Prove the set $\{x \in K : x \text{ is algebraic over }F\}$ is a subfield of $K$, containing $F$

Question

This originates from Pinter's Abstract Algebra, Chapter 29, Exercise G2.

Let F be a field, and K a finite extension of F. Prove the set $\{x \in K : x \text{ is algebraic over }F\}$ is a subfield of $K$, containing $F$.

Let $S=\{x \in K : x \text{ is algebraic over }F\}$. Note $F(x_0,x_1,\dots)$ for $x_i\in S$ is a minimal field extension of $F$, and every element in $F(x_0,x_1,\dots)$ is algebraic over $F$.

1. $a\in S\implies a\in F(x_0,x_1,\dots)$ by construction. Hence $S\subseteq F(x_0,x_1,\dots)$.
2. $F(x_0,x_1,\dots)\subseteq K$, so $a\in F(x_0,x_1,\dots)\implies a\in K$. As $a$ is algebraic over $F$, $a\in F(x_0,x_1,\dots)\implies a\in S$. Hence $F(x_0,x_1,\dots)\subseteq S$.

Therefore $S=F(x_0,x_1,\dots)$ and $F\subseteq S\subseteq K$.

Is this a reasonable argument?

Answer 1

Your argument seems fine, but I think adjoining all the algebraic elements is a bit overkill.

A simpler way to do it is just to use the exercise right before this one, exercise G1:

If $a$ and $b$ are algebraic over $F$, then $a+b$, $a-b$, $ab$, and $a/b$ (when $b\ne 0$) are algebraic over $F$.

We have $F \subset S=\{x \in K : x \text{ is algebraic over }F\}$ since every element of $F$ is algebraic over itself; hence $S \ne \varnothing$. If $a, b \in S$, then they are algebraic, so $a\pm b, ab \in S$ and $b^{-1} = 1/b \in S$ when $b \ne 0$. So, $S$ is a subfield.