# A question about algebraic extension and how to extend an automorphism

Let be $$L/K$$ an algebraic extension, $$\alpha\in L$$ and $$P_\alpha$$ the minimal polynomial of $$\alpha$$ over $$K$$. We denote $$\beta\in L-\{\alpha\}$$ another root of $$P_{\alpha}$$.

I can take an automorphism $$\tau:K(\alpha)\longrightarrow \overline{K}$$ such as $$\tau(\alpha)=\beta$$ and $$\tau_{|_K}=Id$$.

The question is:

why i can extend $$\tau$$ to an automorphism $$\sigma:L\longrightarrow \overline{K}\hspace{0.15cm}$$, i.e, $$\hspace{0.15cm}\sigma_{|_{K(\alpha)}} = \tau$$ ?

The main claim here is that if $$L/K$$ is an algebraic field extension and $$\Omega$$ is any algebraically closed field (for example $$\overline{K}$$) then any homomorphism $$\varphi: K\to \Omega$$ has an extension to a homomorphism $$\hat{\varphi}:L\to\Omega$$. I'll assume here that $$L/K$$ is finite, for infinite algebraic extensions it can be done similarly using Zorn's lemma.

First assume that $$L=K(\alpha)$$, a simple algebraic extension. Let $$\varphi: K\to\Omega$$ be a homomorphism, and we want to extend it to $$L$$. Let $$\hat{K}=\varphi(K)$$. Since any field homomorphism is injective we know $$\varphi: K\to\hat{K}$$ is an isomorphism. Also, if $$P_{\alpha}=\sum_{i=0}^n c_ix^i$$ is the minimal polynomial of $$\alpha$$ over $$K$$ then let $$\hat{P_{\alpha}}=\sum_{i=0}^n \varphi(c_i)x^i\in\hat{K}[x]$$. Since $$\varphi$$ is a bijection and preserves all the field operations it is easy to see that $$\hat{P_{\alpha}}$$ is irreducible over $$\hat{K}$$. So if we let $$\beta\in\Omega$$ be a root of $$\hat{P_{\alpha}}$$ (which exists, since $$\Omega$$ is algebraically closed) then $$\hat{P_{\alpha}}$$ is its minimal polynomial over $$\hat{K}$$. So just like we did in your previous question, there is a chain of isomorphisms:

$$L=K(\alpha)\cong K[x]/(P_{\alpha})\cong \hat{K}[x]/(\hat{P_{\alpha}})\cong \hat{K}(\beta)\subseteq\Omega$$

And if $$k\in K$$ then this chain sends $$k\to k+(P_{\alpha})\to \varphi(k)+\hat{P_{\alpha}}\to \varphi(k)$$. So this homomorphism $$L\to\Omega$$ is indeed an extension of $$\varphi$$.

General case: Now assume $$L/K$$ is any finite extension and let $$\varphi:L\to\Omega$$ be a homomorphism, and we want to prove it has an extension to $$L$$. We use induction on $$[L:K]$$. If $$[L:K]=1$$ there is nothing to prove. Assume the statement is true for extensions of degree up to $$n-1$$ and suppose $$[L:K]=n$$. We split into two cases:

Case 1: $$L/K$$ has no intermediate fields. In that case for any $$\alpha\in L\setminus K$$ we have $$L=K(\alpha)$$. In that case we already know an extension exists.

Case 2: There is an intermediate field $$K\subset M\subset L$$. Then $$[M:K],[L:M]. By induction hypothesis $$\varphi$$ has an extension to $$\varphi_0:M\to\Omega$$. And again by induction hypothesis $$\varphi_0$$ has an extension to $$\hat{\varphi}:L\to\Omega$$.

Conclusion: Finally we can answer your question. $$\tau:K(\alpha)\to\overline{K}$$ is a homomorphism, so it can be extended to $$L$$. (since $$L$$ is an extension of $$K(\alpha)$$).

• Thank you very much !!!! Commented May 4, 2020 at 11:36