# Field Adjoined Algebraic Elements is Algebraic

## Definition

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

If each $x \in S$ is algebraic over $K$ then $K \left({S}\right)$ is algebraic over $K$.

## Proof

Let $S \subseteq L$ be arbitrary, and $x \in K \left({S}\right)$.

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

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

Moreover a Finite Field Extension is Algebraic.

Therefore $x$ is algebraic over $K$.

$\blacksquare$