Discussion
Let L|K be an algebraic field extension and \(\sigma : L \to L\) be a K-algebra homomorphism. Then :
Answer: A
\(\sigma\) is bijective.
A ring homomorphism between fields that sends 1 to 1 is injective, because its kernel is an ideal of a field not containing 1, hence zero. For surjectivity, take \(\alpha \in L\) with minimal polynomial \(p\) over K, and let \(R\) be the finite set of roots of \(p\) in L. Since \(\sigma\) fixes K, it maps \(R\) into \(R\), and being injective on a finite set it maps \(R\) onto \(R\). So \(\alpha = \sigma(\beta)\) for some \(\beta \in R\).
Injectivity rules out C and D. B can happen only for transcendental extensions, such as \(x \mapsto x^{2}\) on \(K(x)\), not for algebraic ones.
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