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It is very intuitive to think that since modular arithmetic partitions the integers into equivalence classes (or residue classes), the sentence

...the Chinese remainder theorem can be generalized to ideals.

is true.

Here is what I imagine: Say the residue classes mod 3 are

\begin{align}[0]&=\{...,-6,-3,0,3,6,...\}\\ [1]&=\{...,-5,-2,1,4,7,...\}\\ [2]&=\{...,-4,-1,2,5,8,...\} \end{align}

Are the ideals (subsets) of integers generalized through the CRT mod $N= n_1\cdot n_2\cdot n_3= 3\cdot 4\cdot 5=60$ all the $x$ integers for any given $3$ integers $a_1,a_2,a_3$ that fulfill

\begin{align} x\equiv a_1 \mod 3\\ x\equiv a_2 \mod4\\ x\equiv a_3 \mod 5 \end{align}

?

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  • $\begingroup$ An ideal is a subobject in a ring which is an abelian group under addition and which is closed under multiplication by elements of the ring. Ideals in the integers are always "principal", meaning they're always generated by one element, and they all look like $(m) = \lbrace km : k \in \Bbb Z \rbrace$ for some $m \in \Bbb Z$. The statement of the Chinese Remainder Theorem is that the map $\Bbb Z/(mn) \to \Bbb Z/(m) \times \Bbb Z/(n)$ sending $x \bmod mn$ to $(x \bmod m, x \bmod n)$ is an isomorphism for coprime $m, n \in \Bbb Z$. $\endgroup$ – ÍgjøgnumMeg Mar 1 '18 at 0:21
  • $\begingroup$ @ÍgjøgnumMeg Incredibly, I think I follow, but I wonder if there's a way of identifying the cosets of this product group with the $x$'s and $a_i$'s in the formulation of the CRT. A concrete example, if you will... $\endgroup$ – Antoni Parellada Mar 1 '18 at 0:24
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Assuming you know what rings and groups are, let $R$ be a ring and let $I$ be a subset of that ring such that $I$ is an Abelian group under the addition of $R$, and with the property that for any $r \in R$ and $a \in I$, we have $ra \in I$ (that is, $I$ is closed under multiplication by ring elements). Then $I$ is called an $\textbf{ideal}$ in $R$.

In general, suppose there are ideals $I_1, \dots, I_n$ in the ring $R$ such that for any pair of ideals $I_i, I_j$, with $i \neq j$, we have $I_i + I_j = R$. Then we call the ideals pairwise $\textbf{coprime}$ (often also pairwise $\textbf{comaximal}$). The statement of the Chinese Remainder Theorem for arbitrary rings is then that

$$R/I_1\cdots I_n \cong R/I_1 \times \cdots \times R/I_n$$

or, more explicitly, that the map $x \bmod I_1\cdots I_n \mapsto (x \bmod I_1, \dots, x \bmod I_n)$ is an isomorphism.

If we let $R = \Bbb Z$, then every ideal has the property of being $\textbf{principal}$ (that is, every ideal in $\Bbb Z$ is generated by exactly one ring element). Suppose $n_1, \dots, n_r$ are pairwise coprime integers. Then it follows that the ideals they generate are pairwise coprime, and so we can apply the Chinese Remainder Theorem; that is, letting $N = n_1\cdots n_r$, we have

$$\Bbb Z/(N) \cong \Bbb Z/(n_1) \times \cdots \times \Bbb Z/(n_r)$$

with the map given by $x \bmod N \mapsto (x \bmod n_1, \dots, x \bmod n_r)$.

An "explicit" example is as follows; let $n_1 = 7$ and $n_2 = 8$. Then $N = 56$, and so $$\Bbb Z/(56) \cong \Bbb Z/(7) \times \Bbb Z/(8).$$

This is a more abstract reformulation of the Chinese remainder theorem you probably know from elementary number theory.

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  • $\begingroup$ This is crystal clear. Thank you for you detailed elaboration. $\endgroup$ – Antoni Parellada Mar 1 '18 at 1:14
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No. The generalised Chinese remainder theorem is an abstract version in the context of commutative rings, which states this:

Let $R$ be a commutative ring, $I_1,\dots, I_n$ pairwise relatively prime ideals (i.e. $I_k+I_\ell=R\;$ for any $k\ne \ell$). Then

  1. $I_1\cap\dots\cap I_n=I_1\dotsm I_n$.
  2. The canonical homomorphism: \begin{align} R&\longrightarrow R/I_1\times\dotsm \times R/I_n,\\ x&\longmapsto (x+I_1,\dots ,x+I_n), \end{align} induces an isomorphism: $$R/I_1\cap\dots\cap I_n=R/I_1\dotsm I_n\simeq R/I_1\times\dotsm \times R/I_n.$$
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