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  "body": "\n## Proof\n\nFirst we show that every [[field|0QA8CHH8MG]] $F$ has an extension $F_1$ in which each nonconstant polynomial over $F$ has a root, with $F_1/F$ algebraic.\n\n1. Let $S$ be the set of [[monic|0BDX4MBKT4]] polynomials $f \\in F[x]$ with $\\deg f \\ge 1$. Take one variable $X_f$ for each $f \\in S$.\n2. Let $R = F[X_f : f \\in S]$ be the union of the [[polynomial rings|0CES91M79A]] in finitely many of these variables. Two elements of $R$ lie in a common such ring, where we add and multiply them. Let $I$ be the [[ideal generated|030TZ0DQWT]] by the elements $f(X_f)$ for $f \\in S$.\n3. We show $I \\ne R$, by [[contradiction|0J0XBKG64S]]. Suppose $1 \\in I$. Then $1 = \\sum_{i=1}^n g_i \\, f_i(X_{f_i})$ for some distinct $f_1, \\dots, f_n \\in S$ and some $g_i \\in R$.\n4. The $g_i$ involve finitely many variables. So there are distinct $f_{n+1}, \\dots, f_N \\in S$ with all $g_i \\in R_0 = F[X_{f_1}, \\dots, X_{f_N}]$.\n5. Set $E_0 = F$. For $1 \\le i \\le n$, [[some extension|0VG02M1FEZ]] $E_i$ of $E_{i-1}$ contains a [[root|00EHX6FWC0]] $a_i$ of $f_i$. By [[induction|0ASJVEFA9J]] this gives a field $E = E_n$ containing $F$ and $a_1, \\dots, a_n$.\n6. Applying the [[universal property of the polynomial ring|0DQMDSVSXJ]] one variable at a time gives a [[ring homomorphism|0GGE0Z0JQ1]] $\\varphi \\colon R_0 \\to E$. It is the identity on $F$, sends $X_{f_i}$ to $a_i$ for $i \\le n$, and sends $X_{f_i}$ to $0$ for $i \u003e n$.\n7. Applying $\\varphi$ to Step 3 gives $1 = \\sum_{i=1}^n \\varphi(g_i) \\, f_i(a_i) = 0$ in $E$. This is impossible in a field, so $I \\ne R$.\n8. By the [[existence of maximal ideals|02HJYT53F7]], some maximal ideal $M$ of $R$ contains $I$. Then $F_1 = R/M$ is a field, because [[R/M is a field iff M is maximal|0WD3SE7S1X]].\n9. The map $F \\to F_1$, $c \\mapsto c + M$, is a ring homomorphism into the [[quotient ring|0VFW4Y69BT]]. It is [[injective|09N7DZ7SDB]], so we identify $F$ with its image, a subfield of $F_1$.\n10. For $f \\in S$ let $x_f = X_f + M$. Then $f(x_f) = f(X_f) + M = 0$, since $f(X_f) \\in I \\subseteq M$. So each $f \\in S$ has a root in $F_1$.\n11. Let $g \\in F[x]$ with $\\deg g \\ge 1$ and leading coefficient $c$. Then $c^{-1} g \\in S$ has a root in $F_1$, which is also a root of $g$.\n12. Each $x_f$ is [[algebraic|0NPW1DWECM]] over $F$, since $f \\ne 0$ and $f(x_f) = 0$.\n13. Let $u \\in F_1$. Then $u = h + M$ with $h \\in R$, and $h$ involves finitely many variables $X_{f_1}, \\dots, X_{f_k}$. So $u = h(x_{f_1}, \\dots, x_{f_k})$ lies in the [[generated field|0MA5B744K3]] $G_k = F(x_{f_1}, \\dots, x_{f_k})$.\n14. Let $G_0 = F$ and $G_j = G_{j-1}(x_{f_j})$. The element $x_{f_j}$ is algebraic over $G_{j-1}$, since its polynomial $f_j$ has coefficients in $F \\subseteq G_{j-1}$.\n15. So $[G_j : G_{j-1}]$ is finite by the [[degree of a simple extension|0N5QW9K7MH]]. By the [[tower law|0A7KJ68XHT]], $G_k/F$ is a [[finite extension|0NRYPM9Y27]].\n16. [[Finite extensions are algebraic|0DFYTNZ6AM]], so $u$ is algebraic over $F$. Hence $F_1/F$ is [[algebraic|09EYMKMSB9]].\n\nNow we iterate this construction for the field $K$ of the statement.\n\n17. Let $K_0 = K$ and $K_{i+1} = (K_i)_1$, the field built from $F = K_i$ in Steps 1 to 16. Replacing each $K_{i+1}$ by an isomorphic copy, we may assume that $K_i$ is a [[subfield|07QY5RPFBR]] of $K_{i+1}$.\n18. Let $\\Omega = \\bigcup_{i \\ge 0} K_i$. Any two elements of $\\Omega$ lie in a common $K_i$, since the $K_i$ increase. Define their sum and product in that $K_i$.\n19. This does not depend on the choice of $i$, because each $K_i$ is a subfield of the later ones. Each field axiom involves finitely many elements, which lie in a common $K_i$. So $\\Omega$ is a field containing $K$.\n20. By induction on $i$, $K_i/K$ is algebraic. For $i = 0$, each $c \\in K$ is a root of $x - c$. If $K_i/K$ is algebraic, then $K_{i+1}/K_i$ is algebraic by Step 16, so $K_{i+1}/K$ is algebraic by [[transitivity|0TPWS0K1BC]].\n21. Every element of $\\Omega$ lies in some $K_i$, so it is algebraic over $K$. Hence $\\Omega/K$ is algebraic.\n22. Let $g \\in \\Omega[x]$ with $\\deg g \\ge 1$. Its finitely many coefficients lie in some $K_N$. By Step 11 for $F = K_N$, $g$ has a root in $K_{N+1} \\subseteq \\Omega$.\n23. So $\\Omega$ is [[algebraically closed|02SJG1HC32]]. With Step 21, $\\Omega$ is an [[algebraic closure|0MCXX29HWY]] of $K$.\n\n## Idea\n\nAdjoin a root of every polynomial at once, using one variable per polynomial and a maximal ideal. The ideal of relations is proper because any finite part of it can be solved in a finite extension. One step gives roots for polynomials over the old field. Repeating countably often and taking the union gives a field that is algebraically closed and algebraic over $K$.\n",
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