In algebraic extension, field homomorphism induces isomorphism.

Question

I read this page's first answer. But I'm curious about why $\varphi$ induces injective map $S \to S$. Isn't it possible to make $\varphi(\alpha)= k$ such that $k$ is not root of $f(X)$?

Answer 1

$\varphi$ fixes $\mathbb{Q}$, hence

$$0 = \varphi(0) = \varphi(f(\alpha)) = \varphi\left(\sum_{k=0}^n c_k\alpha^k\right) = \sum_{k=0}^n \varphi(c_k)\varphi(\alpha)^k = \sum_{k=0}^n c_k\varphi(\alpha)^k = f(\varphi(\alpha)),$$

if $f(X) = \sum\limits_{k=0}^n c_k X^k\in \mathbb{Q}[X]$. Thus $\varphi$ must map zeros of $f$ to zeros of $f$.