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  "body": "\n## Counterexample\n\nIn the integral domain $\\mathbb{Z}[\\sqrt{-5}] = \\{a + b\\sqrt{-5} : a, b \\in \\mathbb{Z}\\}$, the element $2$ is [[irreducible|052SDEPZH7]] but not [[prime|0MNHXBSX3M]].\nLet $N(a + b\\sqrt{-5}) = a^2 + 5b^2$. This norm is multiplicative, and the units are the elements of norm $1$, that is $\\pm 1$.\nIf $2 = x y$ with nonunits $x, y$, then $N(x) N(y) = 4$, so $N(x) = 2$. But $a^2 + 5b^2 = 2$ has no integer solution. So $2$ is irreducible.\nThe element $2$ divides $(1 + \\sqrt{-5})(1 - \\sqrt{-5}) = 6$.\nIt does not divide $1 \\pm \\sqrt{-5}$, because $(1 \\pm \\sqrt{-5})/2$ is not in $\\mathbb{Z}[\\sqrt{-5}]$. So $2$ is not prime.\n\n## Remarks\n\nThe same ring shows that factorization is not unique: $6 = 2 \\cdot 3 = (1 + \\sqrt{-5})(1 - \\sqrt{-5})$. See [[$\\mathbb{Z}[\\sqrt{-5}]$ is not a UFD|0BSJYCG7NM]].\nIn a [[unique factorization domain|06YMXJMXYS]], every irreducible element is prime.\n\n## Explanations\n\n### L1 human\n\nIn the integers, \"cannot be factored\" and \"divides a factor of each product it divides\" are the same property.\nIn $\\mathbb{Z}[\\sqrt{-5}]$, the number $2$ cannot be factored, but it divides $6 = (1 + \\sqrt{-5})(1 - \\sqrt{-5})$ and neither factor.\nSo the two properties are different in general rings.\n",
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