Exercise 3.B.26

Proof. Let $T:P_m(R)\to P_{m-1}(R)$ such that $\mathit{Tp}=\mathit{Dp}$. Suppose $p\in \mathrm{null}<!--\; FUNCTION\; APPLICATION\; -->T$. Since $T$ reduces the degree of every nonconstant polynomial by $1$ and $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->0=-\infty$, it follows that $p$ can only be constant. Hence $\dim <!--\; FUNCTION\; APPLICATION\; -->\mathrm{null}<!--\; FUNCTION\; APPLICATION\; -->T=1$. By the Fundamental Theorem of Linear Maps $T$ is surjective. Because $m$ was arbitrary, so is $D$. □