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A Hopf-Galois extension of commutative rings, as defined by Montgomery here, is a morphism of commutative rings $\phi:A\to B$ with a Hopf-algebra $H$ coacting on $B$ by a ring map $c:B\to B\otimes H$ such that the following two maps are bijections:

  1. The canonical inclusion of cofixed points $A\hookrightarrow B^{coH}$ is a bijection, where $B^{coH}=\{b\in B:c(b)=b\otimes 1_H\}$.
  2. The "Galois" or "torsor" map $\tau:B\otimes_AB\to B\otimes H$ given by $(b_1\otimes b_2)\mapsto\mu_B(b_1,c(b_2))$ is a bijection, where $\mu_B(-,-)$ is the $A$-algebra structure morphism on $B$.

Question: Should any of these maps have more structure? For instance, should the coaction be a map of $A$-algebras rather than just rings? Should the map in 2. be an isomorphism of comodules or corings or something along these lines?

The description I've given above is the way these definitions are given in the sources I've seen, but it seems like they probably could be strengthened in a lot of cases. Does anyone know about this or have a lot of experience with this structure to possibly confirm this?

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  • $\begingroup$ I think any extra structure you could ask of the torsor map will come automatically from its definition (and all of the other structure in the problem). Not sure though. $\endgroup$ – Qiaochu Yuan Dec 14 '15 at 7:00
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Unless I am very mistaken:

  1. If the coaction of $H$ on $B$ is not $A$-linear, then there is not a canonical map $A \to B^H$, and so condition 1 would not make sense.

  2. Commutativity of $B$ together with the fact that the coaction is a ring map together imply that the map $B\otimes B \to B\otimes H$ IS a ring homomorphism. What coring structure did you have in mind? If $B$ is a bialgebra, then both sides are corings, but the map IS NOT a bialgebra homomorphism in most examples. In general, if you drop commutativity on $B$ these notions are still meaningful, and then the map $B\otimes B \to B\otimes H$ IS NOT a ring homomorphism in most examples. In any case, for getting the definition right what type of homomorphism it is doesn't matter, since all you care is that it be an isomorphism, and you can check that on underlying vector spaces.

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  • $\begingroup$ Thanks Theo! I'd like to use as few commutativity conditions as possible, but it seems like there's a bit of give and take. The coring structure I had in mind was that $B\otimes_AB$ is a $B$-coring by $B\otimes_A A\otimes_AB\to B\otimes_A B\otimes_AB$, and $B\otimes H$ should be a coring by the diagonal map of $H$, i.e. $B\otimes H\to B\otimes H\otimes H\cong B\otimes H\otimes_B B\otimes H$. $\endgroup$ – Jonathan Beardsley Dec 14 '15 at 3:51
  • $\begingroup$ Honestly I guess I can write down a whole bunch of diagrams and say the necessary things about them (good catch on the $A$-linearity by the way), but I kind of wonder why they don't seem to be built into the definitions. Also, I'm not necessarily working over a field (re: your statement about vector spaces). $\endgroup$ – Jonathan Beardsley Dec 14 '15 at 3:55
  • $\begingroup$ @JonBeardsley Just find-change "vector space" for "object in your underlying category" (e.g. "abelian group") and you get the same statement --- for pretty much any world you might hope to work in, isomorphism of rings/modules/... is detected on underlying objects. $\endgroup$ – Theo Johnson-Freyd Dec 15 '15 at 5:01
  • $\begingroup$ So it really doesn't matter what coring structure you have in mind: either the map is a morphism, or it isn't, and in neither case is asking that it be an iso any better than asking that (it be a morphism and) it be a bijection. $\endgroup$ – Theo Johnson-Freyd Dec 15 '15 at 5:02
  • $\begingroup$ Your coring structure on $B\otimes H$ is the coring structure on $H$ base-changed to $B$. Your "coring" structure on $B \otimes_A B$ is funny, though. You say it is a "$B$-coring structure", but you use two different $B$-module structures on $B \otimes_A B$. It is a coalgebra object in the monoidal category of $B$-$B$-bimodules, but not in the category of $B$-modules (which is monoidal if $B$ is commutative). The category of bimodules is never symmetric (except in trivial cases), and so for example it is not meaningful to ask that this coring be cocommutative. $\endgroup$ – Theo Johnson-Freyd Dec 15 '15 at 5:06

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