algebraic element over a field

Question

Let $F$ be a subfield of a field $K$, $a$ an element of $K$.

Prove that $a$ is algebraic over $F$ iff $F[a]=F(a)$.

For the first direction I've tried this: $a$ is algebraic over $F$ $\Rightarrow$ there is a polynomail $f$ over $F$ such that $f(a)=0$ $\Rightarrow$ $1,a,a^2,...,a^m$ are linearly dependent $\Rightarrow$ $\deg(F[a])<m$. but it contains $f(a)$ so they are equal..

is that correct? about the other direction.. i simply have no clue

Answer 1

Hints:

Since clearly $\;F[a]\subset F(a)\;$ , we concentrate on the other inclusion:

$$F(a)\subset F[a]\iff \forall\,g(x)\in F[x]\;,\;\;g(a)\neq 0\implies \frac1{g(a)}\in F[a] \iff\;$$

$$\exists\, h(x)\in F[x]\;\;s.t.\;\; \frac1{g(a)}=h(a)\iff h(a)g(a)=1$$

But now, using the fact that $\;F[x]\;$ is an Euclidean domain, we can use divide $\;f(x)\;$ by $\;g(x)\;$ with residue, and since $\;f(x)\;$ is irreducible (and thus prime), we get that

$$g(a)\neq 0\iff f(x)\nmid g(x)\;\;\text{in}\;\;F[x]\iff (f(x),g(x))=1\iff $$

$$\iff \exists\,m(x),n(x)\in F[x]\;\;s.t.\;\;f(x)m(x)+g(x)n(x)=1$$

and now just substitute $\;x\rightarrow a\;\ldots$

Answer 2

For the other direction use that $a$ is invertible in $F[a]=F(a)$, i.e. there is $x = c_0+c_1a+...+c_ma^m$ such that $xa = 1$. Hence, $a(c_0+c_1a+...+c_ma^m)-1=0$.

To show that $F[a]=F(a)$ you could show that the homomorphism of rings $F[X] \rightarrow F[a], x \mapsto a$ is onto and induces a ring homomorphism $F[x]/(f) \cong F[a]$ where f is the minimal polynomial of $a$, showing that $F[a]$ is a field.