Let $F$ be a field, let $E=F(a)$ be a simple extension field of $F$, and let $b\in E-F$. Prove that $a$ is algebraic over $F(b)$.

Question

Question: Let $F$ be a field, let $E=F(a)$ be a simple extension field of $F$, and let $b\in E-F$. Prove that $a$ is algebraic over $F(b)$.

I think I have the right idea, but I am getting hung up on the definition of the field of fractions. So, as $E=F(a)$, we have that $E=\{\frac{f(a)}{g(a)}:f(x),g(x)\in F[x], g(a)\neq 0\}$. So then as $b\in E-F$, $b=\frac{f(x)}{g(x)}$, so if we let $h(x)=bg(x)-f(x)\in F(b)[x]$, then if $h(a)=0$, we'd be done. But, by the definition of field of fractions, isn't $g(a)\neq 0$?

Answer 1

To avoid confusion, let $f,g\in F[x]$ be such that $g(a)\neq0$ and $b=\frac{f(a)}{g(a)}$. To show that $a$ is algebraic over $F(b)$ you need to show that $h(a)=0$ for some nonzero $h\in(F(b))[y]$.

Because $b\notin F$ we have either $\deg f>0$ or $\deg g>0$, or both. Hence $h=bg(y)-f(y)$ is nonzero and does the trick.