Let $R$ be an integral domain with infinite many elements and $\forall a\in R$ let $T(a) =\{b\in R: b\mid a\}$ be the set of dividers which ist for all $a\neq 0_R$ finite. Additionally, let $f\in R[X]$ be a polynomial with degree $n>1$ and $m=\max\{r\in \mathbb{N}:2r\leq n\}$
Show that there are pairwise different elements $a_0,...,a_m\in R$ so that $T_i :=T(f(a_i))$ is finite for $ i=0,...,m$
My attempt:
I know that if $R$ is integral domain, every polynomial $p\in R[X]\setminus \{0_{R[X]}\}$ has a maximum of $\deg(p)$ roots in $R$. That means, there are finite many elements $z_1,...,z_n$, so that $f(z_i) = 0_R$ and $T(f(z_i))$ infinite.
But since $R$ is infinite, I can surely find $a_0,...,a_m$ so that $f(a_i)\neq 0_R$ and $T(f(a_i))$ finite (according to definition).
But I feel like I'm missing something because the question asks specifically for $m$-elements so maybe I understood the question wrong?
I agree. Firstly, the question has told you that all nonzero elements of $R$ have finitely many divisors.
So, the second part is simply asking you to find $m + 1$ elements of $R$ which are not roots of $f (\neq 0)$.
But this follows immediately since $R$ is infinite and so, $f$ has at most $\deg(f) < \infty$ many roots.