Isomorphism of D-modules implied by vanishing of first Lie algebra cohomology

Question

I'm reading the paper Invariant differential operators on a reductive Lie algebra and Weyl group representations by Nolan Wallach.

First I recall the notation used in the paper: $V$ is a finite dimensional complex vector space, $S(V)$ is the symmetric algebra on $V$, endowed with the usual grading, and $\mathscr{P}(V)$ is the $\mathbb{C}$-algebra of polynomial functions on $V$. We write $S_+(V)$ for the irrelevant ideal of $S(V)$. Let $\mathbf{D}(V)$ the algebra of differential operators on $V$ (that is, the Weyl algebra on $V$). Recall that we see the elements of $V$ as directional derivatives and that the multiplication map $$ \mathscr{P}(V)\otimes S(V)\to \mathbf{D}(V) $$ is an isomorphism.

In Appendix 1, Wallach proves the following result:

Lemma 3. Let $M$ be a finitely generated $\mathbf{D}(V)$-module such that if $m\in M$ and $p\in S_+(V)$ then there exists $k$ such that $p^k m= 0$. Then $M$ is isomorphic as a $\mathbf{D}(V)$-module to a finite multiple of $\mathscr{P}(V)$.

The proof goes as follows: We can see $V$ as an abelian Lie algebra and $M$ as a $V$-module via $v\cdot m = \partial_v\cdot m$. Then he proves that $H^1(V,M)=0$ (this is an easy computation, but it is not needed for my question).

Now, it is well known that $$ H^0(V,M) = \{m\in M \; | \; Vm = 0\}=:M^V. $$ If $0\neq m \in M^V$ then the simplicity of $\mathscr{P}(V)$ as a $\mathbf{D}(V)$-module implies that $\mathbf{D}(V)m = \mathscr{P}(V)m$ is isomorphic to $\mathscr{P}(V)$ as $\mathbf{D}(V)$-module.

Finally, Wallach says that the vanishing of the first cohomology easily implies that, as $\mathbf{D}(V)$-module, $M$ is isomorphic to $\mathscr{P}(V)\otimes M^V$, where the action of $\mathbf{D}(V)$ on $\mathscr{P}(V)\otimes M^V$ is on the first factor. And this finishes the proof.

I have two questions regarding this last statement:

1. Why the fact that $H^1(V,M)=0$ implies that $M\cong \mathscr{P}(V)\otimes M^V$ as $\mathbf{D}(V)$-modules? I can see that there is a natural morphism $$ \mathscr{P}(V)\otimes M^V\to M, \quad f\otimes m\mapsto fm $$ but I cannot see from the vanishing of the first cohomology that this is injective nor surjective. How we can deduce this fact?
2. How this finishes the proof? I can see that the lemma is proved if $M^V$ is finite dimensional as a vector space, but I cannot see why this is true.

Remark. I think I can prove the lemma by using the fact that $M$ belongs to the category $\mathcal{O}$ for the algebra $\mathbf{D}(V)$, using thickened Verma modules. But the reason I'm posting this question is because I'm interested in understanding the Lie algebra cohomology argument given in the paper.