Alex Youcis

@Et- Hmm. I guess it's conceivable, but I would be surprised. I will try to think about it more some other time. My guess is it comes more from us making a silly mistake about translating everything into coordinates, but I suppose I could be wrong.

@Et- I don't quite understand -- you think that it's not true that $G$ is glued from gluing the trivial extension to itself over a cover?

@Et- What do you mean 'in the abstract construction of $G$ itself'?

Will do.

@Et- Huh, OK. I will need to think on it. I am not sure where the error is -- it's probably another silly thing. Sorry for the trouble -- doing these nitty-gritty calculations can sometimes be a bit difficult to get right (even though conceptually they are simple). When I have more time I will try to sort this out if you haven't by then.

to be clear, our thing cannot be reduced, so it better not be a product of fields.

@Et- Yeah, I'm not sure I see what could be wrong here -- are you confident that you really get a product of fields -- this would confirm that it's an error in our set-up.

Let me think what could be going wrong here.

@Et- Hmm, I see a mistake in my calculation unfortunately.

I can write out my calculations if that's helpful

So it is the $k$-point (which is expected.

@Et- I could have miscalculated,but I just did the basic case when $p=2$ and for $f_1$ I got that it has to be in $k$.

OK, let me think about this for a second.

(sorry for not seeing this earlier)

this is an automorphism over $k[T,S]/(T^p-a,S^p-a)$ after all

Remember we are multiplying ty $(T/s)^i$ 'on $\mu_p$'

Shouldn't it be more like the second one sends $(f_i(X,Y_i))$ to $(f_i(T,(T/s)^i Y_i)$?

@Et- I just realized there is a mistake above.

Nice. Ping me on here with @ if you finish it.

Sure, I think we get a different thing a priori, but it just shifts the isomorphism classes around (via $x\mapsto x^{-1}$) or something.

If you're invested, it would be nice to actually compute what $\mathcal{O}(G)$ looks like with this gluing data, and make sure we get a $k$-Hopf algebra whose base change to $k(\sqrt[p]{a})$ is $\mu_p\times \mathbb{Z}/p\mathbb{Z}$.

does that seem reasonable to you?

and the other is $(f_i(X,Y_i))$ maps to $f((T/S)^i T,Y_i)$

sending $(f_i(X,Y_i))$ to $f(S,Y_i)$

I think one is the map $$\prod_{i\in\mathbb{Z}/p\mathbb[Z}}k[X][Y_i]/(Y_i^p-1)$

does coprojection mean the map on rings induced by the projections?

Does that sound right to you?

$(0,...,0,(T/S)^i x_i,0,...,0)$

Then, I think the automorphism corresponds to sending $(0,...,0,x_i,0,...0)$ to

Let's choose the generator $T/S$.

right so either $T/S$ or $S/T$.

I think we need to choose a generator (but I think this doesn't matter much) for $\mu_p(k[T,S]/(T^p-a,S^p-a)$$

I think it's just, as before, $\prod_{i\in\mathbb{Z}/p\mathbb{Z}}k[T,S][Y]/(S^p-a,T^p-a,Y^p-1)$.

We need to write down $\mu_p\times\mathbb{Z}/p\mathbb{Z}$ as a co-algebra over $k[T,S]/(T^p-a,S^p-a)$.

@Et- So let's think here.

Exactly right

Does that make sense?

The trivial gluing then basically identify $S$ and $T$, but our non-tivial gluing data modifies this in the way I think I wrote above.

like I said I think that it's helpful to understand the 'symmetry' of $U\times_X U$ as $k[S,T]/(T^p-a,S^p-a)$ and our two projection maps then correspond to the map $k[X]/(X^p-a)\to k[T,S]/(T^p-a,S^p-a)$ given by sending $X$ to $S$ or $T$.

I just think it was that we were not being careful about what projection maps looked like

@Et- Yo. I'm at a computer now if you wanna try to talk this out real quick.

@Et- Ah, you are right! I was hasty last night and made a computational mistake as I worried. I cannot fix it right now, but I will fix it later. But, perhaps you can sort it out for yourself before then and let me know what you get here, so we can compare answers. The issue is that the two projection maps $U\times X U$ correspond to the two two natural maps $k(\sqrt{a})\to k(\sqrt{a})\otimes_k k(\sqrt{a})$. To make this extra clear, let us write the first $k(\sqrt{a})$ as $k[S]/(S^p-a)$ and the second copy as $k[T]/(T^p-a)$. Then, this tensor product is $k[S,T]/(S^p-a,T^p-a)$.

Before we were writing this as $k(\sqrt[p]{a})[T]/(T^p-a)$ which led to the confusion. Finally, let us write the source copy of $k(\sqrt[p]{a})$ as $k[X]/(X^p-a)$. Then, the two projection maps (no gluing yet!!) are given by the one sending $X$ to $S$, and the other sending $X$ to $T$. So, I think the gluing data probably looks like something $f_i(S)=f_i((\tfrac{T}{S}-1)^i T)$ in $k[S,T]/(S^p-a,T^p-a)$ (but again I am on mobile so I cannot check carefully). I will look again later. Let me know what you get.

@Et- Oops, you're right -- it's the comultiplication that shifts -- thanks for the catch!

@Et- Glad it helped. See the above edit (but heed the warning about checking my work).

Best of luck!

anyways, I'm sure you've heard enough

you can't find a neighborhood of the Guass point where it's an isomorphism

and it very much 'acts like the disk mapping to the disk by squaring'

the Guass point is the only point which maps to the Gauss point