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Hey @AlexYoucis, if you don't mind I have a question: Why does (in general) gluing together two extensions give an extension? After all, in general equalisers don't preserve right exactness (and it seems like in our case something similar is happening with G(k) not surjecting onto Z/pZ(k))

Alright, keep me updated

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.

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

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

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?

@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?

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

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

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

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.

Yea, that would be helpful

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

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

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

Yea, I caught that too, no worries

alright

Alright, I'll try working it out

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

Yea, I think this looks more reasonable

Yea

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

p root of a*

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

like the one from Z/pZ to \mu_p

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

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?

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

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)

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?

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.