3

Let $K\subseteq E\subseteq F$ be fields. Prove that if $F$ is algebraic over $K$ then $F$ is algebraic over $E$ and $E$ is algebraic over $K$.

Proof: Suppose that $F$ is algebraic over $K$. We want to show that $F$ is algebraic over $E$. It follows from assumption that there exists $a\in F$ such that $f(a)=0$ for some $f(x)\in K[x]$. Since $K\subseteq E$ it follows that $f(x)\in E[x]$ this $F$ is algebraic over $E$. Now we show that $E$ is algebraic over $K$. That is if $b\in E$ then $g(b)=0$ for some $g(x)\in K[x]$. But I don't get how to show this part. Thanks.

For the second part to show $E$ is algebraic over $K$ I let $b\in E$. It follows that $b\in F$. Since $F$ is algebraic over $K$ there exists a polynomial $g(x)\in K[x]$ such that $g(b)=0$. Since $K\subseteq E$ it follows that $E$ must be algebraic over $K$.

user60887
  • 2,935
  • I think you might be confused about the definition of an algebraic extension. $F/K$ is algebraic if ~every~ element of $F$ is algebraic over $K.$ –  Dec 18 '13 at 06:36

1 Answers1

2

Hint: $b \in E \implies b \in F$, so since, $F$ is algebraic over $K$, so...

Note the reverse implication is also true, that $F$ is algebraic over $E$ and $E$ is algebraic over $K$ implies $F$ is algebraic over $K$. You may want to try that as well.

Aman
  • 411