '11 at 12:18

The connection between the two is explained in an old blog post of mine. The Wikipedia article has more details.

share|cite|improve this answer

i would say, Lagrange interpolation is the extention of the CRT for polynomials. This paper gives some examples and tries to answer the question about the connection. I think it's quite understandable.


share|cite|improve this answer
The cited paper says "Moreover, it is still unknown to us if anybody has contemplated this idea [Lagrange interpolation via CRT] before the late 1960’s and early 1970’s when it occurred to Phil Kutzko, who at the time was a graduate student at the University of Wisconsin". I don't recall the precise history but I'm sure this was known long before 1960. – Bill Dubuque Jul 28 '11 at 15:59
@Bill: None of the following mention it: Birkhoff–MacLane Algebra (1967), van der Waerden Algebra (1966 rev), Zariski–Samuel Commutative Algebra (1959), and Weber Lehrbuch der Algebra (1899). B–M seems the most telling, as it explicitly has Lagrange interpolation with the direct proof, then CRT for PIDs, then CRT for Z as a corollary, and asks the reader to give a direct proof. It mentions CRT with ideals in the exercises, etc. I also checked 4 or 5 old algebra textbooks from the 50s and 60s without any luck. It may have been known, but maybe it was not written down in a textbook. – Jack Schmidt Jul 28 '11 at 17:20
@Bill: According to this article by Schoenberg it seems to have occurred to Marcel Riesz in the 1950's. Two passages: 1. "Sometime in the 1950's the late Hungarian-Swedish mathematician Marcel Riesz visited the University of Pennsylvania and told us informally that the CRP (1) can be thought of as an analogue of the interpolation by polynomials" 2. "My colleague Richard Askey tells me that Riesz' remark is well known to computer scientists, but apparently not to mathematicians." – t.b. Jul 29 '11 at 2:28
@Jack: see my comment to Bill. So the connection was there before. Interestingly, Schoenberg works at Wisconsin. I don't really know what to make of this. – t.b. Jul 29 '11 at 2:33
@Theo Yes, I know the Schoenberg article. I'll post some others later. – Bill Dubuque Jul 29 '11 at 2:40

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.