Algebraic Closure

(1) $\Rightarrow$ (2). We induct on $\deg f$. If $\deg f = 1$, $f$ is already linear. Suppose $\deg f = n > 1$ and the claim holds for all smaller degrees. By (1), $f$ has a root $\alpha \in K$. Polynomial division in $K[x]$ gives

with $r \in K$ constant; substituting $x = \alpha$ and using $f(\alpha)=0$ gives $r = 0$, so $f(x) = (x-\alpha)g(x)$ with $\deg g = n-1$. By the inductive hypothesis $g$ is a product of linear factors, hence so is $f$.

(2) $\Rightarrow$ (3). Suppose $L/K$ is algebraic and $\alpha \in L$. Let $m(x) \in K[x]$ be the minimal polynomial of $\alpha$ over $K$; it is irreducible. By (2), any nonconstant polynomial over $K$, in particular any monic multiple of $m$ that is degree $\deg m$ — namely $m$ itself if we normalize it monic — factors into linear factors over $K$. Since $m$ is irreducible and factors into linear factors over $K$, $m$ itself must already be linear (an irreducible polynomial cannot be a product of two or more linear factors of positive degree unless it has degree $1$, since a factorization into $\ge 2$ nontrivial factors contradicts irreducibility). Hence $\deg m = 1$, so $m(x) = x - \alpha$ with $\alpha \in K$, forcing $\alpha \in K$. Thus $L = K$.

(3) $\Rightarrow$ (1). Suppose toward contradiction some nonconstant $f \in K[x]$ has no root in $K$. Let $p(x)$ be an irreducible factor of $f$ in $K[x]$ with $\deg p \ge 1$. If $\deg p = 1$ then $p(x) = c(x - \alpha)$ for some $\alpha \in K$, giving a root of $f$ in $K$, contradiction. So $\deg p \ge 2$. Form the quotient ring

which is a field because $p$ is irreducible in the PID $K[x]$, so $(p(x))$ is a maximal ideal. The composite $K \hookrightarrow K[x] \twoheadrightarrow L$ is an injective ring homomorphism (it is nonzero, hence injective, since $K$ is a field), so we may regard $L$ as a field extension of $K$ with $[L:K] = \deg p \ge 2$. This extension is algebraic (it is finite, and every finite extension is algebraic: if $[L:K]=n<\infty$, the powers $1,\alpha,\dots,\alpha^n$ of any $\alpha \in L$ are $n+1$ elements of the $n$-dimensional $K$-vector space $L$, hence linearly dependent over $K$, giving a nonzero polynomial over $K$ killing $\alpha$). This contradicts (3), which asserts no proper algebraic extension of $K$ exists. Hence every nonconstant $f \in K[x]$ has a root in $K$. $\blacksquare$

We now prove the main theorem, in two parts.

Part 1: Existence. Step 1 (Adjoin one root of every polynomial simultaneously). Let $S$ be the set of all nonconstant polynomials in $F[x]$. For each $f \in S$ introduce a formal indeterminate $x_f$, and form the polynomial ring

in these (possibly infinitely many) variables. Let $I \subseteq R$ be the ideal generated by all the elements $f(x_f)$ for $f \in S$ (here $f(x_f)$ means substitute the variable $x_f$ into $f$). Claim: $I$ is a proper ideal of $R$. If not, $1 \in I$, so there is a finite expression

for finitely many $f_1,\dots,f_n \in S$ and $g_i \in R$, with all other variables not appearing (only finitely many variables can occur in a finite sum). We may build a field extension $E/F$ in which each $f_i$ has an actual root $\alpha_i$: iterate the standard construction $K[x]/(p(x))$ for an irreducible factor $p$ of $f_i$ (as in the previous lemma's proof of (3) $\Rightarrow$ (1)) finitely many times, once for each $f_i$, to obtain a finite extension $E/F$ containing roots $\alpha_1,\dots,\alpha_n$ of $f_1,\dots,f_n$ respectively — this uses only the finite, iterated version of Splitting Fields existence. Now specialize the identity above by mapping $x_{f_i} \mapsto \alpha_i \in E$ (and any other variable, if it appeared, to $0$); this is a well-defined ring homomorphism $R \to E$ extending $F \hookrightarrow E$ since $R$ is a polynomial ring, so it sends the identity $1 = \sum g_i f_i(x_{f_i})$ to

in $E$, since $f_i(\alpha_i) = 0$ by construction. This says $1 = 0$ in $E$, contradicting that $E$ is a field. This contradiction proves $I \ne R$. Since $I$ is proper, by Zorn's Lemma $I$ is contained in a maximal ideal $\mathfrak{m}$ of $R$ (the union of a chain of proper ideals is proper, since $1$ lies in none of them, giving the hypothesis of Zorn's Lemma on the poset of proper ideals containing $I$, ordered by inclusion). Set

a field. The composite $F \hookrightarrow R \twoheadrightarrow F_1$ is injective (nonzero ring map out of a field), so $F_1$ is a field extension of $F$. By construction, for every nonconstant $f \in F[x]$, the image of $x_f$ in $F_1$ is a root of $f$ (since $f(x_f) \in I \subseteq \mathfrak{m}$, so $f(x_f) \equiv 0$ in $F_1$). Thus every nonconstant polynomial over $F$ has a root in $F_1$. Step 2 (Iterate and take a union). $F_1$ need not itself be algebraically closed (a polynomial with coefficients in $F_1 \setminus F$ might lack a root in $F_1$). Repeat Step 1 with $F_1$ in place of $F$ to produce $F_2 \supseteq F_1$ such that every nonconstant polynomial over $F_1$ has a root in $F_2$. Iterating countably often produces a chain

Let $L = \bigcup_{n \ge 0} F_n$. This is a field: given $a,b \in L$, both lie in some common $F_n$ (as the $F_n$ form a chain), so $a+b, ab, a^{-1}$ (for $a \ne 0$) are computed inside $F_n \subseteq L$, and the field axioms for $L$ follow from those for each $F_n$. Claim: $L$ is algebraically closed. Let $f(x) = \sum_{i=0}^d c_i x^i \in L[x]$ be nonconstant. Each coefficient $c_i$ lies in some $F_{n_i}$; since the chain is totally ordered, all coefficients lie in a single $F_n$ for $n$ large enough. Then $f \in F_n[x]$, so by the construction of $F_{n+1}$, $f$ has a root in $F_{n+1} \subseteq L$. Hence every nonconstant polynomial over $L$ has a root in $L$; by the Lemma (1)$\Leftrightarrow$(2)$\Leftrightarrow$(3), $L$ is algebraically closed. Claim: $L/F$ is algebraic. We show each $F_{n+1}/F_n$ is algebraic, and conclude by transitivity that $L/F$ is a union of finite steps of algebraic extensions. Fix $\alpha \in F_{n+1} = R/\mathfrak{m}$ (with $R$ the big polynomial ring over $F_n$). Every element of $R/\mathfrak m$ is a polynomial expression in finitely many of the images $\overline{x_f}$, and each $\overline{x_f}$ is algebraic over $F_n$ (it is a root of $f \in F_n[x]$). A field generated over $F_n$ by finitely many algebraic elements is a finite, hence algebraic, extension of $F_n$ — this is the standard fact about Finitely Generated Algebraic Extensions: if $\beta_1,\dots,\beta_k$ are each algebraic over $F_n$, then $[F_n(\beta_1,\dots,\beta_k):F_n] < \infty$, and every element of a finite extension is algebraic over the base (as shown in the Lemma's proof of (3)$\Rightarrow$(1)). Hence $\alpha$, lying in such a finitely-generated subfield, is algebraic over $F_n$. As $\alpha$ was arbitrary, $F_{n+1}/F_n$ is algebraic. Now if $\beta \in L$, then $\beta \in F_n$ for some $n$, and by transitivity of algebraicity through the finite tower $F = F_0 \subseteq F_1 \subseteq \cdots \subseteq F_n$ (an element algebraic over a field algebraic over $F$ is algebraic over $F$, since it satisfies a polynomial with finitely many coefficients, each algebraic over $F$, generating a finite extension of $F$ containing $\beta$), $\beta$ is algebraic over $F$. Hence $L/F$ is algebraic. Set $\overline{F} := L$. We have shown $\overline{F}$ is algebraically closed and algebraic over $F$, i.e. $\overline{F}$ is an algebraic closure of $F$. This proves existence.

Part 2: Uniqueness. Let $\overline{F}_1, \overline{F}_2$ be algebraic closures of $F$. Consider the set $\mathcal{P}$ of pairs $(K, \tau)$ where $F \subseteq K \subseteq \overline{F}_1$ is an intermediate field and $\tau : K \to \overline{F}_2$ is a field embedding with $\tau|_F = \mathrm{id}_F$. This set is nonempty since $(F, \mathrm{id}_F) \in \mathcal{P}$ (where we regard $F \subseteq \overline F_2$). Partially order $\mathcal P$ by

Chains have upper bounds. Given a chain $\{(K_i, \tau_i)\}_{i \in \Lambda}$, set $K = \bigcup_i K_i$ (a field, by the same chain argument as above) and define $\tau : K \to \overline{F}_2$ by $\tau(a) = \tau_i(a)$ for any $i$ with $a \in K_i$; this is well-defined because the $\tau_i$ agree on overlaps (by compatibility in the chain), and it is a field homomorphism because it agrees locally with each $\tau_i$. So $(K,\tau) \in \mathcal{P}$ is an upper bound. By Zorn's Lemma, $\mathcal{P}$ has a maximal element $(M, \sigma)$. Claim: $M = \overline{F}_1$. Suppose not; pick $\alpha \in \overline{F}_1 \setminus M$. Since $\overline{F}_1/F$ is algebraic, $\alpha$ is algebraic over $F$, hence over $M \supseteq F$; let $m(x) \in M[x]$ be its minimal polynomial over $M$, which is irreducible in $M[x]$. Apply $\sigma$ to the coefficients of $m$ to get an irreducible polynomial $m^\sigma(x) \in \sigma(M)[x]$ (irreducibility is preserved since $\sigma$ is a field isomorphism onto its image, inducing an isomorphism $M[x] \cong \sigma(M)[x]$). Since $\overline{F}_2$ is algebraically closed, by the Lemma every nonconstant polynomial over $\sigma(M) \subseteq \overline{F}_2$, in particular $m^\sigma$ (viewed as a polynomial over $\overline F_2$, or directly noting $\overline{F}_2$ algebraically closed implies $m^\sigma$ splits into linear factors over $\overline F_2$), has a root $\beta \in \overline{F}_2$. There is a standard isomorphism

where the first isomorphism sends $x \mapsto x$ compatibly with $\sigma$ on coefficients (i.e. is induced by $\sigma$), and the second sends $x \mapsto \beta$ (using that $\beta$ is a root of $m^\sigma$, so this is well-defined by the universal property of the quotient, and it is an isomorphism onto $\sigma(M)(\beta)$ because $m^\sigma$, being irreducible over $\sigma(M)$, is the minimal polynomial of $\beta$). Composing gives an embedding $\tau : M(\alpha) \to \overline{F}_2$ extending $\sigma$, with $\tau(\alpha) = \beta$. Then $(M(\alpha), \tau) \in \mathcal{P}$ strictly exceeds $(M,\sigma)$, since $M(\alpha) \supsetneq M$ (as $\alpha \notin M$) and $\tau|_M = \sigma$. This contradicts maximality of $(M,\sigma)$. Hence $M = \overline{F}_1$, and $\sigma : \overline{F}_1 \to \overline{F}_2$ is a field embedding fixing $F$ pointwise. Claim: $\sigma$ is surjective. The image $\sigma(\overline{F}_1)$ is algebraically closed: given a nonconstant $g \in \sigma(\overline{F}_1)[x]$, applying $\sigma^{-1}$ to coefficients gives $g' \in \overline{F}_1[x]$ nonconstant, which has a root $\gamma \in \overline{F}_1$ since $\overline{F}_1$ is algebraically closed; then $\sigma(\gamma)$ is a root of $g$ in $\sigma(\overline{F}_1)$ (apply $\sigma$ to the identity $g'(\gamma) = 0$, using that $\sigma$ is a ring homomorphism, to get $g(\sigma(\gamma)) = 0$). So $\sigma(\overline{F}_1)$ is an algebraically closed subfield of $\overline{F}_2$. Now let $\beta \in \overline{F}_2$. Since $\overline{F}_2/F$ is algebraic, $\beta$ is algebraic over $F$, hence its minimal polynomial $p(x) \in F[x]$ has coefficients fixed by $\sigma$ (as $\sigma|_F = \mathrm{id}_F$), so $p$ splits completely over the algebraically closed field $\sigma(\overline{F}_1) \supseteq F$ — indeed $p$ splits over $\sigma(\overline F_1)$ because $\sigma(\overline F_1)$ is algebraically closed and contains $F$, hence contains all roots of $p$ by the Lemma applied to $\sigma(\overline F_1)$. In particular $\beta$, being a root of $p$ in $\overline{F}_2$, must actually already lie in $\sigma(\overline{F}_1)$: this needs justification since $p$ could a priori have multiple roots in $\overline F_2$ only some of which lie in $\sigma(\overline F_1)$. To see all roots of $p$ in $\overline F_2$ coincide with the (full) root set of $p$ inside $\sigma(\overline F_1)$, note $p$ has degree $d = \deg p$ and factors into exactly $d$ linear factors over $\sigma(\overline F_1)\subseteq \overline F_2 \subseteq \overline F_2$; since $\overline F_2[x]$ is a UFD (a polynomial ring over a field), this factorization into linear factors is the same factorization when $p$ is viewed as an element of $\overline F_2[x]$, up to reordering and units, so every root of $p$ in $\overline F_2$ already appears among the linear factors coming from $\sigma(\overline F_1)$. Hence $\beta \in \sigma(\overline{F}_1)$. Since $\beta \in \overline{F}_2$ was arbitrary, $\sigma(\overline{F}_1) = \overline{F}_2$, so $\sigma$ is an isomorphism $\overline{F}_1 \xrightarrow{\sim} \overline{F}_2$ fixing $F$. $\blacksquare$