Let $k$ be a field of characteristic zero.
Let $f=f(t), g=g(t) \in k[t]$ and assume that $\deg(f)=dn$, $\deg(g)=dm$, where $d=\gcd(\deg(f),\deg(g)) \geq 2$, $\gcd(n,m)=1$ and $n < m$.
Then we can write $f=\sum_{i=0}^{dn}a_it^i$ and $g=\sum_{j=0}^{dm}b_jt^j$, with $a_i,b_j \in k$, $a_{dn} \neq 0$, $b_{dm} \neq 0$. Further assume that $a_{dn-1}=b_{dm-1}=0$.
Remark: I am not sure if the assumption that $\deg(f)=dn$ and $\deg(g)=dm$ is important or not. However, I wish to assume that the 'second leading coefficients' of $f$ and $g$ are zeros, namely, $a_{\deg(f)-1}=b_{\deg(g)-1}=0$.
What additional conditions are required in order to guarantee that $\deg(\gcd(f,g)) \in \{0,1\}$?
According to this, we need one of the following cases to happen: (i) $s_0 \neq 0$ (for the case $\deg(\gcd(f,g)) =0$). (ii) $s_0= 0$ and $s_1 \neq 0$ (for the case $\deg(\gcd(f,g)) =1$).
In other words, we wish to show that it is not possible to have $s_0=s_1=0$ (which is equivalent to $\deg(\gcd(f,g)) \geq 2$).
Actually, I wish to find the 'exact' form of $s_0$ and $s_1$ (these are expressions in $a_i, b_j$), and then I can see when $s_0=s_1=0$ is not possible. Unfortunately, I am afraid that for arbitrary $f$ and $g$, $s_0$ and $s_1$ are very complicated expressions in $a_i,b_j$, so it will be difficult to determine whether $s_0=s_1=0$ or not. Are there additional conditions on the coefficients or degrees that would help? (Is my condition $a_{\deg(f)-1}=b_{\deg(g)-1}=0$ helpful?).
Thank you very much!