Wonderful Answer! Thank you very much, your explanation of the use of cohomology was very illuminating and clear, and the answer gave me quite a bit to think about (and will likely give me even more as I'll progress through the book). I've tried to work on the exercise you gave at the end, but I'm finding it quite challenging (which mainly shows that I need more experience with sites and stacks haha :) ). To be more specific, I'm not quite famillar yet with gluing schemes over the fpqc (or any non zariski) site. Some extra help would be appreciated.

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

Thank you! This helped a lot, I did the calculations myself too and we got the same answer so it's probably fine. I only have one small question however: why is the multiplication defined by $x_i^r=x_{ir}$? Wouldn't the product be self contained in each coordinate?

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

Hey, I am sorry but I have one more thing that bugs me... Isn't the definition we arrived at for $\mathscr{O}(G)$ just a product of fields? Because $f_i(x_i)$ has no $T$ factor in any monomial, all the terms of exponent $1$ up to $p-1$ must vanish for the equality to hold, which would mean $f_i$ lies in $k(\sqrt[p]{a})$. (And of course, we can't have $G$ be a product of fields as then it would have no k points)

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

Alright, I'm on one too. So you think the problem comes from wrong gluing?

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

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 $T$.

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.

Does that make sense?

Oh, I think I see it now... So like we treated the extension $\sqrt([p]{a}) as fixed which was wrong as the second projection isn't a map over this copy of the extension, is that right?

Exactly right

What would be the group automorphism over $k[T,S]/(T^p-a,S^p-a)$ in this case?

like the one from Z/pZ to \mu_p

@Et- So let's think here.

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

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

In the old solution it was T/S, right? (identifying S^p with the p of a)

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)$$

p root of a*

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

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

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

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

Does that sound right to you?

wait... what are the two coprojection maps in this case?

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

Yea

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

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

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

does that seem reasonable to you?

Yea, I think this looks more reasonable

Though it may be more convenient in this case to use the generator S/T

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 $\mu_p\times \mathbb{Z}/p\mathbb{Z}$.

Alright, I'll try working it out

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.

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

alright

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

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

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

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

(sorry for not seeing this earlier)

Yea, I caught that too, no worries

@AlexYoucis I am running into the same problem as before, which is that we get a product of field extensions. More specific

More specifically, writing f_i = \Sigma c_nm * Y^n X^m, where the summation runs for n,m in (Z/pZ)^2, we see that we must have m=ni mod p, which means that the ring associated to each coordinate is p dimensional and is generated by a monomial of the form XY^k for k s.t ki=m mod p

For example for i=1 this is just the polynomial ring in the variable XY, and we have (XY)^p=a, so this is just the field extension given by adjoining a pth root of a

OK, let me think about this for a second.

@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 $k$.

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

I can write out my calculations if that's helpful

Yea, that would be helpful

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

Let me think what could be going wrong here.

@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.

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

Well, it should certainly be a product right? As the elements of the form (0,...1,...0) lie in the algebra. And for i=1 we have the element XY in the ring, with (XY)^p = a.

My argument for each coordinate ring being a field is as follows: we can write \Sigma c_nm * Y^n X^m as a sum of single variable polynomials \Sigma p_n(X)Y^n, there cant be an cancellations happening between different polynomials due to a different Y^n exponenet, so it reduces to show that for p_n(X)Y^n to satisfy the necessary, we must have p_n(X) a monomial, which is immediate

"so it reduces to show that for p_n(X)Y^n to satisfy the necessary equality condition" *

@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.

Alright, thank you! Ping me if you find out where the error lies

Will do.

@AlexYoucis Just a question for sanity check, the functor of points G, as a functor to sets, should be a sheaf in the fpqc site, right? That means we should at the very least have an equaliser sequence of sets which should give card(G(k))=5, do we have that here?

Aside from being a nice sanity check, I just realized that in the solution we now tried the case i=0 actually gives the ring k. So G(k) is trivial, which is also what I get when I try to compute the equaliser sequence, So maybe the mistake all along was actually with the abstract construsction of G itself?

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

In the construction of $G$ by gluing two groups over an automorphism

What comes before the "Exercise" section in the answer you posted

@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?

Not necessarily that G isn't constructed like that (I really don't have enough expertise in cohomology and stacks to say anything definitevely) maybe just the way it's glued should be different? I dont know. I'm suggesting that it may be good to see wheter the set G(k), as computed by the sheaf condition it must satisfy, has the necessary cardinality of 5.

@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.