Chinese Remainder Theorem (Commutative Algebra)/Proof 2

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Theorem

Let $A$ be a commutative and unitary ring.


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Let $I_1, \ldots, I_n$ for some $n \ge 1$ be ideals of $A$.

Then the ring homomorphism $\phi: A \to A / I_1 \times \cdots \times A / I_n$ defined as:

$\map \phi x = \tuple {x + I_1, \ldots, x + I_n}$

has the kernel $\ds I := \bigcap_{i \mathop = 1}^n I_i$, and is surjective if and only if the ideals are pairwise coprime, that is:

$\forall i \ne j: I_i + I_j = A$

Hence in that case, it induces an ring isomorphism:

$A / I \to A / I_1 \times \cdots \times A / I_n$

through the First Isomorphism Theorem.


Proof

Consider $\pi$ only as a homomorphism of groups.

Then Chinese Remainder Theorem (Groups) is applicable as Subgroup of Abelian Group is Normal.

It remains to demonstrate that the condition $I_i + I_j = R$ for all $i \ne j$ assumed here is equivalent to:

$\ds \forall k \le n - 1: I_{k + 1} + \bigcap_{i \mathop = 1}^k I_i = R$

The implication from the latter condition is immediate.

For the converse, let $i$ be arbitrary.

The result follows from:

\(\ds R\) \(=\) \(\ds \prod_{j \mathop \ne i} \paren {I_i + I_j}\) $R$ has a unit and $I_i + I_j = R$ for $j \ne i$
\(\ds \) \(\subseteq\) \(\ds I_i + \prod_{j \mathop \ne i} I_j\) Ring Axiom $\text D$: Distributivity of Product over Addition, Definition of Ideal of Ring
\(\ds \) \(\subseteq\) \(\ds I_i + \bigcap_{j \mathop \ne i}^n I_j\) Intersection of Ideals of Ring contains Product

$\blacksquare$


Sources


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