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Given a split Hopf algebroid $(S,\Sigma)=(S,S\otimes B)$ over $K$, Ravenel leaves as an exercise the proof of the following:

An ideal $J\subset S$ is invariant under the action of the group $\mathrm{Hom}(B,K)$ if and only if $\eta_R(J)\subset J\Sigma$, where $\eta_R$ is the right unit.

Is it clear where the group action of $\mathrm{Hom}(B,K)$ on $S$ or $J$ comes from? Obviously, by the fact that the algebroid is split, there is an action of $\mathrm{Hom}(B,K)$ on $\mathrm{Hom}(S,K)$.

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When the Hopf algebroid is $(MU_*, MU_*(MU))$, the action of Hom$(B, \mathbb{Z})=\Gamma$ is the following: given $f \in \Gamma$, let $f$ act on the universal formal group law. This new formal group law gives a map $MU_* \rightarrow MU_*$. I am not sure how to do this for an arbitrary Hopf algebroid. – Vitaly Lorman Jul 3 '12 at 18:55
The map $\operatorname{Hom}(B,K) \to \operatorname{End} S$ comes from sending $\alpha \in \operatorname{Hom}(B,K)$ to the composite $$S \xrightarrow{\eta} S \otimes_K B \xrightarrow{1 \otimes \alpha} S \otimes_K K \xrightarrow{\simeq} S,$$ where $\eta$ is either $\eta_R$ or $\eta_L$, depending upon convention. – Eric Peterson Jul 3 '12 at 23:26
Oh man of course! Thanks! :) – Jon Beardsley Jul 5 '12 at 0:31
@Eric I'm going to write up the (now obvious) proof to the proposition, unless you want to write an answer or something. Also, I've been looking at the notes you wrote up on the geometry of formal varieties, and they've been super helpful! – Jon Beardsley Jul 5 '12 at 18:15
No, go right ahead. And I'm glad to hear someone has found them helpful. :) – Eric Peterson Jul 7 '12 at 22:38

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