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text: '[summary(Technical): An [-5md] $R$ is said to be a *unique factorisation domain* if every nonzero non-[5mg unit] element of $R$ may be written as a product of [5m1 irreducibles], and moreover this product is unique up to reordering and multiplying by units.]\n\n[summary: A [3gq ring] $R$ is a *unique factorisation domain* if an analog of the [5rh] holds in it. The condition is as follows: every nonzero element, which does not have a multiplicative inverse, must be expressible as a product of [5m1 irreducibles], and the expression must be unique if we do not care about ordering or about multiplying by elements which have multiplicative inverses.]\n\n[3gq Ring theory] is the art of extracting properties from the [48l integers] and working out how they interact with each other.\nFrom this point of view, a *unique factorisation domain* is a ring in which the integers' [5rh] holds.\n\nThere have been various incorrect "proofs" of Fermat's Last Theorem throughout the ages; it turns out that if we assume the false "fact" that all subrings of [4zw $\\mathbb{C}$] are unique factorisation domains, then FLT is not particularly difficult to prove.\nThis is an example of where abstraction really is helpful: having a name for the concept of a UFD, and a few examples, makes it clear that it is not a trivial property and that it does need to be checked whenever we try and use it.\n\n# Formal statement\n\nLet $R$ be an [-5md].\nThen $R$ is a *unique factorisation domain* if every nonzero non-[5mg unit] element of $R$ may be expressed as a product of [5m1 irreducibles], and moreover this expression is unique up to reordering and multiplying by units.\n\n# Why ignore units?\n\nWe must set things up so that we don't care about units in the factorisations we discover.\nIndeed, if $u$ is a unit %%note:That is, it has a multiplicative inverse $u^{-1}$.%%, then $p \\times q$ is always equal to $(p \\times u) \\times (q \\times u^{-1})$, and this "looks like" a different factorisation into irreducibles.\n($p \\times u$ is irreducible if $p$ is irreducible and $u$ is a unit.)\nThe best we could possibly hope for is that the factorisation would be unique if we ignored multiplying by invertible elements, because those we may always forget about.\n\n## Example\n\nIn $\\mathbb{Z}$, the units are precisely $1$ and $-1$.\nWe have that $-10 = -1 \\times 5 \\times 2$ or $-5 \\times 2$ or $5 \\times -2$; we need these to be "the same" somehow.\n\nThe way we make them be "the same" is to insist that the $5$ and $-5$ are "the same" and the $2$ and $-2$ are "the same" (because they only differ by multiplication of the unit $-1$), and to note that $-1$ is not irreducible (because irreducibles are specifically defined to be non-unit) so $-1 \\times 5 \\times 2$ is not actually a factorisation into irreducibles.\n\nThat way, $-1 \\times 5 \\times 2$ is not a valid decomposition anyway, and $-5 \\times 2$ is just the same as $5 \\times -2$ because each of the irreducibles is the same up to multiplication by units.\n\n# Examples\n\n- Every [-5r5] is a unique factorisation domain. ([pid_implies_ufd Proof.]) This fact is not trivial! Therefore $\\mathbb{Z}$ is a UFD, though we can also prove this directly; this is the [5rh].\n- $\\mathbb{Z}[-\\sqrt{3}]$ is *not* a UFD. Indeed, $4 = 2 \\times 2$ but also $(1+\\sqrt{-3})(1-\\sqrt{-3})$; these are both decompositions into irreducible elements. (See the page on [5m1 irreducibles] for a proof that $2$ is irreducible; the same proof can be adapted to show that $1 \\pm \\sqrt{-3}$ are both irreducible.)\n\n# Properties\n\n- If it is hard to test for uniqueness up to reordering and multiplying by units, there is an easier but equivalent condition to check: an integral domain is a unique factorisation domain if and only if every element can be written (not necessarily uniquely) as a product of irreducibles, and all irreducibles are [5m2 prime]. ([alternative_condition_for_ufd Proof.])',
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