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  "body": "\n## Proof\n\n1. Let $K$ be a [[field|0QA8CHH8MG]] and let $I$ be an [[ideal|0MYSA6MYNA]] of $K$.\n2. Since $I$ is an additive subgroup of $K$, $0 \\in I$. So if $I$ has no nonzero element, then $I = \\set{0}$.\n3. Otherwise, let $a \\in I$ with $a \\neq 0$. Since $K$ is a field, there is $b \\in K$ with $b a = 1$.\n4. Since $I$ is an ideal, $b a \\in I$. So $1 \\in I$.\n5. Let $r \\in K$. Since $I$ is an ideal and $1 \\in I$, $r = r \\cdot 1 \\in I$. So $K \\subseteq I$, and $I = K$.\n6. Hence $I = \\set{0}$ or $I = K$.\n\n## Idea\n\nA nonzero element of an ideal of a field can be multiplied by its inverse. So the ideal contains $1$, and then it contains every multiple of $1$, which is every element.\n",
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  "name": "Proof that the only ideals of a field are zero and the whole field",
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