By M. Kisin, let $k$ be an algebraically closed field of characteristic $p$, and $K$ be a totally ramified extension of $B(k)$, the fraction field of the Witt vector ring $W(k)$, the category of finite flat group schemes over $\mathcal{O}_K$ which are killed by $p$ is equivalent to the category of Kisin modules over $\mathfrak{S}_1=k[[u]]$, whose objects are finite free $k[[u]]$ modules endowed with a Frobenius $\phi$ such that the cokernel is killed by the Eisenstein polynomial of a uniformizer in $K$.

My question is, there are a lot of submodules of a Kisin module whose quotient is not a free $k[[u]]$ module. However on the finite group scheme side, the quotient of a subgroup scheme is always again a finite flat group scheme. Where am I wrong in this inconsistency argument?

Thanks a lot!

  • $\begingroup$ Are these sub-modules $\phi$-stable though? $\endgroup$ – Keerthi Madapusi Pera Jan 5 '13 at 5:10
  • $\begingroup$ Moreover, they have to satisfy the condition that the cokernel of $\phi$ is killed by $u^e$ (that's the reduction of the Eisenstein; $e$ is the ramification index of $K$). $\endgroup$ – Keerthi Madapusi Pera Jan 5 '13 at 5:11

I get it. A submodule only corresponds to a monomorphism on the group scheme side, hence it may not admit a quotient, as is well known that the category of finite locally free group schemes over a general base is not abelian. That's why the stated equivalence only makes sense as exact categories. Thank you all the same!

| cite | improve this answer | |
  • $\begingroup$ By "monomorphism on the group scheme side" do you mean "monic map in the category of finite flat $O_K$-groups"? Note that if $e < p-1$ then such monic maps are always closed immersions and so do admit quotients within the category. So do you mean to say something other than what you have written? (That is, do you understand why whatever is confusing you also does not confuse you when $e < p-1$?) $\endgroup$ – user30379 Jan 5 '13 at 6:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.