Rational functions in C[t] I was solving one of my questions in algebraic geometry when I encountered this:
Let $f(t)$ and $g(t)$ be rational functions in $\mathbb{C}[t]$.
Show that $f(t)$ and $g(t)$ are in $\mathbb{C}$ provided that $f^4(t)+g^4(t)=1$ in $\mathbb{C}(t)$. Deduce that
$f^4(α)+g^4(α)\neq 1$
for some $α ∈ \mathbb{C}$ such that $f(α)$ and $g(α)$ are both well defined.
My question is, we can show that $f^4(t)+g^4(t)=1$ implies, by using $f(t)=\frac{a(t)}{b(t)}$ and $g(t)=\frac{c(t)}{d(t)}$ where $a(t),b(t)$ and $c(t),d(t)$ are coprime, that
$f(t),g(t)$ are both constant. If this is the case, why would there be $\alpha \in \mathbb{C} $ where it does not hold? 
 A: Let $C\subset \mathbb P^2(\mathbb C)$ be the smooth curve $x^4+y^4=z^4$ and let $U\subset \mathbb C$ be the open Zariski subset where both rational functions are regular.
The affine curve $ C\cap \mathbb C^2$ has equation $x^4+y^4=1$ and we  have a regular map $$F_0:U\to C\cap \mathbb C^2:t\mapsto (f(t),g(t))=[f(t):g(t):1]$$ This regular map  extends to a morphism (Hartshorne, Chap.I, Prop. 6.8, page 43)$$F:\mathbb P^1(\mathbb C)\to C$$ Since $C$ is not rational (it has degree $4$ and thus genus $g=\frac {(4-1)(3-1)}{2}=3$), such a regular map must be a constant so that both $f(t)$ and $g(t)$ are constant: $f(t), g(t)\in \mathbb C$.
A: Technically don't we want a two-way relation here? We need to show - $f(t),g(t) \in \mathbb{C} \iff f(t)^{4}+g^{4}(t) =1 $. The above proves $(\impliedby)$, but not ($\implies$)? Is this trivial because divisions of functions by elements of $\mathbb{C}$ leads to an equivalent function? [as a member of the ring, since prime ideals are just these] i.e it's obvious that: 
$$ f(t),g(t) \in \mathbb{C} \implies f^{4}(t)+g^{4}(t) =\lambda : \lambda \in \mathbb{C}$$ 
But why is it necessarily true that this is equivalent to: 
$$ f^{4}(t)+g^{4}(t) =1 $$ 
Since $\mathbb{C}[t]$ has to map to field $\mathbb{C}$ and not to $\mathbb{P}^{1}_{\mathbb{C}}$? 
