 Solution

For the contrapositive of $(b)\Rightarrow (c)$, suppose $\mathrm{char}k=p>0$. ConsiderLet $x_i$ act by multiplication and $D_i$ by formal differentiation. Differentiation preserves the defining ideal, since $D_j(x_i^{p}f)=x_i^p{D}_{j}f$ in characteristic $p$. The relations $[D_i,x_j]={\delta }_{ij}$ hold before quotienting and hence hold on $M$. Its basis consists of the monomials $x^{\alpha }$ with $0\le {\alpha }_i<p$, so ${\dim }_{k}M=p^n>0$. This is the explicit finite-dimensional module for the Weyl algebra in positive characteristic.

Therefore condition (b) fails in positive characteristic and implies characteristic zero. Combining this with the preceding two implications proves $(a)⟺(b)⟺(c)$. The same representation also has nonzero proper kernel, since the infinite-dimensional Weyl algebra cannot inject into a finite-dimensional endomorphism algebra, confirming its failure of simplicity in positive characteristic.