# Why $\mathbb{C}(f(t),g(t))=\mathbb{C}(t)$ implies that $\gcd(f(t)-a,g(t)-b)=t-c$, for some $a,b,c \in \mathbb{C}$?

Assume that $$f(t),g(t) \in \mathbb{C}[t]$$ satisfy the following two conditions:

(1) $$\deg(f) \geq 2$$ and $$\deg(g) \geq 2$$.

(2) $$\mathbb{C}(f(t),g(t))=\mathbb{C}(t)$$.

In this question it was mentioned that in that case, there exist $$a,b,c \in \mathbb{C}$$ such that $$\gcd(f(t)-a,g(t)-b)=t-c$$.

Unfortunately, I do not see why this is true.

Perhaps Theorem 2.1 (about resultants) or this question (about subresultants) may somehow help (perhaps no).

Edit: Just to make sure:

Is it true that there exist infinitely many $$a \in \mathbb{C}$$ and infinitely many $$b \in \mathbb{C}$$ such that $$\gcd(f(t)-a,g(t)-b)=t-c$$, for (infinitely many) $$c \in \mathbb{C}$$?

Choose $$c \in \mathbb{C}$$ such that $$f'(c) \neq 0$$ etc. (as in the answer). Clearly, there are infinitely many such $$c$$'s. Let $$a:=f(c)$$ and $$b:=g(c)$$.

Asumme that there exist finitely many $$a \in \mathbb{C}$$ or finitely many $$b \in \mathbb{C}$$ such that $$\gcd(f(t)-a,g(t)-b)=t-c$$, $$c \in \mathbb{C}$$.

W.l.o.g., there exist finitely many $$a \in \mathbb{C}$$ such that $$\gcd(f(t)-a,g(t)-b)=t-c$$, $$c \in \mathbb{C}$$.

By the pigeon hole principle, there exist $$a_0$$ (among those finitely many $$a$$'s), such that for infinitely many $$c$$'s, we have $$a_0=f(c)$$.

This is impossible from the following reason: Let $$h(t):=f(t)-a_0$$. Then $$h(c)=f(c)-a_0=0$$, so $$c$$ is a root of $$h(t)$$, and trivially every polynomial can have only finitely many different roots.

So after all, I think that I have proved that there exist infinitely many $$a \in \mathbb{C}$$ and infinitely many $$b \in \mathbb{C}$$ such that $$\gcd(f(t)-a,g(t)-b)=t-c$$, $$c \in \mathbb{C}$$.

Any hints are welcome!

We may assume that $$f$$ and $$g$$ are monic. There exists some nonzero two-variable polynomials $$P,Q$$ such that $$P(f(t),g(t))=tQ(f(t),g(t))$$, and $$Q(f,g)(t)=0$$ only finitely many times (else the composition $$P/Q (f,g)$$ is not defined because $$Q(f,g)=0$$).

Let $$c$$ be such that $$f’(c) \neq 0$$, and there exists no $$d$$ such that $$Q(f(d),g(d))=0$$ and $$f(d)=f(c)$$.

Then $$f(t)-f(c)$$ and $$g(t)-g(c)$$ have only $$c$$ as a common root, because any root $$d$$ satisfies $$(f,g)(d)=(f,g)(c)$$, thus $$d=(P/Q)(f(d),g(d))=(P/Q)(f(c),g(c))=c$$. Moreover, $$c$$ is a simple root of $$f-f(c)$$. So the gcd of the polynomials is $$t-c$$.

• Thank you very much! I understand that there exist two-variable polynomials $P$ and $Q$ such that $\frac{P(f(t),g(t))}{Q(f(t),g(t))}=t$. Please, why $Q \in \mathbb{C}^{\times}$? – user237522 Dec 15 '18 at 22:51
• I mistook the parentheses for brackets. The main point should hold though: $(f,g)$ is injective because of that relation. – Mindlack Dec 15 '18 at 22:54