---- > [!theorem] Theorem. ([[Chinese remainder theorem]]) > Let $I_{1},\dots,I_{k}$ be [[ideal|ideals]] of a [[ring]] $R$ such that $I_{i}$ and $I_{j}$ are [[relatively prime integers|coprime]] for all $i \neq j$.[^1] Then the natural [[ring homomorphism|homomorphism]] $\varphi : R \to \frac{R}{I_{1}} \times \dots \times \frac{R}{I_{k}}$is [[surjection|surjective]] with [[kernel of a ring homomorphism|kernel]] $I_{1}\cdots I_{k}$, [[first isomorphism theorem|inducing]] an [[isomorphism]] $\tilde{\varphi}: \frac{R}{I_{1} \cdots I_{k}} \xrightarrow{\sim} \frac{R}{I_{1}} \times \dots \times \frac{R}{I_{k}}.$ > > [!proof]- Proof. ([[Chinese remainder theorem]]) > ~ We will proceed via induction relying on the following lemma. > [!proposition] Lemma. > Let $I_{1},\dots,I_{k}$ be [[ideal|ideals]] of a [[ring]] $R$ such that $I_{i}$ and $I_{k}$ are [[relatively prime integers|coprime]][^2] for all $i=1,\dots,k-1$. Then $(I_{1} \cdots I_{k-1})+I_{k}=(1)$. > [!proof]+ Lemma Proof. > By hypothesis, for $i=1,\dots,k-1$ there exists $a_{i} \in I_{k}$ such that $(1-a_{i}) \in I_{i}$. Then $(1-a_{1}) \cdots (1-a_{k-1}) \in I_{1} \cdots I_{k-1},$ and $1-(1-a_{1}) \cdots (1-a_{k-1}) \in I_{k},$ because it is a combination of $a_{1},\dots,a_{k-1} \in I_{k}$. ^proof The kernel of $\varphi$ is clearly $I_{1} \cap \dots \cap I_{k}$. Since [[product of coprime ideals is their intersection]], this is $I_{1}\cdots I_{k}$. So really we just need to show [[surjection|surjectiveness]]. Argue by induction on $k$. For $k=1$ there is nothing to show. For $k>1$, assume the statement is known for fewer [[ideal|ideals]]. Thus, we may assume that the natural projection induces an [[isomorphism]] $\frac{R}{I_{1} \cdots I_{k-1}} \cong \frac{R}{I_{1}} \times \dots \times \frac{R}{I_{k-1}};$ and all that we have to prove is that the natural homomorphism $R \to \frac{R}{I_{1} \cdots I_{k-1}} \times \frac{R}{I_{k}}$ is [[surjection|surjective]]. By the above lemma, $(I_{1} \cdots I_{k-1})+I_{k}=(1)$, thus we are reduced to the case of *two* [[ideal|ideals]]. Let then $I,J$ be [[ideal|ideals]] of a [[commutative ring]] $R$, such that $I + J=(1)$, and let $r_{I},r_{J} \in R$; we have to verify that there exists $r \in R$ such that $r \equiv r_{I} \text{ mod }I$ and $r \equiv r_{J} \text{ mod }J$. Since $I+J=(1)$, there are $a \in I$, $b \in J$, such that $a+b=1$. Let $r=ar_{J}+br_{I}$: then $r=ar_{J}+(1-a)r_{I}=r_{I}+a(r_{J}-r_{I}) \equiv r_{I} \text{ mod }I$ as $a \in I$, and $r=(1-b)r_{J}+ br_{I}=r_{J}+b(r_{I}-r_{J}) \equiv r_{J} \text{ mod }J$ as $b \in J$, as needed, completing the proof. ---- #### [^1]: That is, $I_{i}+I_{j}=(1)$ for all $i \neq j$. [^2]: That is, $I_{i}+I_{k}=(1)$. ----- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```