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text: '[summary: An irreducible element of a [3gq ring] is one which cannot be written as a nontrivial product of two other elements of the ring.]\n\n[summary(Technical): Let $(R, +, \\times)$ be a [3gq ring] which is an [5md integral domain]. We say $x \\in R$ is *irreducible* if, whenever we write $r = a \\times b$, it is the case that (at least) one of $a$ or $b$ is a [5mg unit] (that is, has a multiplicative inverse).]\n\n[3gq Ring theory] can be viewed as the art of taking the integers [48l $\\mathbb{Z}$] and extracting or identifying its essential properties, seeing where they lead.\nIn that light, we might ask what the abstracted notion of "[4mf prime]" should be.\nConfusingly, we call this property *irreducibility* rather than "primality"; "[5m2 prime]" in ring theory corresponds to something closely related but not the same.\n\nIn a ring $R$ which is an [-5md], we say that an element $x \\in R$ is *irreducible* if, whenever we write $r = a \\times b$, it is the case that (at least) one of $a$ or $b$ is a [5mg unit] (that is, has a multiplicative inverse).\n\n# Why do we require $R$ to be an integral domain?\n[todo: this]\n\n# Examples\n[todo: lots]\n\n# Relationship with primality in the ring-theoretic sense\n\nIt is always the case that [5m2 primes] are irreducible in any integral domain.\n%%hidden(Show proof):\nIf $p$ is prime, then $p \\mid ab$ implies $p \\mid a$ or $p \\mid b$ by definition.\nWe wish to show that if $p=ab$ then one of $a$ or $b$ is invertible.\n\nSuppose $p = ab$.\nThen in particular $p \\mid ab$ so $p \\mid a$ or $p \\mid b$.\nAssume without loss of generality that $p \\mid a$; so there is some $c$ such that $a = cp$.\n\nTherefore $p = ab = cpb$; we are working in a commutative ring, so $p(1-bc) = 0$.\nSince the ring is an integral domain and $p$ is prime (so is nonzero), we must have $1-bc = 0$ and hence $bc = 1$.\nThat is, $b$ is invertible.\n%%\n\nHowever, the converse does not hold (though it may in certain rings).\n\n- In $\\mathbb{Z}$, it is a fact that irreducibles are prime. Indeed, it is a consequence of [5mp Bézout's theorem] that if $p$ is "prime" in the usual $\\mathbb{Z}$-sense (that is, irreducible in the rings sense) %%note:I'm sorry about the notation. It's just what we're stuck with. It is very confusing.%%, then $p$ is "prime" in the rings sense. ([5mh Proof.])\n- In the ring $\\mathbb{Z}[\\sqrt{-3}]$ of [4zw complex numbers] of the form $a+b \\sqrt{-3}$ where $a, b$ are integers, the number $2$ is irreducible but we may express $4 = 2 \\times 2 = (1+\\sqrt{-3})(1-\\sqrt{-3})$. That is, we have $2 \\mid (1+\\sqrt{-3})(1-\\sqrt{-3})$ but $2$ doesn't divide either of those factors. Hence $2$ is not prime.\n\n%%hidden(Proof that $2$ is irreducible in $\\mathbb{Z}[\\sqrt{-3}]):\nA slick way to do this goes via the [norm_complex_number norm] $N(2)$ of the complex number $2$; namely $4$.\n\nIf $2 = ab$ then $N(2) = N(a)N(b)$ because [norm_of_complex_number_is_multiplicative the norm is a multiplicative function], and so $N(a) N(b) = 4$.\nBut $N(x + y \\sqrt{-3}) = x^2 + 3 y^2$ is an integer for any element of the ring, and so we have just two distinct options: $N(a) = 1, N(b) = 4$ or $N(a) = 2 = N(b)$.\n(The other cases follow by interchanging $a$ and $b$.)\n\nThe first case: $N(x+y \\sqrt{3}) = 1$ is only possible if $x= \\pm 1, y = \\pm 0$.\nHence the first case arises only from $a=\\pm1, b=\\pm2$; this has not led to any new factorisation of $2$.\n\nThe second case: $N(x+y \\sqrt{3}) = 2$ is never possible at all, since if $y \\not = 0$ then the norm is too big, while if $y = 0$ then we are reduced to finding $x \\in \\mathbb{Z}$ such that $x^2 = 2$.\n\nHence if we write $2$ as a product, then one of the factors must be invertible (indeed, must be $\\pm 1$).\n%%\n\nIn fact, in a [-principal_ideal_domain], "prime" and "irreducible" are equivalent. ([5mf Proof.])\n\n# Relationship with unique factorisation domains\n\nIt is a fact that an integral domain $R$ is a [unique_factorisation_domain UFD] if and only if it has "all irreducibles are [5m2 prime] (in the sense of ring theory)" and "every $r \\in R$ may be written as a product of irreducibles". ([alternative_condition_for_ufd Proof.])\nThis is a slightly easier condition to check than our original definition of a UFD, which instead of "all irreducibles are prime" had "products are unique up to reordering and multiplying by invertible elements".\n\nTherefore the relationship between irreducibles and primes is at the heart of the nature of a unique factorisation domain.\nSince $\\mathbb{Z}$ is a UFD, in some sense the [fundamental_theorem_of_arithmetic Fundamental Theorem of Arithmetic] holds precisely because of the fact that "prime" is the same as "irreducible" in $\\mathbb{Z}$.',
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