Let $K/\mathbb{Q}\_p$ be a finite field extension and $\mathcal{O}\_K\subseteq K$ be its ring of integral elements. Let also $G/\mathcal{O}\_K$ be a finite flat $p$-group scheme that is also an $\mathcal{O}\_E$-module where $E/\mathbb{Q}\_p$ is a finite field extension and $\mathcal{O}\_E\subseteq E$ is its ring of integral elements. Let $H \subseteq G$ be any subgroup scheme (it does not have to be $\mathcal{O}\_E$-stable). Let $G' = G \otimes\_{\mathcal{O}\_E} \mathcal{O}\_{E'}$ be the Serre tensor construction, where $E'/E$ is a totally ramified extension of degree $r$. Then it is true that $\text{lg}(\omega\_{G}) = \frac{1}{r}\text{lg}(\omega\_{G'})$, where $\omega\_G = s^\*\Omega^1\_{G/\mathcal{O}\_K}$ and $s:\text{Spec}(\mathcal{O}\_K)\to G$ is the identity section. Let $H' \subseteq G'$ be a subgroup such that it pulls back to $H$ under the closed embedding $G \to G'$. Then it should be true that $\text{lg}(\omega\_{H}) \ge \frac{1}{r}\text{lg}(\omega\_{H'})$ for every such group $H'$. My question is when equality holds.


When $H$ is an $\mathcal{O}\_E$-module (i.e., $\mathcal{O}\_E$-stable in $G$), then I can prove when $H'$ is the Serre tensor construction of $H$, which is a subgroup of $G'$, that we have an equality as we do with $G$ (same proof). When it is not $\mathcal{O}\_E$-stable, I don't think I can "Serre tensor construct" $H$ to be a subgroup of $G$, but I can, for example, take the image of $H$ in $G'$ under the embedding $G \to G'$ which pulls back to $H$. This is an example of a strict inequality.



Serre tensor construction on finite flat group schemes
Thinking about this more, it might be easier than you think. You don't seem to even care about the actions of $\mathcal{O}\_E$ or $\mathcal{O}\_{E'}$ in your question so you should consider what happens if you just forget them. Maybe $G'$ is complicated as an $\mathcal{O}\_{E'}$-group scheme, but if you just consider $G'$ as a group scheme with $\mathcal{O}\_E$-action then you can construct it as $G\otimes\_{\mathcal{O}\_E}\mathcal{O}\_{E'}$ where $\mathcal{O}\_{E'}$ is just considered as a $\mathcal{O}\_E$-module. As a $\mathcal{O}\_E$-module $\mathcal{O}\_{E'}$ is simply free of rank $r$, and so it should just follow that $G'$ is isomorphic to $G^r$. Now your questions become easy because if we fix such an isomorphism (e.g. by picking an $\mathcal{O}\_E$-basis of $\mathcal{O}\_{E'}$) then we can set $H'=H^r$ and this gives you the construction you want. 


NB I am assuming that by $\#\omega\_G=\frac{1}{r}\#\omega\_{G'}$ you mean actually that the *length* of $\omega$ gets divided by $r$; the orders themselves are powers of $p$ and so of course usually not multiples of $r$.


What do you think?


