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  "body": "\n## Proof\n\nLet $L$ and $L'$ be [[algebraic closures|0MCXX29HWY]] of the [[field|0QA8CHH8MG]] $K$. First we extend embeddings by one element.\n\n1. Let $E$ be a [[subfield|07QY5RPFBR]] of $L$ containing $K$, and $\\varphi \\colon E \\to L'$ a [[ring homomorphism|0GGE0Z0JQ1]] with $\\varphi(c) = c$ for $c \\in K$. Let $\\alpha \\in L$.\n2. Since $L/K$ is [[algebraic|09EYMKMSB9]], some nonzero $g \\in K[x]$ has $g(\\alpha) = 0$. As $K \\subseteq E$, $\\alpha$ is [[algebraic|0NPW1DWECM]] over $E$.\n3. Let $m$ be the [[minimal polynomial|0GK1WB9R98]] of $\\alpha$ over $E$. It is [[irreducible|0NFEG48RDR]] in $E[x]$, so $\\deg m \\ge 1$.\n4. Let $\\varphi m \\in L'[x]$ be the polynomial obtained by applying $\\varphi$ to the coefficients of $m$. It is monic of degree $\\deg m \\ge 1$.\n5. Since $L'$ is [[algebraically closed|02SJG1HC32]], $\\varphi m$ has a [[root|00EHX6FWC0]] $\\beta \\in L'$.\n6. By the [[universal property of the polynomial ring|0DQMDSVSXJ]], $\\psi \\colon E[x] \\to L'$, $h \\mapsto (\\varphi h)(\\beta)$, is a ring homomorphism. Its [[kernel|0AAHE7GEW6]] contains $m$, so it contains $(m)$.\n7. The kernel is not $E[x]$, since $\\psi(1) = 1$. The ring $E[x]$ is a [[Euclidean domain|0DQG39K2NH]], hence a [[PID|0QKVE1GD9A]].\n8. Since $m$ is irreducible, $(m)$ is a [[maximal ideal|09J6KGT65G]] by the [[theorem on principal ideal domains|095ZHQ7WCJ]]. So the kernel of $\\psi$ equals $(m)$.\n9. By the [[first isomorphism theorem|0ZMYX7GMH0]], $h + (m) \\mapsto (\\varphi h)(\\beta)$ is an injective ring homomorphism $E[x]/(m) \\to L'$.\n10. By the [[structure of simple extensions|0ZN39QDTP9]], $h(\\alpha) \\mapsto h + (m)$ is a ring isomorphism $E(\\alpha) \\to E[x]/(m)$, where $E(\\alpha)$ is the [[generated field|0MA5B744K3]].\n11. Composing, $\\varphi' \\colon E(\\alpha) \\to L'$, $h(\\alpha) \\mapsto (\\varphi h)(\\beta)$, is a ring homomorphism. On constants $h = c \\in E$ it gives $\\varphi(c)$, so $\\varphi'$ extends $\\varphi$.\n\nNow we apply Zorn's lemma.\n\n12. Let $P$ be the set of pairs $(E, \\varphi)$ as in Step 1. Write $(E_1, \\varphi_1) \\le (E_2, \\varphi_2)$ when $E_1 \\subseteq E_2$ and $\\varphi_2$ restricts to $\\varphi_1$ on $E_1$.\n13. This relation is reflexive, antisymmetric and transitive, so $P$ is a [[partially ordered set|011C29Q1S1]]. It is nonempty, since it contains $K$ with its inclusion map into $L'$.\n14. Let $C$ be a [[chain|0GP9ASVQ1X]] in $P$. If $C$ is empty, then $K$ with its inclusion map is an [[upper bound|0H5J02P8S9]]. Assume $C$ is nonempty, and let $U$ be the union of the fields $E$ with $(E, \\varphi) \\in C$.\n15. Any two elements of $U$ lie in a common $E$ from $C$, since $C$ is a chain. So $U$ is a subfield of $L$ containing $K$.\n16. For $u \\in U$, set $\\varphi_U(u) = \\varphi(u)$ for any $(E, \\varphi) \\in C$ with $u \\in E$. Two such pairs are comparable, so $\\varphi_U(u)$ does not depend on the pair.\n17. Any two elements of $U$ lie in one $E$ from $C$, so $\\varphi_U$ preserves sums and products. So $(U, \\varphi_U) \\in P$, and it is an upper bound of $C$.\n18. By [[Zorn's lemma|04RXTQA5ZA]], $P$ has a [[maximal element|0TZEEFTNCC]] $(E, \\sigma)$.\n19. Suppose $E \\ne L$, for [[contradiction|0J0XBKG64S]], and pick $\\alpha \\in L \\setminus E$. Steps 1 to 11 give $(E(\\alpha), \\sigma') \\in P$ above $(E, \\sigma)$ and different from it. This contradicts maximality, so $E = L$.\n\nFinally we show that $\\sigma \\colon L \\to L'$ is an isomorphism.\n\n20. The map $\\sigma$ is [[injective|09N7DZ7SDB]], so it is a ring isomorphism from $L$ onto the subfield $\\sigma(L)$ of $L'$. Also $\\sigma(c) = c$ for every $c \\in K$.\n21. The field $\\sigma(L)$ is algebraically closed. Indeed, let $g \\in \\sigma(L)[x]$ with $\\deg g \\ge 1$. Applying $\\sigma^{-1}$ to its coefficients gives $g_0 \\in L[x]$ of the same degree.\n22. Then $g_0$ has a root $r \\in L$, and applying $\\sigma$ to $g_0(r) = 0$ gives $g(\\sigma(r)) = 0$.\n23. Every nonzero $q \\in \\sigma(L)[x]$ is a product of a constant and factors $x - c_j$ with $c_j \\in \\sigma(L)$. This follows by [[induction|0ASJVEFA9J]] on $\\deg q$ from Steps 21 and 22 and the [[factor theorem|0K3MWA7DMC]].\n24. Let $b \\in L'$. Since $L'/K$ is algebraic, $b$ has a minimal polynomial $m_b \\in K[x]$, and $K \\subseteq \\sigma(L)$.\n25. By Step 23, $m_b = \\prod_j (x - c_j)$ with $c_j \\in \\sigma(L)$, since $m_b$ is monic. Evaluating at $b$ in the field $L'$ gives $\\prod_j (b - c_j) = 0$.\n26. So $b = c_j$ for some $j$, and $b \\in \\sigma(L)$. Thus $\\sigma$ is surjective, and $\\sigma \\colon L \\to L'$ is a ring isomorphism fixing $K$.\n\n## Idea\n\nAn embedding of a subfield of $L$ into $L'$ over $K$ extends to one more element, by sending it to a root of the image of its minimal polynomial. Zorn's lemma gives an embedding of all of $L$. Its image is algebraically closed, and every element of $L'$ is a root of a polynomial that splits over the image, so the embedding is onto.\n",
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