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  "body": "\n## Counterexample\n\nThe integral domain $\\mathbb{Z}[\\sqrt{-5}] = \\{a + b\\sqrt{-5} : a, b \\in \\mathbb{Z}\\}$ is not a [[unique factorisation domain|06YMXJMXYS]].\nIn this ring,\n$$6 = 2 \\cdot 3 = (1 + \\sqrt{-5})(1 - \\sqrt{-5}).$$\nLet $N(a + b\\sqrt{-5}) = a^2 + 5b^2$. Then $N$ is multiplicative, and $N(2) = 4$, $N(3) = 9$, $N(1 \\pm \\sqrt{-5}) = 6$.\nNo element has norm $2$ or $3$, because $a^2 + 5b^2 \\notin \\{2, 3\\}$ for integers $a, b$. So the four factors are [[irreducible|052SDEPZH7]].\nThe only units are $\\pm 1$, so $2$ is not an [[associate|053E86NWVK]] of $1 + \\sqrt{-5}$ or of $1 - \\sqrt{-5}$. The two factorisations are different.\n\n## Remarks\n\nThe element $2$ is [[irreducible but not prime|0CHE3J7QMX]] in this ring.\nEvery nonzero nonunit of $\\mathbb{Z}[\\sqrt{-5}]$ is still a product of irreducibles. Only the uniqueness fails.\nUnique factorisation holds again for ideals: $\\mathbb{Z}[\\sqrt{-5}]$ is a Dedekind domain.\n\n## Explanations\n\n### L1 human\n\nThe number $6$ has two essentially different factorisations into irreducible elements in this ring.\nThe norm shows that none of the four factors can be split further.\nSo unique factorisation, which we take for granted in the integers, can fail.\n",
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  "name": "Z[√-5] is not a UFD",
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