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  "body": "\n## Proof\n\n1. Let $R$ be a [[commutative ring|0GYGHKX834]] and let $P$ be an [[ideal|0MYSA6MYNA]] of $R$. The [[quotient ring|0VFW4Y69BT]] $R/P$ is commutative, since $(a + P)(b + P) = a b + P = b a + P = (b + P)(a + P)$.\n2. For $a \\in R$, $a + P = 0 + P$ if and only if $a \\in P$. Indeed, if $a + P = P$, then $a = a + 0 \\in P$. If $a \\in P$, then $a + P \\subseteq P$ and $P = a + (P - a) \\subseteq a + P$, since $P$ is an additive subgroup.\n3. Claim: $1 + P \\neq 0 + P$ if and only if $P \\neq R$. By step 2, $1 + P \\neq 0 + P$ means $1 \\notin P$.\n4. If $1 \\in P$, then $r = r \\cdot 1 \\in P$ for every $r \\in R$, so $P = R$. If $P = R$, then $1 \\in P$. So $1 \\notin P$ exactly when $P \\neq R$, which proves the claim.\n5. Every element of $R/P$ is $a + P$ for some $a \\in R$, and $(a + P)(b + P) = a b + P$.\n6. So by step 2, the condition \"for all $a, b \\in R$, $(a + P)(b + P) = 0 + P$ implies $a + P = 0 + P$ or $b + P = 0 + P$\" says exactly \"for all $a, b \\in R$, $a b \\in P$ implies $a \\in P$ or $b \\in P$\".\n7. By definition, $R/P$ is an [[integral domain|09S3PFBEPM]] when it is commutative, $1 + P \\neq 0 + P$, and the condition of step 6 holds in $R/P$.\n8. By definition, $P$ is a [[prime ideal|06HC6HK0Z8]] when $P \\neq R$ and the condition of step 6 holds for $P$.\n9. By steps 1, 3 and 6, the two definitions match term by term. So $R/P$ is an integral domain if and only if $P$ is prime.\n\n## Idea\n\nAn element of $R/P$ is zero exactly when its representative lies in $P$. Under this translation, $1 \\neq 0$ becomes $P \\neq R$, and \"no zero divisors\" becomes \"a product in $P$ has a factor in $P$\". These are the two conditions of a prime ideal.\n",
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