Kernels of commuting linear operators on infinite dimensional vector space If $S$ and $T$ are commuting operators on an infinite dimensional vector space $V$, it is in general true that
$$\ker S + \ker T \subseteq \ker(ST),$$
but in general equality does not hold. A simple example is given by $S = T = \frac{d}{dx}$ on $C^\infty(\mathbb{R})$. I am looking for conditions on $S$ and $T$ that will give equality in the above equation, ie:
$$\ker S + \ker T = \ker (ST)$$
Writing $\ker T^\infty$ for $\cup_n \ker T^n$, I am currently trying to show that the conditions

*

*$\mathrm{im} S = \mathrm{im} T = V$,

*$\ker S^\infty \cap \ker T^\infty = \{ 0 \}$,

*$\dim \ker S < \infty$ and $\dim \ker T < \infty$,

*$ST = TS$
imply that $\ker S + \ker T = \ker(ST)$. I think the second condition can be weakened to $\ker S^2 \cap \ker T^2 = \{ 0 \}$, but I have this stronger condition for some operators I am interested in. Any help would be appreciated, thanks.
-edit- I am not confident that all these conditions are necessary.
 A: It suffices to assume that $ST=TS$, $\ker T$ is finite dimensional, and $\ker S\cap \ker T=0$.  Given these assumptions, suppose $v\in \ker(ST)$.  Let $A$ be the span of $\{v,Sv,S^2v,\dots\}$ and let $B$ be the span of $\{Sv,S^2v,\dots\}$.  Since $TS^nv=S^{n-1}STv=0$ for any $n>0$, $B\subseteq \ker T$.  Since $\ker T$ is finite-dimensional, $B$ must be finite dimensional, and hence so is $A$.  Let $p$ be the minimal polynomial of $S$ on $A$.  If $p$ has a nonzero constant term, then $v\in B$ and hence $v\in\ker T$.  Thus we may assume the constant term of $p$ is $0$ and write $p(x)=xq(x)$ for some polynomial $q$.
Observe now that $q(S)v\in\ker S$, and is nonzero by the minimality of $p$.  If the constant term of $q$ is zero, then $q(S)v\in B\subseteq \ker T$, but that is impossible since $\ker S\cap \ker T=0$.  Thus $q$ has nonzero constant term, and we may multiply by a scalar to assume the constant term is $1$.  But then $v-q(S)v\in B\subseteq \ker T$, and so $v=q(S)v+(v-q(S)v)\in\ker S+\ker T$.
