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  "body": "\n## Proof\n\n1. Let $R$ be a [[commutative ring|0GYGHKX834]], and let $I_1, \\dots, I_n$ be [[ideals|0MYSA6MYNA]] of $R$ that are pairwise [[coprime|0MDJP82TEH]]. Put $K = I_1 \\cap \\dots \\cap I_n$ and let $P = R/I_1 \\times \\dots \\times R/I_n$ be the [[product of the rings|0MNZCN4BQK]] $R/I_k$.\n2. For an ideal $L$ and $a, b \\in R$, $a + L = b + L$ if and only if $a - b \\in L$. Indeed, if the cosets are equal, then $a \\in b + L$. If $a - b = x \\in L$, then $a + y = b + (x + y)$ and $b + y = a + (y - x)$ for each $y \\in L$.\n3. Define $\\varphi \\colon R \\to P$ by $\\varphi(a) = (a + I_1, \\dots, a + I_n)$. The operations of $P$ and of each [[quotient ring|0VFW4Y69BT]] $R/I_k$ are componentwise and come from $R$. So $\\varphi$ preserves sums and products, and $\\varphi(1) = (1 + I_1, \\dots, 1 + I_n)$ is the identity of $P$. Thus $\\varphi$ is a [[ring homomorphism|0GGE0Z0JQ1]].\n4. By step 2, $\\varphi(a) = 0$ if and only if $a \\in I_k$ for every $k$. So the [[kernel|0AAHE7GEW6]] of $\\varphi$ is $K$.\n5. Fix $k$. We build $e_k \\in R$ with $e_k - 1 \\in I_k$ and $e_k \\in I_j$ for all $j \\neq k$. If $n = 1$, take $e_1 = 1$.\n6. Let $n \\ge 2$. For each $j \\neq k$, the ideals $I_k$ and $I_j$ are coprime. So there are $u_j \\in I_k$ and $v_j \\in I_j$ with $u_j + v_j = 1$. Put $e_k = \\prod_{j \\neq k} v_j$.\n7. For $j \\neq k$, the product $e_k$ has the factor $v_j \\in I_j$. Since $I_j$ is an ideal and $R$ is commutative, $e_k \\in I_j$.\n8. For each $j \\neq k$, $v_j - 1 = -u_j \\in I_k$, so $v_j + I_k = 1 + I_k$ by step 2. Multiplying in $R/I_k$ gives $e_k + I_k = \\prod_{j \\neq k} (v_j + I_k) = 1 + I_k$.\n9. Now let $(r_1 + I_1, \\dots, r_n + I_n)$ be any element of $P$, with $r_1, \\dots, r_n \\in R$. Put $a = r_1 e_1 + \\dots + r_n e_n$.\n10. Fix $k$. In $R/I_k$, $a + I_k = \\sum_{j} (r_j + I_k)(e_j + I_k)$. By step 7, $e_j + I_k = 0 + I_k$ for $j \\neq k$. By step 8, $e_k + I_k = 1 + I_k$. So $a + I_k = r_k + I_k$.\n11. Thus $\\varphi(a) = (r_1 + I_1, \\dots, r_n + I_n)$, and $\\varphi$ is surjective.\n12. By the [[first isomorphism theorem|0ZMYX7GMH0]] applied to $\\varphi$, with steps 4 and 11, the map $R/K \\to P$, $a + K \\mapsto \\varphi(a) = (a + I_1, \\dots, a + I_n)$, is a [[ring isomorphism|0B6QC0C0G1]].\n\n## Idea\n\nThe map $a \\mapsto (a + I_1, \\dots, a + I_n)$ has kernel $I_1 \\cap \\dots \\cap I_n$, so only surjectivity needs work. Coprimality gives, for each $k$, an element that is $1$ modulo $I_k$ and $0$ modulo every other $I_j$. Combinations of these elements hit every target, just as in the integer version.\n",
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