I have a few questions about a proof of the statement that if $u$ is transcendental over $K$ then $K(u)\cong K(x)$. My questions are marked as red and stated below the proof.
The proof goes like this:
Assume $u$ is transcendental over $K$. For every non-zero polynomial $g(x) \in K[x]$ we have $g(u) = \phi_u(g(x)) \neq 0$ must be an invertible element of $K(u)$. Hence for every $f,g \in K[x]$ with $g(x) \neq 0$
we have $f(u)/g(u) = f(u) \times g(u)^{-1} \in K(u)$. Therefore we can define a homomorphism $\theta:K(x) \rightarrow K(u)$ given by $\theta(f/g) = f(u)/g(u)$ for all $f/g \in K(x)$
Since $\theta$ is non-trivial it's injective. $\color{red}{(1)}$. Therefore $Im \theta$ is a subfield $\color{red}{(2)}$ containing $u$ and $K$ and hence must be equal to K(u). $\color{red}{(3)}$. So our result follows from the first isomorphism theorem. $\color{red}{(4)}$
$\color{red}{(1)}$ I don't understand why this means that $\theta$ is injective.
$\color{red}{(2)}$ Why is $Im \theta$ necessarily a field?
$\color{red}{(3)}$ Is this because $K(u)$ is the smallest field containing $u$ and $K$?
$\color{red}{(4)}$ I guess this is because we've shown that in fact $\theta$ is both injective and surjective and hence an isomorphism? Can't I argue that $ker\theta$ is in fact the zero-element of $K(x)$ and hence trivial and so $\theta$ is an isomorphism?
Really grateful for any help with this!
(1) The kernel of the ring-homomorphism $\theta$ must be an ideal of the field $K(x)$. The only ideals of a field are $0$ and the field itself.
(2) The image of an injective homomorphism is isomorphic to the domain (via that homomorphism)
(3) by definition
(4) Or we already noted in (2) that it is an isomorphism with the image, which happens to be $K(u)$