Consider the $D\_{\mathbb{A}^1}$-module $M:=D\_{\mathbb{A}^1}/(x)$, and the map $f:z\mapsto z^k$. I want to know $f^\*(M)$. I believe it only has a single non-zero cohomology, namely in degree $0$, which is equal to
$$\mathbb{C}[z]\otimes\_{\mathbb{C}[z^k]}(D/(x)),$$
with connection
$$\partial(z^q\otimes \partial^w)=qz^{q-1}\otimes \partial^w+kz^{q+k-1}\otimes \partial^{w+1}.$$
Now by general theory this should be holonomic, and in particular finitely generated. However, I have some trouble finding a generating set. For example, it is not clear to me how the elements $1\otimes \partial^{w}, w=1,2,\dots,$ can be generated from a finite generating set.


Any help or hints would be appreciated.



Explicit computation of D-modules pullback
If you write the degree of $z^q \otimes \delta^w$ as $ kw -q$ then there is a one-dimensional vector space of elements of each degree $\geq (1-k)$ (since if $q \geq k$ we can reduce by bring $z^q$ over to the other side) and $\partial$ takes elements of a given degree to elements of the same degree plus one.


So to check that every element can be generated from a finite generating set, it suffices to check that there are only finitely many degrees where $\partial$ applied to the generator in that degree produces the zero multiple of the generator in the next degree (instead of a nonzero multiple).


In fact, I claim there are no such degrees, and the whole thing is generated by $z^{k-1} \otimes 1$. I calculated this by passing to constructible sheaves, computing the pullback there, and going back but let's check this explicitly in D-modules.


To check this, note that for $q\geq 1$,


$$\partial(z^q\otimes \partial^w)=qz^{q-1}\otimes \partial^w+kz^{q+k-1}\otimes \partial^{w+1} $$ $$= qz^{q-1}\otimes \partial^w+kz^{q-1}\otimes x\partial^{w+1}$$ $$ = qz^{q-1}\otimes \partial^w+kz^{q-1}\otimes (\partial^{w+1} x - (w+1) \partial^w) $$ $$ = (q- k (w+1)) z^{q-1}\otimes \partial^w $$


and we have $q \leq k-1$ for the generator $q- k (w+1)>0$.


For the $q=0$ case, the argument is simpler - the first term vanishes and the second term is nontrivial.


