Congruences (not mod ideals) and CRT in non-UFD (examples).

Asked Dec 25 '16 at 02:25

Active Dec 25 '16 at 03:11

Viewed 46 times

Is there a version of Chinese Remainder Theorem for non-UFD?

Can someone give an example?

Say $n=ab=cd$ holds in the domain where each of $a,b,c,d$ are irreducibles and we know $m\bmod a$, $m\bmod b$ and $m\bmod c$, $\bmod d$.

So the queries are:

suppose we know $m\bmod a$, $m\bmod b$ and $m\bmod c$, $\bmod d$ can we get $m\bmod ab$, $m\bmod ac$ and $m\bmod bc$, $\bmod bd$, $m\bmod ad$ and $m\bmod cd$?

Can we also get $m\bmod abc$, $m\bmod abd$, $m\bmod acd$ and $m\bmod bcd$?

Can we also get $m\bmod abcd$?

edited Dec 25 '16 at 03:11

asked Dec 25 '16 at 02:25

Turbo

6,221

What do you mean by mod in this context? – Ashwin Trisal Dec 25 '16 at 02:36
The wikipedia page implies it generalizes to arbitrary rings. – Mark Schultz-Wu Dec 25 '16 at 02:37
@Mark So my query is valid? (but the general CRT is for ideals). – Turbo Dec 25 '16 at 02:37
1

The normal statement of the CRT isn't in terms of congruances, but in the following way. If $R$ is a commutative ring with identity, and $I = I_1I_2\dots I_n$ is a product of irreducible ideals, then there exists an isomorphism $R/I\cong R/I_1\times R/I_2\times\dots R/I_n$.
The wikipedia page refers to a generalization past commutative rings, but I'm less familiar with that.
– Mark Schultz-Wu Dec 25 '16 at 02:39
@Mark so my query is not valid? – Turbo Dec 25 '16 at 02:42
@AJ. I'm not trying to say it's valid or not valid, I'm just trying to mention how it's normally presented. Someone else may have more to about this. – Mark Schultz-Wu Dec 25 '16 at 02:44
The domains satisfying CRT are known as Prüfer domians. They have many interesting characterizations, e.g. see this answer for $27$ characterizations. – Bill Dubuque Dec 25 '16 at 03:20

Congruences (not mod ideals) and CRT in non-UFD (examples).

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