When $\mathbb{C}(f(t),g(t))=\mathbb{C}(t)$ imply that there exist $a,b \in \mathbb{C}$ such that $\deg(\gcd(f(t)-a,g(t)-b))=2$?

Question

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

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

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

After asking this and then this and getting nice answers, I would like to ask the following question:

What additional conditions on $f$ and $g$ are required in order to guarantee the existence of $a,b \in \mathbb{C}$ such that $\deg(\gcd(f-a,g-b))=2$?

Notice that if $f=t^2$ and $g=t^2-2t$, then there are no such $a,b$.

A plausible(?) additional condition: What if we assume, in addition, that all monomials of degrees $\geq 1$ in $f$ and $g$ have non-zero coefficients?

Remarks: (1) Take $f=t^2, g=t^3$. For $a=b=0$ we have $\deg(\gcd(f-a,g-b))=\deg(\gcd(t^2-0,t^3-0))=\deg(\gcd(t^2,t^3))=\deg(t^2)=2$. If I am not wrong, $a=b=0$ is the only option for $\deg(\gcd(f-a,g-b))=2$.

If $(a,b) \neq (0,0)$, then $\deg(\gcd(f-a,g-b)) \in \{0,1\}$.

Therefore, it is not reasonable to expect the existence of infinitely many such $(a,b)$. In contrast to the case $\deg(\gcd(f-a,g-b))=1$, where we had infinitely many such $a,b \in \mathbb{C}$.

(2) Observe that if $\deg(\gcd(f-a,g-b))=2$, and if it happens that $\gcd(f-a,g-b)=(t-c)^2$ for some $c \in \mathbb{C}$, then $\deg(\gcd(f',g'))\geq 1$ (since $t-c$ divides $\gcd(f',g')$).

Then the ideal in $k[t]$ generated by $f',g'$ cannot equal $k[t]$, since otherwise there would exist $u,v \in k[t]$ such that $1=uf'+vg'= u(t-c)F+v(t-c)G$, which is impossible. ($F$ is such that $f'=(t-c)F$ and $G$ is such that $g'=(t-c)G$).

We could be more precise and require either $\gcd(f-a,g-b)=(t-c)^2$ or $\gcd(f-a,g-b)=(t-c)(t-d)$, where $c,d \in \mathbb{C}$ with $d \neq c$; both cases are interesting for me.

Please see also this question and its nice answer; so a necessary and sufficient condition is that $s_0=s_1=0$ and $s_2 \neq 0$, and then $s_2=\gcd(f-a,g-b)$. However, it seems quite difficult to compute $s_0,s_1,s_2$ for high degree $f$ and $g$.

Is there something interesting that we can say about the coefficients of $f$ and $g$ that will guarantee $s_0=s_1=0, s_2 \neq 0$? In particular, is the plausible condition I have mentioned above (all coefficients of monomials of degrees $geq 1$ are non-zero) may help?

Of course, my condition is not necessary, since for $f=t^3-4t$, $g=t^2+1$, $t$ does not appear in $g$, but still for $a=0,b=5$ we have: $f-0=t^3-4t=t(t^2-4)$ and $g-5=t^2+1-5=t^2-4$, so $\gcd(f-0,g-5)=t^2-4$.

Any hints and comments are welcome!

Answer 1

The $\mathbb{C}(f(t),g(t))=\mathbb{C}(t)$ condition means the polynomials are coprime, if they are not it suffices to replace $f(t)$ by $f(t)+c$, so this is not really a problem.

Thus we are in the following situation : $f(t)-f(\alpha),\ g(t)-g(\alpha)$ have a common root at $\alpha$ and you are asking if for some $\alpha$ they have another common root.

• With $f(t) = t^2, g(t) = t^2+t$ the answer is no.

• The case of cubic polynomials. Wlog we can assume they are monic, shifting $t$ to $t+\alpha$ and substracting $f(0),g(0)$ we obtain the generic monic cubic polynomials with a common root at $0$ : $F(t) = t^3+at^2+bt, G(t)=t^3+At^2+Bt$. $$\gcd(t^2+at+b,t^2+At+B) = \gcd(t^2+at+b,(A-a)t+(B-b))$$ The $\gcd$ is not $1$ iff $(A-a)=(B-b) = 0$ or $(\frac{B-b}{a-A})^2+a (\frac{B-b}{a-A}) + b = 0$. Whence

  Given two cubic polynomials $f(t),g(t)$, there exists $\alpha$ such that $\gcd(f(t)-f(\alpha),g(t)-g(\alpha))$ is of degree $2$ iff $$f(t) = r((t+\alpha)^3+a(t+\alpha)^2+b(t+\alpha)+c), \\ g(t) = s((t+\alpha)^3+A(t+\alpha)^2+B(t+\alpha)+C), \\ A-a \ne 0, \quad (B-b)^2 + a (B-b)(a-A)+b(a-A)^2 = 0$$

• For higher degrees there are similar algebraic equations defining the polynomials satisfying your problem.