Let $\mathbb{F}$ be an arbitrary field. For a polynomial $f\in\mathbb{F}[x]$, we use $Z(f)$ to denote set of roots of $f$ in $\mathbb{F}$. Let $S$ and $T$ be sets of elements of $\mathbb{F}$ of size $s$ and $t$ respectively. Let $V$ be subspace of $\mathbb{F}[x]$ of polynomials of degree at most $d$. Let $U$ be subspace of $V$ of polynomials $f$ such that $S\subseteq Z(f)$ and $T\subseteq Z(f')$. Here $f'$ is the first formal derivative of $f$. What can be said about dimension of $U$? If $T=\emptyset$ then $\dim(U)=d-s+1$. Also if $T\subseteq S$ then $\dim(U)=d-(s+t)+1$. I am interested in lower/upper bounds on $\dim(U)$ for general case when $S$ and $T$ can be anything. Even if we can prove these bounds for special fields $\mathbb{F}$, it might be interesting. One of the first interesting cases is when $S\cap T = \emptyset$.

## 1 Answer

Isn't this just linear algebra? We know that $f(z) = s(z) g(z),$ where $s$ is the polynomial whose roots are in $s,$ and $g$ has degree $d-s$ (so $d-s+1$ undetermined coefficients). On the other hand, $f^\prime(z) = t(z) h(z),$ where $h$ has $d-t$ undetermined coefficients. Since $f$ itself has $d+1$ undetermined coefficients, we have $2d+1$ (inhomogeneous) *linear* equations in $3d - s - t + 4$ unknowns (unless my arithmetic is off), so the dimension of the solution space is $d-s-t+4$ in the general case.