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  "body": "\n## Proof\n\nLet $R$ be an [[integral domain|09S3PFBEPM]] and let $p \\in R$ be a [[prime element|0MNHXBSX3M]].\n\n1. By the definition of a prime element, $p \\neq 0$ and $p$ is not a [[unit|019NFGFPR5]]. These are the first two conditions for $p$ to be [[irreducible|052SDEPZH7]].\n2. Let $a, b \\in R$ with $p = a b$. Then $p$ [[divides|04GNK7K4VP]] $a b$, because $a b = p \\cdot 1$.\n3. Since $p$ is prime, $p \\mid a$ or $p \\mid b$. The roles of $a$ and $b$ are symmetric, so we may assume $p \\mid a$.\n4. Write $a = p c$ with $c \\in R$. Then $p \\cdot 1 = p = a b = p (c b)$.\n5. Since $p \\neq 0$, [[cancellation|0GJM24EXP9]] gives $c b = 1$. Since $R$ is commutative, also $b c = 1$, so $b$ is a unit.\n6. Thus in every factorization $p = a b$, one factor is a unit. With step 1, $p$ is irreducible.\n\n## Idea\n\nIf $p = a b$, then $p$ divides the product $a b$, so $p$ divides one factor, say $a = p c$. Then $p = p c b$, and cancelling $p$ shows that $b$ is a unit.\n",
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