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  "body": "\n## Proof\n\nWe prove both directions directly.\n\n1. Let $R$ be a [[commutative ring|0GYGHKX834]] and let $M$ be an [[ideal|0MYSA6MYNA]] of $R$. The [[quotient ring|0VFW4Y69BT]] $R/M$ is commutative, since $(a + M)(b + M) = a b + M = (b + M)(a + M)$.\n2. For $a, b \\in R$, $a + M = b + M$ if and only if $a - b \\in M$. Indeed, if the cosets are equal, then $a \\in b + M$. If $a - b = x \\in M$, then $a + y = b + (x + y)$ and $b + y = a + (y - x)$ for each $y \\in M$.\n3. We have $1 + M \\neq 0 + M$ if and only if $M \\neq R$. By step 2, $1 + M = 0 + M$ means $1 \\in M$. If $1 \\in M$, then $r = r \\cdot 1 \\in M$ for all $r$, so $M = R$. If $M = R$, then $1 \\in M$.\n4. Suppose first that $M$ is a [[maximal ideal|09J6KGT65G]]. Then $M \\neq R$, so $1 + M \\neq 0 + M$ by step 3.\n5. Let $a + M \\neq 0 + M$. Then $a \\notin M$ by step 2. Put $L = \\set{m + r a : m \\in M, r \\in R}$.\n6. The set $L$ is an ideal of $R$. It contains $0 = 0 + 0 a$. It is closed under differences, since $(m + r a) - (m' + r' a) = (m - m') + (r - r') a$. For $s \\in R$, $s (m + r a) = s m + (s r) a \\in L$, since $s m \\in M$.\n7. Taking $r = 0$ shows $M \\subseteq L$. Taking $m = 0$ and $r = 1$ shows $a \\in L$. Since $a \\notin M$, $L \\neq M$.\n8. By maximality of $M$, $L = R$. So $1 = m + r a$ for some $m \\in M$ and $r \\in R$.\n9. Then $r a - 1 = -m \\in M$. By step 2, $(r + M)(a + M) = r a + M = 1 + M$. So $a + M$ has an inverse.\n10. By steps 1, 4 and 9, $R/M$ is a [[field|0QA8CHH8MG]].\n11. Conversely, suppose that $R/M$ is a field. Then $1 + M \\neq 0 + M$, so $M \\neq R$ by step 3.\n12. Let $L$ be an ideal of $R$ with $M \\subseteq L$ and $L \\neq M$. Choose $a \\in L$ with $a \\notin M$. Then $a + M \\neq 0 + M$ by step 2.\n13. Since $R/M$ is a field, there is $b \\in R$ with $(a + M)(b + M) = 1 + M$, that is, $a b + M = 1 + M$. By step 2, $1 - a b \\in M \\subseteq L$.\n14. Also $a b = b a \\in L$, since $L$ is an ideal and $a \\in L$. So $1 = (1 - a b) + a b \\in L$.\n15. Then $r = r \\cdot 1 \\in L$ for every $r \\in R$, so $L = R$.\n16. By steps 11 to 15, every ideal containing $M$ equals $M$ or $R$, and $M \\neq R$. So $M$ is maximal.\n\n## Idea\n\nTo invert $a + M$, enlarge $M$ by the multiples of $a$. Maximality forces this larger ideal to contain $1$, which writes $1$ as a multiple of $a$ modulo $M$. Conversely, an inverse modulo $M$ puts $1$ into any ideal that strictly contains $M$.\n",
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