11
$\begingroup$

This is a followup to Analog to the Chinese Remainder Theorem in groups other than Z_n. . I shouldn't have used the comments to ask a new question, in fact...

Here is the statement of the Chinese Remainder Theorem, as it occurs in most books and websites:

(1) Let $R$ be a commutative ring with unity, and $I_1$, $I_2$, ..., $I_n$ be finitely many ideals of $R$ such that ($I_i+I_j=R$ for any two distinct elements $i$ and $j$ of $\left\lbrace 1,2,...,n\right\rbrace$). Then, $I_1I_2...I_n=I_1\cap I_2\cap ...\cap I_n$, and the canonical ring homomorphism $R/\left(I_1I_2...I_n\right)\to R/I_1 \times R/I_2 \times ... \times R/I_n$ is an isomorphism.

But there seems to be another, even more general form of (1) which doesn't get even half of the attention:

(2) Let $R$ be a commutative ring with unity, and $I_1$, $I_2$, ..., $I_n$ be finitely many ideals of $R$ such that ($I_i+I_j=R$ for any two distinct elements $i$ and $j$ of $\left\lbrace 1,2,...,n\right\rbrace$). Let $A$ be an $R$-module. Then, $I_1I_2...I_n\cdot A=I_1A\cap I_2A\cap ...\cap I_nA$, and the canonical $R$-module homomorphism $A/\left(I_1I_2...I_n\cdot A\right)\to A/I_1A \times A/I_2A \times ... \times A/I_nA$ is an isomorphism.

I am wondering: is (2) a trivial corollary of (1)? Because otherwise I don't see any reason why (2) shouldn't appear in literature as "the" Chinese Remainder Theorem, with (1) being but a corollary. Or is (2) wrong? The only way I see to get (2) from (1) is to apply (1) to the ring $R\oplus A$ (with multiplication on $R\oplus 0$ inherited from $R$, multiplication between $R\oplus 0$ and $0\oplus A$ given by the $R$-module structure on $A$, and multiplication on $0\oplus A$ given by $0$), which seems quite artificial to me. Am I missing something very obvious?

$\endgroup$
  • 8
    $\begingroup$ Tensor the map $R/(I_1...I_n)\rightarrow R/I_1\times...\times R_I_n$ over $R$ with $A$. $\endgroup$ – Rebecca Bellovin Mar 21 '10 at 23:15
  • $\begingroup$ So yes, I was missing something very obvious. Thanks. $\endgroup$ – darij grinberg Mar 21 '10 at 23:30
  • $\begingroup$ I guess you added your comment while I was typing my response. $\endgroup$ – Keenan Kidwell Mar 21 '10 at 23:36
14
$\begingroup$

The second result you're talking about is also sometimes called the Chinese remainder theorem, and can be derived from the Chinese remainder theorem for rings by "tensoring the CRT isomorphism" with $A$. Explicitly, (1) gives

$R/\prod_{k=1}^n I_k\cong\prod_{k=1}^n R/I_k$

via the natural map. This is an isomorphism of rings as well as an isomorphism of $R$-modules. Therefore, upon tensoring with $A$, it becomes

$A/\big(\prod_{k=1}^nI_k\big)A\simeq \prod_{k=1}^n A/I_kA$

via the natural map, using the canonical isomorphism $R/I\otimes_RA\cong A/IA$ as well as the fact that tensor product commutes with finite direct products. It follows that the kernel of $A\rightarrow\prod_{k=1}^n A/I_kA$, which is clearly $\bigcap_{k=1}^n I_kA$, is equal to $\big(\prod_{k=1}^n I_k\big)A$. So, you've derived (2) from (1). Keep in mind that (2) is an isomorphism of $R$-modules, while (1) is an isomorphism of rings (as well as $R$-modules).

$\endgroup$
  • 1
    $\begingroup$ Thanks to you too (particularly for keeping your answer half a page long in order not to embarrass me). $\endgroup$ – darij grinberg Mar 21 '10 at 23:40
  • $\begingroup$ tensoring with $A$... Does CRT also hold for arbitrary (not necessarily commutative or unital) algebras? Wiki seems to suggest that yes, but then why is CRT never mentioned in noncommutative algebra, or formulated in its full generality? $\endgroup$ – Leo Feb 1 '16 at 17:50
  • 1
    $\begingroup$ @Leon, Alfred Foster studied general algebras in which a form of the CRT held. I do not recall the details but I have the impression that idempotent elements were important in finding such a representation. I don't remember how general the construction was. Gerhard "Still Has A Few Preprints" Paseman, 2017.07.03. $\endgroup$ – Gerhard Paseman Jul 4 '17 at 5:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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