Proof of the uniqueness of the algebraic closure

Proof

Let $L$ and $L'$ be algebraic closures of the field $K$. First we extend embeddings by one element.

1. Let $E$ be a subfield of $L$ containing $K$, and $\varphi \colon E \to L'$ a ring homomorphism with $\varphi(c) = c$ for $c \in K$. Let $\alpha \in L$.
2. Since $L/K$ is algebraic, some nonzero $g \in K[x]$ has $g(\alpha) = 0$. As $K \subseteq E$, $\alpha$ is algebraic over $E$.
3. Let $m$ be the minimal polynomial of $\alpha$ over $E$. It is irreducible in $E[x]$, so $\deg m \ge 1$.
4. Let $\varphi m \in L'[x]$ be the polynomial obtained by applying $\varphi$ to the coefficients of $m$. It is monic of degree $\deg m \ge 1$.
5. Since $L'$ is algebraically closed, $\varphi m$ has a root $\beta \in L'$.
6. By the universal property of the polynomial ring, $\psi \colon E[x] \to L'$, $h \mapsto (\varphi h)(\beta)$, is a ring homomorphism. Its kernel contains $m$, so it contains $(m)$.
7. The kernel is not $E[x]$, since $\psi(1) = 1$. The ring $E[x]$ is a Euclidean domain, hence a PID.
8. Since $m$ is irreducible, $(m)$ is a maximal ideal by the theorem on principal ideal domains. So the kernel of $\psi$ equals $(m)$.
9. By the first isomorphism theorem, $h + (m) \mapsto (\varphi h)(\beta)$ is an injective ring homomorphism $E[x]/(m) \to L'$.
10. By the structure of simple extensions, $h(\alpha) \mapsto h + (m)$ is a ring isomorphism $E(\alpha) \to E[x]/(m)$, where $E(\alpha)$ is the generated field.
11. 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 $\varphi$.

Now we apply Zorn's lemma.

12. 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 $E_1$.
13. This relation is reflexive, antisymmetric and transitive, so $P$ is a partially ordered set. It is nonempty, since it contains $K$ with its inclusion map into $L'$.
14. Let $C$ be a chain in $P$. If $C$ is empty, then $K$ with its inclusion map is an upper bound. Assume $C$ is nonempty, and let $U$ be the union of the fields $E$ with $(E, \varphi) \in C$.
15. 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 $K$.
16. 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.
17. 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 $C$.
18. By Zorn's lemma, $P$ has a maximal element $(E, \sigma)$.
19. Suppose $E \ne L$, for contradiction, 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 $E = L$.

Finally we show that $\sigma \colon L \to L'$ is an isomorphism.

20. The map $\sigma$ is injective, so it is a ring isomorphism from $L$ onto the subfield $\sigma(L)$ of $L'$. Also $\sigma(c) = c$ for every $c \in K$.
21. 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.
22. Then $g_0$ has a root $r \in L$, and applying $\sigma$ to $g_0(r) = 0$ gives $g(\sigma(r)) = 0$.
23. 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 on $\deg q$ from Steps 21 and 22 and the factor theorem.
24. Let $b \in L'$. Since $L'/K$ is algebraic, $b$ has a minimal polynomial $m_b \in K[x]$, and $K \subseteq \sigma(L)$.
25. 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 $\prod_j (b - c_j) = 0$.
26. 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 $K$.