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  "body": "\n## Proof\n\nWe use [[Zorn's lemma|04RXTQA5ZA]]. We write the proof for left [[ideals|0MYSA6MYNA]]. For [[two-sided ideals|0DDA6CY7GJ]], read \"two-sided ideal\" for \"left ideal\" throughout; step 6 explains the one extra check.\n\n1. Let $R$ be a [[ring|0894J980RK]] and let $I$ be a left ideal of $R$ with $I \\neq R$.\n2. A left ideal $L$ of $R$ equals $R$ if and only if $1 \\in L$. Indeed, if $1 \\in L$, then $r = r \\cdot 1 \\in L$ for every $r \\in R$.\n3. Let $\\mathcal{P}$ be the set of left ideals $L$ of $R$ with $I \\subseteq L$ and $L \\neq R$. Inclusion is a [[partial order|011C29Q1S1]] on $\\mathcal{P}$. The set $\\mathcal{P}$ is nonempty, since $I \\in \\mathcal{P}$.\n4. Let $\\mathcal{C} \\subseteq \\mathcal{P}$ be a [[chain|0GP9ASVQ1X]]. If $\\mathcal{C}$ is empty, then $I$ is an [[upper bound|0H5J02P8S9]] of $\\mathcal{C}$ in $\\mathcal{P}$. So assume that $\\mathcal{C}$ is nonempty, and let $U$ be the [[union|05GK32H72H]] of the members of $\\mathcal{C}$.\n5. The set $U$ is a left ideal. It contains $0$, since some $L \\in \\mathcal{C}$ exists and $0 \\in L$. Let $a, b \\in U$, say $a \\in L_1$ and $b \\in L_2$ with $L_1, L_2 \\in \\mathcal{C}$. Since $\\mathcal{C}$ is a chain, we may assume $L_1 \\subseteq L_2$. Then $a - b \\in L_2 \\subseteq U$. For $r \\in R$, $r a \\in L_1 \\subseteq U$.\n6. In the two-sided case, also $a r \\in L_1 \\subseteq U$ for $a \\in L_1$ and $r \\in R$. So $U$ is then a two-sided ideal.\n7. Every member of $\\mathcal{C}$ contains $I$, so $I \\subseteq U$. No member contains $1$ by step 2, so $1 \\notin U$, and $U \\neq R$.\n8. So $U \\in \\mathcal{P}$, and $U$ contains each member of $\\mathcal{C}$. Thus every chain in $\\mathcal{P}$ has an upper bound in $\\mathcal{P}$.\n9. By Zorn's lemma, $\\mathcal{P}$ has a [[maximal element|0TZEEFTNCC]] $M$. Then $M$ is a left ideal with $I \\subseteq M$ and $M \\neq R$.\n10. Let $L$ be a left ideal of $R$ with $M \\subseteq L$ and $L \\neq R$. Then $I \\subseteq L$, so $L \\in \\mathcal{P}$. By the maximality of $M$ in $\\mathcal{P}$, $L = M$.\n11. So every left ideal $L \\supseteq M$ satisfies $L = M$ or $L = R$. Hence $M$ is a [[maximal ideal|09J6KGT65G]] of $R$ that contains $I$.\n12. Now let $R$ be a nonzero ring, so $1 \\neq 0$. The set $\\set{0}$ is an ideal, and $\\set{0} \\neq R$ since $1 \\notin \\set{0}$. Steps 1 to 11 with $I = \\set{0}$ give a maximal ideal of $R$.\n\n## Idea\n\nA proper ideal is one that avoids $1$. A union of a chain of ideals that avoid $1$ is again an ideal that avoids $1$. So Zorn's lemma applies and gives a proper ideal containing $I$ that cannot be enlarged without reaching the whole ring.\n",
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