Let $k$ be a field of characteristic zero. Let $f_1(t),f_2(t) \in k[t]$ and $f: k \to k^2$ defined by $f(t):=(f_1(t),f_2(t))$.
First case $k=\mathbb{C}$: According to A. van den Essen (page 2), the following claim holds: $\mathbb{C}[f_1(t),f_2(t)]=\mathbb{C}[t]$ if and only if $f'(t)\neq (0,0)$ for all $t \in \mathbb{C}$ and $f$ is injective.
(1) Can one please sketch a proof for this claim?
Second case $k=\mathbb{R}$: It is not true that if $f'(t)\neq (0,0)$ for all $t \in \mathbb{R}$ and $f$ is injective, then $\mathbb{R}[f_1(t),f_2(t)]=\mathbb{R}[t]$, as the following example shows: $f_1(t)=t^2$, $f_2(t)=t+t^3$.
(2) Is there an additional differential geometry condition, call it $C$, such that: $\mathbb{R}[f_1(t),f_2(t)]=\mathbb{R}[t]$ if and only if $f'(t)\neq (0,0)$ for all $t \in \mathbb{R}$, $f$ is injective, and $C$.
Remarks: Concerning question (2): (i) I am interested in a 'not too strong' additional condition, namely, not something like $f_1'(t)=1$. (ii) Perhaps the additional condition $C$ will involve the second derivative $f''(t)$? Concerning both questions: (iii) Please see the comments in this question, especially, how page 8, claim b is relevant to my question? (iv) Considering the fields of fractions $k(f_1(t),f_2(t))=k(t)$ instead of $k[f_1(t),f_2(t)]=k[t]$ seem also interesting.
Edit: This paper is somewhat relevant.
Any comments and hints are welcome!
