# No minimal polynomial for differentiation operator

Let $D$ be the Differentation operator of the of polynomials over $R.$ Prove that there is no polynomial $g(t),$ such that $g(D)=T_0.$

But characteristic polynomials satisfies it's operator.

I dnt understand the phenomenon.

If $g$ is a polynomial of degree $n$, then $g(D)$ is the differentiation of order $n$. If you take a polynomial $p$ of degree $n+1$, then $g(D)p\ne0$, hence $g(D)\ne0$.
On the other hand, if you consider the space of polynomials of maximum degree $n$, which is finite-dimensional, then $g(t)=t^{n+1}$ is the minimal polynomial of $D$.