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text: '[summary: An integral domain is a [3gq ring] in which the only way to make $0$ as a product is to multiply $0$ by something. For instance, in an integral domain like [48l $\\mathbb{Z}$], $2 \\times 3$ is not equal to $0$ because neither $2$ nor $3$ is.]\n\n[summary(Technical): An integral domain is a [3gq ring] in which $ab=0$ implies $a=0$ or $b=0$. (We exclude the ring with one element: that is conventionally not considered an integral domain.)]\n\nIn keeping with [3gq ring theory] as the attempt to isolate each individual property of [48l $\\mathbb{Z}$] and work out how the properties interplay with each other, we define the notion of **integral domain** to capture the fact that if $a \\times b = 0$ then $a=0$ or $b=0$.\nThat is, an integral domain is one which has no "zero divisors": $0$ cannot be nontrivially expressed as a product.\n(For uninteresting reasons, we also exclude the ring with one element, in which $0=1$, from being an integral domain.)\n\n# Examples\n\n- $\\mathbb{Z}$ is an integral domain.\n- Any [481 field] is an integral domain. (The proof is an exercise.)\n\n%%hidden(Show solution):\nSuppose $ab = 0$, but $a \\not = 0$. We wish to show that $b=0$.\n\nSince we are working in a field, $a$ has an inverse $a^{-1}$; multiply both sides by $a^{-1}$ to obtain $a^{-1} a b = 0 \\times a^{-1}$.\nSimplifying, we obtain $b = 0$.\n%%\n\n- When $p$ is a [4mf prime] integer, the ring $\\mathbb{Z}_p$ of integers [modular_arithmetic mod] $p$ is an integral domain.\n- When $n$ is a [composite_number composite] integer, the ring $\\mathbb{Z}_n$ is *not* an integral domain. Indeed, if $n = r \\times s$ with $r, s$ positive integers, then $r s = n = 0$ in $\\mathbb{Z}_n$.\n\n# Properties\n\nThe reason we care about integral domains is because they are precisely the rings in which we may cancel products: if $a \\not = 0$ and $ab = ac$ then $b=c$.\n%%hidden(Proof):\nIndeed, if $ab = ac$ then $ab-ac = 0$ so $a(b-c) = 0$, and hence (in an integral domain) $a=0$ or $b=c$.\n\nMoreover, if we are not in an integral domain, say $r s = 0$ but $r, s \\not = 0$.\nThen $rs = r \\times 0$, but $s \\not = 0$, so we can't cancel the $r$ from both sides.\n%%\n\n## Finite integral domains\n\nIf a ring $R$ is both finite and an integral domain, then it is a [481 field].\nThe proof is an exercise.\n%%hidden(Show solution):\nGiven $r \\in R$, we wish to find a multiplicative inverse.\n\nSince there are only finitely many elements of the ring, consider $S = \\{ ar : a \\in R\\}$.\nThis set is a subset of $R$, because the multiplication of $R$ is [3gy closed].\nMoreover, every element is distinct, because if $ar = br$ then we can cancel the $r$ (because we are in an integral domain), so $a = b$.\n\nSince there are $|R|$-many elements of the subset $S$ (where $| \\cdot |$ refers to the [-4w5]), and since $R$ is finite, $S$ must in fact be $R$ itself.\n\nTherefore in particular $1 \\in S$, so $1 = ar$ for some $a$.\n%%\n',
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