Find a polynomial $P_{n}\in{\mathbb{C}[X]}$ such as $S_{n}=\ker P_{n}(D)$

Find a polynomial $P_{n}\in{\mathbb{C}[X]}$ such as $S_{n}=\ker P_{n}(D)$

-With $S_{n}$ $(n\geq 1)$ all of the functions such as $y^{(n)}=y $
(where $y^{(n)}$ is the nth derivative of $y$)

-D the endomorphism sending the functions $C^{\infty}$ on their derivatives D : y -> y'

I already proved that D is an endomorphism and $S_{n}$ a vector space containing the functions $ke^x$

But my probleme is that i dont really understand the meaning of $P_{n}(D)$, and so dont see what $S_{n}=ker P_{n}(D)$ could be. It probably have to do with eigenvalues, eigenvectors...But really I dont see anything.

Thanks for your help

Let $P_n = X^n - 1$.

Saying $x \in \ker(P_n(D))$ is precisely saying $P_n(D)(x) = 0$.

Then, we have $D^n(x) - Id(x) = x^{(n)} - x = 0$, which is exactly $x^{(n)} = x$.

Also, $P_n(D)$ is a polynomial evaluated at an endomorphism.

If $f$ is an endomorphism of a $\mathbb{K}$-vector space $E$ and $P = a_nX^n + \cdots + a_1X + a_0 \in \mathbb{K}[X]$ a polynomial, then $P(f)$ is the endomorphism $a_nf^n + \cdots + a_1f + a_0$ where $f^i$ is $f$ composed with itself $i$ times.