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  "body": "\n## Proof\n\nLet $R$ be a [[unique factorisation domain|06YMXJMXYS]] and let $p \\in R$. In particular $R$ is an [[integral domain|09S3PFBEPM]].\n\n1. If $p$ is [[prime|0MNHXBSX3M]], then $p$ is [[irreducible|052SDEPZH7]], because [[prime elements are irreducible in an integral domain|0F0K8DB9TD]].\n2. For the converse, let $p$ be irreducible. Then $p \\neq 0$ and $p$ is not a [[unit|019NFGFPR5]]. It remains to show the divisibility condition.\n3. Let $a, b \\in R$ with $p$ [[dividing|04GNK7K4VP]] $a b$, say $a b = p c$ with $c \\in R$.\n4. If $a = 0$, then $a = p \\cdot 0$ and $p \\mid a$. In the same way, $b = 0$ gives $p \\mid b$. So we may assume $a \\neq 0$ and $b \\neq 0$.\n5. If $a$ is a unit, then $b = a^{-1} a b = p (a^{-1} c)$, so $p \\mid b$. In the same way, if $b$ is a unit, then $p \\mid a$. So we may assume that $a$ and $b$ are not units.\n6. By unique factorisation, $a = a_1 \\cdots a_k$ and $b = b_1 \\cdots b_l$ with all $a_i, b_j$ irreducible and $k, l \\ge 1$.\n7. Since $R$ is an integral domain, $a b \\neq 0$. So $p c \\neq 0$ and $c \\neq 0$.\n8. Suppose that $c$ is a unit. The element $p c$ is not zero, and it is not a unit, since otherwise $p = (p c) c^{-1}$ would be a unit.\n9. If $p c = x y$ with $x, y \\in R$, then $p = x (y c^{-1})$. So $x$ is a unit, or $y c^{-1}$ is a unit and then $y$ is a unit. Hence $p c$ is irreducible.\n10. Then $p c = a_1 \\cdots a_k b_1 \\cdots b_l$ are two factorisations into irreducibles, of lengths $1$ and $k + l \\ge 2$. This contradicts uniqueness, so $c$ is not a unit.\n11. By unique factorisation, $c = c_1 \\cdots c_r$ with all $c_s$ irreducible and $r \\ge 1$. Then $p c_1 \\cdots c_r = a_1 \\cdots a_k b_1 \\cdots b_l$.\n12. Both sides are products of irreducibles. By uniqueness, the factor $p$ on the left is an [[associate|053E86NWVK]] of some factor on the right.\n13. Suppose that $p$ is an associate of $a_i$, so $a_i = u p$ with $u$ a unit. Then $a = p \\cdot u \\prod_{s \\neq i} a_s$, so $p \\mid a$.\n14. In the same way, if $p$ is an associate of some $b_j$, then $p \\mid b$. So $p \\mid a$ or $p \\mid b$, and $p$ is prime.\n\n## Idea\n\nIf $p$ divides $a b$, write $a b = p c$ and factor $a$, $b$ and $c$ into irreducibles. Uniqueness of factorisation says that $p$ must match, up to a unit, one of the irreducible factors of $a$ or of $b$. That factor is then a multiple of $p$, so $p$ divides $a$ or $b$.\n",
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