Tip_pay wrote:

nguyenalex wrote:

I have the following exercise:

Let $E$ be an algebraic extension of $K$ , and let $F$ be an algebraic closure of $K$ containing $E$ . Prove that if $\sigma : E \to F$ is an embedding such that $\sigma(c) = c$ for all $c \in K$ , then $\sigma$ extends to an automorphism of $F$ .

Consider the set of all pairs $(L, \tau)$ , where $L$ is an intermediate field such that $E \subseteq L \subseteq F$ , and $\tau : L \to F$ is a $K$ -embedding extending $\sigma$ , that is, $\tau|_E = \sigma$ . This set is nonempty, since the pair $(E, \sigma)$ satisfies the conditions. We define a partial order $(L_1, \tau_1) \leq (L_2, \tau_2)$ if $L_1 \subseteq L_2$ and $\tau_2|_{L_1} = \tau_1$ . Given a chain ${(L_i, \tau_i)}$ , the union $L = \bigcup L_i$ is a field, and we can define $\tau : L \to F$ as $\tau(x) = \tau_i(x)$ if $x \in L_i$ . Since the chain is constraint-compatible, $\tau$ is a well-defined $K$ -embedding extending $\sigma$ . Thus, every chain has an upper bound. By Zorn's lemma, there exists a maximal element $(M, \rho)$ , where $M \subseteq F$ and $\rho : M \to F$ is a $K$ -embedding such that $\rho|_E = \sigma$

Suppose that $M \neq F$ , i.e., there exists $\alpha \in F \setminus M$ . Since $F$ is algebraic over $K$ and $K \subseteq M$ , the element $\alpha$ is algebraic over $M$ . Let $p(x) \in M[x]$ be the minimal polynomial of $\alpha$ over $M$ , it is irreducible over $M$ . Applying $\rho$ to the coefficients of $p(x)$ , we obtain the polynomial $\rho(p)(x) \in \rho(M)[x]$ . Since $F$ is algebraically closed, $\rho(p)(x)$ has a root $\beta \in F$ , that is, $\rho(p)(\beta) = 0$ . Since $\rho : M \to \rho(M)$ is an isomorphism and $p(x)$ is irreducible over $M$ , the polynomial $\rho(p)(x)$ is irreducible over $\rho(M)$ . Therefore, there is an embedding $\tau : M(\alpha) \to F$ such that $\tau|_M = \rho$ and $\tau(\alpha) = \beta$ (e.g., $M(\alpha) \cong \rho(M)(\beta)$ ). This means that $\rho$ extends to $\tau$ on a larger field $M(\alpha)$ , where $M \subsetneq M(\alpha) \subseteq F$ , which contradicts

It remains to show that $\tau': F \to F$ is an automorphism. Since $\tau'$ is an embedding, it is injective. The image of $\tau'(F)$ is an algebraically closed subfield of $F$ containing $K$ . But $F$ is the minimal algebraic closure, so $\tau'(F) = F$

Thanks, but I already solved it. I'm stuck at finding such $\bar{\sigma}$ .