Let $k$ be a field. Let $a\in k(x)-k$. Then, it is to be shown that $k(x)$ is an algebraic extension of $k(a)$.
It is clear that $k(a)\subset k(x)$. (because $a\in k(x)-k$)
Since $k(a)$ is quotient field of $k[a]$, elements of $k(a)$ look like $f(a)/g(a)$, where $f,g\in k[x]$ and $g(a)\ne 0$. Take any $f/g\in k(x)-k(a)$. It is now to be shown that there exists a non zero polynomial in $k(a)[x]$ which has a root $f/g$.
Claim: $k(x)\cong k(a)$.
Proof: Here, $a$ is transcendental over $k$. If not, then $k(a)=k[a]$. Suppose that $a=p/q, (p,q)=1, q\ne0, p,q\in k(x)$. $$q/p\in k(a)\implies \exists b_i, q/p=b_0+b_1(p/q)+...+b_n(p/q)^n\implies q^{n+1}=(b_0q^n+...+b_np^n)p$$
It follows that $p|q\implies p,q$ are units. $\implies p/q=a\in k$, contradiction. Hence, $a$ is transcendental over $k$.
The following proves the claim.
Define $\phi:k(x)\to k(a)$ by $\phi(f(x)/g(x))=f(a)/g(a) $. (Note that $g(a)\ne 0$ as $a$ is algebraic over $k$).
It's clear that $\phi$ is onto homomorphism. In fact, $\phi$ is $1-1$ too as the following shows: $f(a)/g(a)=0\implies f(a)=0\implies f=0\implies \ker \theta=(0).$ So by the first isomorphism theorem, $k(x)\cong k(a). \square$
But since $k(a)\subset k(x)$, we must have $k(x)=k(a)\implies [k(x):k(a)]=1\implies k(x)$ is a finite extension of $k(a).\implies k(x)$ is an algebraic extension of $k(a)$.
Is this correct? Thanks.