Field Adjoined Algebraic Elements is Algebraic

From ProofWiki
Jump to navigation Jump to search


Let $L / K$ be a field extension and $S \subseteq L$ a subset.

If each $x \in S$ is algebraic over $K$ then $\map K S$ is algebraic over $K$.


Let $S \subseteq L$ be arbitrary, and $x \in \map K S$.

By Field Adjoined Set $x \in \map K S$ if and only if $x \in \map K {\alpha_1, \ldots, \alpha_n}$ for some $\alpha_1, \ldots, \alpha_n \in S$.

We have that $\map K {\alpha_1, \ldots, \alpha_n} / K$ is finite by Finitely Generated Algebraic Extension is Finite.

Moreover a Finite Field Extension is Algebraic.

Therefore $x$ is algebraic over $K$.