# Group scheme Zp, two isomorphic forms

This article provides a direct isomorphism between the two descriptions of ${Z}_{p}$, in the former case considered as an etalè subgroup of ${G}_{a}$, char = p, while in the latter as a finite constant group, giving at the same time canonical idempotents of the second description.

As an etalè subgroup of ${G}_{a}$ it has the following description:${ k[x] }/({ { x }^{ p }-x) }$, $k$ a field, $char(k)=p$, with $\varepsilon (x)=0$ and $\Delta \left( x \right) =x\otimes 1+1\otimes x$ while as a finite constant group it is $k{ e }_{ 0 }\bot k{ e }_{ 1 }\bot k{ e }_{ 2 }\bot \dots \bot k{ e }_{ p-1 }$ where ${ e }_{ i }$ are pairwise orthogonal idempotents such that $\Delta ({ e }_{ k })=\Sigma_ {r=0}^{p-1}{ e }_{ r }\otimes { e }_{ k-r }$ and $\varepsilon ({ e }_{ k }) = \begin{cases}1\quad if\quad k= 0\\0\quad if \quad k\neq 0\end{cases}$

˜
Before, let’s study low values of p:

case ${Z}_{2}$ , char(k)=2:
it is described by $A=k[x]/({ x }^{ 2 }-x)$ with $\varepsilon (x)=0$ and $\Delta (x)=x\otimes 1+1\otimes x$ and we are looking for ${e}_{0}$ and ${e}_{1}$ such that $A=Ae_{ 0 }\bot Ae_{ 1 }$ and ${ e }_{ 0 }^{ 2 }={ e }_{ 0 }\quad$ ,$\quad { e }_{ 1 }^{ 2 }={ e }_{ 1 }\quad$ ,$\quad { e }_{ 0 }+{ e }_{ 1 }=1$.
Then we have to solve $(a+bx)^{ 2 }=a+bx$. Hence ${a}^{2}+2abx+{b}^{2}{x}^{2}={a}^{2}+{b}^{2}{x}$ (because ${x}^{2}-x=0$, and 2=0) $=a+bx$. Hence ${a}^{2}=a$ and ${b}^{2}=b$ that is ${a}^{2}-a=0$ and ${b}^{2}-b=0$. k field implies $a=0$ or 1 and $b=0$ or 1. Their four combinations give $0,1,x,1+x$. Only the last two terms summed give 1 and so they give good candidates. Let’s call them ${ e }_{ 0 }=1+x$ and ${ e }_{ 1 }=x$.
Let’s examine $\Delta({ e }_{ 0 })$:
$\Delta({ e }_{ 0 })=\Delta(1+x)=\Delta(1)+\Delta(x)= 1\otimes 1+ x\otimes 1+ 1\otimes x=({ e }_{ 0 }+{ e }_{ 1 })\otimes ({ e }_{ 0 } + { e }_{ 1 } )+{ e }_{ 1 }\otimes ({ e }_{ 0 } + { e }_{ 1 } )+({ e }_{ 0 } + { e }_{ 1 })\otimes { e }_{ 1 }= { e }_{ 0 }\otimes { e }_{ 0 } + { e }_{ 0 }\otimes { e }_{ 1 }+ { e }_{ 1 }\otimes { e }_{ 0 }+ { e }_{ 1 }\otimes { e }_{ 1 }+{ e }_{ 1 }\otimes { e }_{ 0 }+{ e }_{ 1 }\otimes { e }_{ 1 }+{ e }_{ 0 }\otimes { e }_{ 1 }+{ e }_{ 1 }\otimes { e }_{ 1 }= { e }_{ 0 }\otimes { e }_{ 0 } + { e }_{ 1 }\otimes { e }_{ 1 }$
$\Rightarrow$ the rule of the second description is satisfied!
Let’s examine $\Delta({ e }_{ 1 })$:
$\Delta({ e }_{ 1 })=$$\Delta (x)=x$$\otimes 1+1 \otimes x=$${ e }_{ 1 } \otimes ({ e }_{ 0 } + { e }_{ 1 })+({ e }_{ 0 } + { e }_{ 1 } ) \otimes { e }_{ 1 } =$${ e }_{ 1 } \otimes { e }_{ 0 } + { e }_{ 1 } \otimes { e }_{ 1 }+{ e }_{ 0 } \otimes { e }_{ 1 }+{ e }_{ 1 } \otimes { e }_{ 1 }= { e }_{ 1 } \otimes { e }_{ 0 }+{ e }_{ 0 } \otimes { e }_{ 1 }$
$\Rightarrow$ this rule is satisfied too.

case ${ Z }_{ 3 }$ , char(k)=3:
it is described by $A=k[x]/({x}^{3}-x)$ with $\varepsilon \left( x \right) =0$ and $\Delta \left( x \right) =x\otimes 1+1\otimes x$ and we are looking for ${ e }_{ 0 }$ ,${ e }_{ 1 }$ and ${ e }_{ 2 }$ such that $A=A{ e }_{ 0 }\bot A{ e }_{ 1 }\bot A{ e }_{ 2 }$ where ${ e }_{ 0 }^{ 2 }={ e }_{ 0 },$ ${ e }_{ 1 }^{ 2 }={ e }_{ 1 }$, ${ e }_{ 2 }^{ 2 }={ e }_{ 2 }$, ${ e }_{ 0 }+{ e }_{ 1 }+{ e }_{ 2 }= 1$.
Let’s search for all idempotents:
${(a+bx+c{x}^{2})}^{2}= a+bx+c{x}^{2}$
${a}^{2}+{b}^{2}{x}^{2}+{c}^{2}{x}^{4}+2abx+2ac{x}^{2}+2bcx=a+bx+c{x}^{2}$
(the third term on the left is equal to ${c}^{2}{x}^{2}$ from ${x}^{3}=x$ multiplying both sides by x) so we have
${a}^{2}+{b}^{2}{x}^{2}+{c}^{2}{x}^{2}+2abx+2ac{x}^{2}+2bcx=a+bx+c{x}^{2}$.
Equating coefficients we get the following solutions:
$0,1,{x}^{2}, x-{x}^{2},-x-{x}^{2},1-{x}^{2},1+x+{x}^{2},1-x+{x}^{2}$ that is
$0,1,{x}^{2},-x(x-1),-x(x+1)=-x(x-2),-(x-1)(x+1),1+x+{x}^{2}= 1-2x+{x}^{2}={(x-1)}^{2},{(x+1)}^{2}$
Orthogonality among two of them is assured iff all factors of the minimal relation ${x}^{3}-x=x(x-1)(x-2)=0$ are present in their product! By inspection we have three possibilities:
${e}_{0}=-(x-1)(x-2)=-(x-1)(x+1)=1-{x}^{2}$
${e}_{1}=-x(x-2)=-x(x+1)=-x-{x}^{2}$
${e}_{2}=-x(x-1)=x-{x}^{2}$
Observe that we cannot remove the minus one factor: we haven’t any more idempotence as one can convince himself with a calculation!
Let’s now check comultiplication.
$\Delta {e}_{0} = \Delta (1-{x}^{2})$ =$\Delta 1-\Delta {x}^{2} = 1\otimes 1- {(\Delta x)}^{2} = 1\otimes 1-{(x\otimes 1+1\otimes x)}^{2}$ =
$1\otimes 1-{x}^{2}\otimes 1-2x\otimes x-1 \otimes {x}^{2}=(since -2=1)= 1\otimes 1 - {x}^{2}\otimes 1 +x\otimes x- 1\otimes {x}^{2}$.
And we obtain the same result developing the following expression:${e}_{2}\otimes {e}_{1}+{e}_{0}\otimes {e}_{0}+{e}_{1}\otimes {e}_{2} = (x- {x}^{2})\otimes (-x- {x}^{2})+(1- {x}^{2}) \otimes (1- {x}^{2})+(-x-{x}^{2})\otimes (x-{x}^{2})$

case ${Z}_{5}$, char(k)=5:
here ${x}^{5}-x=x(x-1)(x-2)(x-3)(x-4)=x(x-1)(x-2)(x+1)(x+2)$ and we try with:
${e}_{0}=-(x-1)(x-2)(x-3)(x-4)$
${e}_{1}=-x(x-2)(x-3)(x-4)$
${e}_{2}=-x(x-1)(x-3)(x-4)$
${e}_{3}=-x(x-1)(x-2)(x-4)$
${e}_{4}=-x(x-1)(x-2)(x-3)$
and calculating we see that they work perfectly.

Remarks:
1) we understand that the candidates idempotents are minus the product of all by one factors of ${x}^{p}-x$ ;
2) given an idempotent we can by translation obtain $p-1$ others:
in fact an idempotent is an expression of the form $e = {k}_{0}+{k}_{1}x+$$+{k}_{p-1}{x}^{p-1}$ that satisfies ${ e }^{ 2 }=e$ after subsituting ${x}^{p}= x$. But substituting $x$ with $x+1$ gives the same relations since ${(x+1)}^{p}=(x+1)$ is valid.

In general we have the following:

Theorem: ${e}_{k} = -\cfrac { { x }^{ p }-x }{ x-k }$, $0\le k, are a base of the finite constant group scheme ${Z}_{p}$,that is, they are a vector space base and satisfy the conditions $\Delta ({ e }_{ k })=\Sigma_{r=0}^{p-1} { e }_{ r }\otimes { e }_{ k-r }$ and $\varepsilon ({e}_{k}) = \begin{cases}1\quad if\quad k= 0\\0\quad if \quad k\neq 0\end{cases}$.

Proof:
step 0)
the ${e}_{k}$ are p in number so they are a basis as a vector space iff they are linearly independent. But given a linear combination of them ${A}_{0}{e}_{0}+{ A}_{1}{e}_{1}+$$+{A}_{p-1}{e}_{p-1}= 0$ substituting the value 0 only the first term survives and so ${A}_{0} = 0$, substituting the value 1 only the second term survives and so on…
step 1)
${e}_{p-1}=-x(x-1)(x-2)(x-3)\cdot\cdot (x-(p-2))$ is idempotent:
$[-x(x-1)(x-2)(x-3) \cdot\cdot\cdot {(x-(p-2))]}^{2}-[-x(x-1)(x-2)(x-3)\cdot\cdot\cdot (x-(p-2))]= x(x-1)(x-2)(x-3) \cdot\cdot\cdot (x-(p-2))[x(x-1)(x-2)(x-3)\cdot\cdot\cdot (x-(p-2))+1]$
But the last factor is divisible by $(x-(p-1))$ since substituting in it $(p-1)$ we obtain $(p-1)(p-2)\cdot\cdot\cdot 321 +1 =(p-1)!+1=0$ mod p by Wilson! (here the minus sign was important to have Wilson!). Then the last factor is divisible by $(x-(p-1))$ and the whole expression by $x(x-1)(x-2)(x-3)\cdot\cdot (x-(p-2))(x-(p-1)) ={ x }^{ p }-x=0$.
step 2)
translating, we obtain other $p-1$ idempotents that are all products of all but one factors of ${x}^{p}-x$ changed of sign
step 3)
let’s study $\Delta({ e }_{ 0 })$:
starting with the equality $x{e}_{0} =-( {x}^{p}-x)=0$, by the true definition of ${e}_{0}$, and applying to both sides the $\Delta$ operator we obtain $\Delta (x{e}_{0})=\Delta x\Delta{e}_{0}=0$.
So $\Delta({ e }_{ 0 })$ satisfies the tensorial equation $\left(x\otimes 1+1\otimes x\right)\Delta {e}_{0}=0$.

step 4)
let’s study the solutions of $\left(x\otimes 1+1\otimes x\right)y=0$. Writing $y=\Sigma_{i,j=0}^{i,j=p-1}{a}_{ij}{x}^{i}\otimes {x}^{j}$ , developing the calculations reducing by the relation ${x}^{p}-x = 0$, it is easy to see that there are exactly p independent solutions of the equation.

We show it with $p=5$ and $p = 7$ and then generalize.

$p=5:$

$(x\otimes 1 + 1\otimes x)\begin{pmatrix}{a}_{00}1\otimes 1+{a}_{10}x\otimes 1+{a}_{20}{x}^{2}\otimes 1+{a}_{30}{x}^{3}\otimes 1+{a}_{40}{x}^{4}\otimes 1+\\{a}_{o1}1\otimes x + {a}_{11}x\otimes x + {a}_{21}{x}^{2}\otimes x +{a}_{31}{x}^{3}\otimes x+{a}_{41}{x}^{4}\otimes x+\\{a}_{02}1\otimes {x}^{2}+{a}_{12}x\otimes {x}^{2}+{a}_{22}{x}^{2}\otimes {x}^{2}+{a}_{32}{x}^{3}\otimes {x}^{2}+{a}_{42}{x}^{4}\otimes {x}^{2}+\\{a}_{03}1\otimes{x}^{3}+{a}_{13}x\otimes {x}^{3}+{a}_{23}{x}^{2}\otimes {x}^{3}+{a}_{33}{x}^{3}\otimes{x}^{3}+{a}_{43}{x}^{4}\otimes{x}^{3}+\\{a}_{04}1\otimes {x}^{4}+{a}_{14}x\otimes {x}^{4}+{a}_{24}{x}^{2}\otimes{x}^{4}+{a}_{34}{x}^{3}\otimes {x}^{4}+{a}_{44}{x}^{4}\otimes {x}^{4}\end{pmatrix} = 0;$(this is not a product of matrices but a product of two tensorial polynomials written in this columned way!)

$\begin{pmatrix}{a}_{00}x\otimes 1+{a}_{10}{x}^{2}\otimes 1+{a}_{20}{x}^{3}\otimes 1+{a}_{30}{x}^{4}\otimes 1+{a}_{40}{x}^{5}\otimes 1+\\{a}_{o1}x\otimes x + {a}_{11}{x}^{2}\otimes x + {a}_{21}{x}^{3}\otimes x +{a}_{31}{x}^{4}\otimes x+{a}_{41}{x}^{5}\otimes x+\\{a}_{02}x\otimes {x}^{2}+{a}_{12}{x}^{2}\otimes {x}^{2}+{a}_{22}{x}^{3}\otimes {x}^{2}+{a}_{32}{x}^{4}\otimes {x}^{2}+{a}_{42}{x}^{5}\otimes {x}^{2}+\\{a}_{03}x\otimes{x}^{3}+{a}_{13}{x}^{2}\otimes {x}^{3}+{a}_{23}{x}^{3}\otimes {x}^{3}+{a}_{33}{x}^{4}\otimes{x}^{3}+{a}_{43}{x}^{5}\otimes{x}^{3}+\\{a}_{04}x\otimes {x}^{4}+{a}_{14}{x}^{2}\otimes {x}^{4}+{a}_{24}{x}^{3}\otimes{x}^{4}+{a}_{34}{x}^{4}\otimes {x}^{4}+{a}_{44}{x}^{5}\otimes {x}^{4}\end{pmatrix} +$

$\begin{pmatrix}{a}_{00}1\otimes x+{a}_{10}x\otimes x+{a}_{20}{x}^{2}\otimes x+{a}_{30}{x}^{3}\otimes x+{a}_{40}{x}^{4}\otimes x+\\{a}_{o1}1\otimes {x}^{2} + {a}_{11}x\otimes {x}^{2} + {a}_{21}{x}^{2}\otimes {x}^{2} +{a}_{31}{x}^{3}\otimes {x}^{2}+{a}_{41}{x}^{4}\otimes {x}^{2}+\\{a}_{02}1\otimes {x}^{3}+{a}_{12}x\otimes {x}^{3}+{a}_{22}{x}^{2}\otimes {x}^{3}+{a}_{32}{x}^{3}\otimes {x}^{3}+{a}_{42}{x}^{4}\otimes {x}^{3}+\\{a}_{03}1\otimes{x}^{4}+{a}_{13}x\otimes {x}^{4}+{a}_{23}{x}^{2}\otimes {x}^{4}+{a}_{33}{x}^{3}\otimes{x}^{4}+{a}_{43}{x}^{4}\otimes{x}^{4}+\\{a}_{04}1\otimes {x}^{5}+{a}_{14}x\otimes {x}^{5}+{a}_{24}{x}^{2}\otimes{x}^{5}+{a}_{34}{x}^{3}\otimes {x}^{5}+{a}_{44}{x}^{4}\otimes {x}^{5}\end{pmatrix} = 0$

reducing by ${x}^{5} = x$ and using the linear independence of ${x}^{i}\otimes {x}^{j}$ we obtain the following conditions on coefficients:

$\begin{pmatrix}0&{a}_{40}+{a}_{00}&{a}_{01}&{a}_{02}&{a}_{03}\\{a}_{04}+{a}_{00}&{a}_{41}+{a}_{01}+{a}_{14}+{a}_{10}&{a}_{42}+{a}_{02}+{a}_{11}&{a}_{43}+{a}_{03}+{a}_{12}&{a}_{44}+{a}_{04}+{a}_{13}\\{a}_{10}&{a}_{11}+{a}_{24}+{a}_{20}&{a}_{12}+{a}_{21}&{a}_{13}+{a}_{22}&{a}_{14}+{a}_{23}\\{a}_{20}&{a}_{21}+{a}_{34}+{a}_{30}&{a}_{22}+{a}_{31}&{a}_{23}+{a}_{32}&{a}_{33}+{a}_{24}\\{a}_{30}&{a}_{44}+{a}_{40}+{a}_{31}&{a}_{32}+{a}_{41}&{a}_{33}+{a}_{42}&{a}_{34}+{a}_{43}\end{pmatrix} = \begin{pmatrix}&&&&\\&&&&\\&&0&&\\&&&&\\&&&&\end{pmatrix}$ (these are true matrixes)

these give 25 equations in the 25 unknown ${a}_{ij}$:

we observe at once that ${a}_{10} = {a}_{20} = {a}_{30} = {a}_{01} = {a}_{02}={a}_{03} = 0$ and hence we can suppress them in the middle of the matrix simplyfing it:

$\begin{pmatrix}0&{a}_{40}+{a}_{00}&{a}_{01}&{a}_{02}&{a}_{03}\\{a}_{04}+{a}_{00}&{a}_{41}+{a}_{14}&{a}_{42}+{a}_{11}&{a}_{43}+{a}_{12}&{a}_{44}+{a}_{04}+{a}_{13}\\{a}_{10}&{a}_{11}+{a}_{24}&{a}_{12}+{a}_{21}&{a}_{13}+{a}_{22}&{a}_{14}+{a}_{23}\\{a}_{20}&{a}_{21}+{a}_{34}&{a}_{22}+{a}_{31}&{a}_{23}+{a}_{32}&{a}_{33}+{a}_{24}\\{a}_{30}&{a}_{44}+{a}_{40}+{a}_{31}&{a}_{32}+{a}_{41}&{a}_{33}+{a}_{42}&{a}_{34}+{a}_{43}\end{pmatrix} = \begin{pmatrix}&&&&\\&&&&\\&&0&&\\&&&&\\&&&&\end{pmatrix}$

We notice then that the equations that appear in the following diagonals

$\begin{pmatrix}&0/4&1&2&3\\0/4&1&2&3&0/4\\1&2&3&0/4&1\\2&3&0/4&1&2\\3&0/4&1&2&3\end{pmatrix}$

are self-contained that is the coefficients that appear in them don’t appear elsewhere: we call them cycles. The cycle of numbers $0/4$, of 6 equations, can be written in the following way:

$\begin{pmatrix}{a}_{40}+&{a}_{00}&&&&&\\&{a}_{00}+&{a}_{04}&&&&\\&&{a}_{04}+&{a}_{44}+&{a}_{13}&&\\&&&&{a}_{13}+&{a}_{22}&\\&&&&&{a}_{22}&+{a}_{31}\\{a}_{40}+&&&{a}_{44}&&&+{a}_{31}\end{pmatrix}=\begin{pmatrix}&&&&\\&&&&\\&&0&&\\&&&&\\&&&&\end{pmatrix}$

the last row is a linear combination of the preceeding ones multiplied by $+1,-1,+1,-1,+1$ and so we can cancel it. We obtain:

$\begin{pmatrix}{a}_{40}+&{a}_{00}&&&&&\\&{a}_{00}+&{a}_{04}&&&&\\&&{a}_{04}+&{a}_{44}+&{a}_{13}&&\\&&&&{a}_{13}&{a}_{22}&\\&&&&&{a}_{22}&+{a}_{31}\end{pmatrix}=\begin{pmatrix}&&&&\\&&&&\\&&0&&\\&&&&\\&&&&\end{pmatrix}$

the first slope before the middle three-terms equation means that all coefficients in it are uniquely expressible by ${a}_{04}$ and similarly the coefficients of the second slope are uniquely determined by ${a}_{13}$. We remain with three variables ${a}_{04},{a}_{44},{a}_{13}$ tied by a linear relation: this cycle that we call cycle $0/4$ gives two independent solutions.

The other cycles are simpler and all equal in the form. The cycle $1$ is:

$\begin{pmatrix}{a}_{41}&+{a}_{14}&&&&\\&{a}_{14}&+{a}_{23}&&&\\&&{a}_{23}&+{a}_{32}&&\\&&&{a}_{32}&+{a}_{41}&\\&&&&&{a}_{01}&\\&&&&&&{a}_{10}\end{pmatrix}=\begin{pmatrix}0\\0\\0\\0\\0\\0\end{pmatrix}$

the cycle $2$:

$\begin{pmatrix}{a}_{42}&+{a}_{11}&&&&\\&{a}_{11}&+{a}_{24}&&&\\&&{a}_{24}&+{a}_{33}&&\\&&&{a}_{33}&+{a}_{42}&\\&&&&&{a}_{02}&\\&&&&&&{a}_{20}\end{pmatrix}=\begin{pmatrix}0\\0\\0\\0\\0\\0\end{pmatrix}$

the cycle $3$:

$\begin{pmatrix}{a}_{43}&+{a}_{12}&&&&\\&{a}_{12}&+{a}_{21}&&&\\&&{a}_{21}&+{a}_{34}&&\\&&&{a}_{34}&+{a}_{43}&\\&&&&&{a}_{03}&\\&&&&&&{a}_{30}\end{pmatrix}=\begin{pmatrix}0\\0\\0\\0\\0\\0\end{pmatrix}$

it’s easy to see that each one gives one independent coefficient while all others coefficients are uniquely linearly expressible by it. Then we have four cycles of which one is “double” and gives two independent coefficients while the other three are “simple” cycles all equal in the form and give exactly one solution each. Totally we have five independent solutions.

$p=7:$

the matrix of relations between coefficients is

$\begin{pmatrix}0&{a}_{00}+{a}_{06}&{a}_{01}&{a}_{02}&{a}_{03}&{a}_{04}&{a}_{05}\\{a}_{60}+{a}_{00}&{a}_{61}+{a}_{01}+{a}_{16}+{a}_{10}&{a}_{62}+{a}_{02}+{a}_{11}&{a}_{12}+{a}_{03}+{a}_{63}&{a}_{13}+{a}_{04}+{a}_{64}&{a}_{14}+{a}_{05}+{a}_{65}&{a}_{15}+{a}_{06}+{a}_{66}\\{a}_{10}&{a}_{11}+{a}_{26}+{a}_{20}&{a}_{12}+{a}_{21}&{a}_{13}+{a}_{22}&{a}_{14}+{a}_{23}&{a}_{15}+{a}_{24}&{a}_{16}+{a}_{25}\\{a}_{20}&{a}_{30}+{a}_{36}+{a}_{21}&{a}_{22}+{a}_{31}&{a}_{23}+{a}_{32}&{a}_{33}+{a}_{24}&{a}_{25}+{a}_{34}&{a}_{35}+{a}_{26}\\{a}_{30}&{a}_{40}+{a}_{46}+{a}_{31}&{a}_{32}+{a}_{41}&{a}_{33}+{a}_{42}&{a}_{34}+{a}_{43}&{a}_{44}+{a}_{35}&{a}_{45}+{a}_{36}\\{a}_{40}&{a}_{50}+{a}_{56}+{a}_{41}&{a}_{42}+{a}_{51}&{a}_{43}+{a}_{52}&{a}_{53}+{a}_{44}&{a}_{54}+{a}_{45}&{a}_{55}+{a}_{46}\\{a}_{50}&{a}_{51}+{a}_{60}+{a}_{66}&{a}_{52}+{a}_{61}&{a}_{62}+{a}_{53}&{a}_{63}+{a}_{54}&{a}_{64}+{a}_{55}&{a}_{65}+{a}_{56}\end{pmatrix} = \begin{pmatrix}&&&&\\&&&&\\&&0&&\\&&&&\\&&&&\end{pmatrix}$

the upper and left borders give

${a}_{01}={a}_{02}={a}_{03}={a}_{04}={a}_{05}={a}_{10}={a}_{20}={a}_{30}={a}_{40}={a}_{50}=0$

and suppressing them in the middle of the matrix:

$\begin{pmatrix}&{a}_{00}+{a}_{06}&{a}_{01}&{a}_{02}&{a}_{03}&{a}_{04}&{a}_{05}\\{a}_{60}+{a}_{00}&{a}_{61}+{a}_{16}&{a}_{62}+{a}_{11}&{a}_{12}+{a}_{63}&{a}_{13}+{a}_{64}&{a}_{14}+{a}_{65}&{a}_{15}+{a}_{06}+{a}_{66}\\{a}_{10}&{a}_{11}+{a}_{26}&{a}_{12}+{a}_{21}&{a}_{13}+{a}_{22}&{a}_{14}+{a}_{23}&{a}_{15}+{a}_{24}&{a}_{16}+{a}_{25}\\{a}_{20}&{a}_{36}+{a}_{21}&{a}_{22}+{a}_{31}&{a}_{23}+{a}_{32}&{a}_{33}+{a}_{24}&{a}_{25}+{a}_{34}&{a}_{35}+{a}_{26}\\{a}_{30}&{a}_{46}+{a}_{31}&{a}_{32}+{a}_{41}&{a}_{33}+{a}_{42}&{a}_{34}+{a}_{43}&{a}_{44}+{a}_{35}&{a}_{45}+{a}_{36}\\{a}_{40}&{a}_{56}+{a}_{41}&{a}_{42}+{a}_{51}&{a}_{43}+{a}_{52}&{a}_{53}+{a}_{44}&{a}_{54}+{a}_{45}&{a}_{55}+{a}_{46}\\{a}_{50}&{a}_{51}+{a}_{66}&{a}_{52}+{a}_{61}&{a}_{62}+{a}_{53}&{a}_{63}+{a}_{54}&{a}_{64}+{a}_{55}&{a}_{65}+{a}_{56}\end{pmatrix} = \begin{pmatrix}&&&&\\&&&&\\&&0&&\\&&&&\\&&&&\end{pmatrix}$

the cycle $0/6$, of 8 equations, is

$\begin{pmatrix}{a}_{06}+&{a}_{00}&&&&&&&\\&{a}_{00}+&{a}_{60}&&&&&&\\&&{a}_{60}+&{a}_{66}+&{a}_{51}&&&&\\&&&&{a}_{51}+&{a}_{42}&&&\\&&&&&{a}_{42}+&{a}_{33}&&\\&&&&&&{a}_{33}+&{a}_{24}&\\&&&&&&&{a}_{24}+&{a}_{15}\\{a}_{06}+&&&{a}_{66}+&&&&&{a}_{15}\end{pmatrix}=\begin{pmatrix}&&&&\\&&&&\\&&0&&\\&&&&\\&&&&\end{pmatrix}$

again the last row is linear combination of the others multiplied by $+1,-1,+1,-1,+1,-1,+1$ and suppressing it

$\begin{pmatrix}{a}_{06}+&{a}_{00}&&&&&&&\\&{a}_{00}+&{a}_{60}&&&&&&\\&&{a}_{60}+&{a}_{66}+&{a}_{51}&&&&\\&&&&{a}_{51}+&{a}_{42}&&&\\&&&&&{a}_{42}+&{a}_{33}&&\\&&&&&&{a}_{33}+&{a}_{24}\\&&&&&&&{a}_{24}+&{a}_{15}\end{pmatrix}=\begin{pmatrix}&&&&\\&&&&\\&&0&&\\&&&&\\&&&&\end{pmatrix}$

it differs from the correspondent with $p=5$ only for the length of the second slope

the cycle $1$ is

$\begin{pmatrix}{a}_{16}&+{a}_{61}&&&&&&&\\&{a}_{61}&+{a}_{52}&&&&&&\\&&{a}_{52}&+{a}_{43}&&&&&\\&&&{a}_{43}&+{a}_{34}&&&&\\&&&&{a}_{34}&+{a}_{25}&&&\\{a}_{16}&&&&&+{a}_{25}&&&\\&&&&&&&{a}_{01}&\\&&&&&&&&{a}_{10}\end{pmatrix}=\begin{pmatrix}0\\0\\0\\0\\0\\0\\0\\0\end{pmatrix}$

and differs from the correspondent with $p = 5$ only by length. The same for all other simple cycles that we omit.

Now the general pattern is clear:

writing the relations on coefficients we have a $(p-1)\times(p-1)$ matrix;

its upper and left borders give ${a}_{10}={a}_{20}=$${a}_{p-2,0}={a}_{01}={a}_{02}$${a}_{0,p-2}=0$ and suppressing these in the relations in the middle of the matrix we obtain a reduced one;

these relations split along $(p-1)$ “diagonal cycles”that vary with p only by length:

one “double cycle” $o/(p-1)$, of $(p+1)$ equations, gives two independent solutions:

its last row is linear combination of the others rows multiplied by $+1,-1,+1,-1$$+1,-1,+1$ (this works with every prime p since p is uneven, the case $p=2$ being already treated separately) and can be removed; the two slopes give dependence of all coefficients that don’t appear in the middle 3-term relation from the two extremes of this; this linear relation between three coefficients gives two independent solutions;

$(p-2)$ equal “simple cycles” give one independent solution each;

totally we have p independent solutions

step 5)

now we give p independent solutions of such equation verifying that $\left(x\otimes 1+1\otimes x\right){e}_{k}\otimes {e}_{p-k}$=0, $0\le k :
$\left(x\otimes 1+1\otimes x\right)(-\cfrac{{x}^{p}-x}{x-k}\otimes -\cfrac{{x}^{p}-x}{x-p+k})$ =
$x\cfrac{{x}^{p}-x}{x-k}\otimes \cfrac{{x}^{p}-x}{x-p+k}+\cfrac{{x}^{p}-x}{x-k}\otimes \cfrac{{x}^{p}-x}{x-p+k}x$ =
$((x-k)+k)\cfrac{{x}^{p}-x}{x-k}\otimes \cfrac{{x}^{p}-x}{x-p+k}+\cfrac{{x}^{p}-x}{x-k}\otimes \cfrac{{x}^{p}-x}{x-p+k}(((x-p+k)+(p-k))=$$(x-k)\cfrac{{x}^{p}-x}{x-k}\otimes \cfrac{{x}^{p}-x}{x-p+k}$+$k\cfrac{{x}^{p}-x}{x-k}\otimes \cfrac{{x}^{p}-x}{x-p+k}$+$\cfrac{{x}^{p}-x}{x-k}\otimes \cfrac{{x}^{p}-x}{x-p+k}(x-p+k)$+$\cfrac{{x}^{p}-x}{x-k}\otimes \cfrac{{x}^{p}-x}{x-p+k}(p-k)$=
$({x}^{p}-x)\otimes \cfrac{{x}^{p}-x}{x-p+k}+k\cfrac{{x}^{p}-x}{x-k}\otimes \cfrac{{x}^{p}-x}{x-p+k}+\cfrac{{x}^{p}-x}{x-k}\otimes ({x}^{p}-x)+(p-k)\cfrac{{x}^{p}-x}{x-k}\otimes \cfrac{{x}^{p}-x}{x-p+k}=$
$0\otimes \cfrac{{x}^{p}-x}{x-p+k}+(k+(p-k))\cfrac{{x}^{p}-x}{x-k}\otimes \cfrac{{x}^{p}-x}{x-p+k}+\cfrac{{x}^{p}-x}{x-k}\otimes 0$=
$p\cfrac{{x}^{p}-x}{x-k}\otimes \cfrac{{x}^{p}-x}{x-p+k}$=0 (char = p).

step 6)
we have found exactly p independent solutions of $\left( x\otimes 1+1\otimes x\right)y=0$ that is ${ e }_{ k }\otimes { e }_{ p-k }$ ,$0\le k, and then we can write $\Delta({ e }_{ 0 })$ as a linear combination of them:
$\Delta ({ e }_{ 0 })=\Sigma_{0}^{p-1} {A}_{k}{ e }_{ k }\otimes { e }_{ p-k }$ .
step 7)
we calculate ${ A }_{ k }$:
${e}_{0}=-(x-1)(x-2)\cdot\cdot (x-(p-1))$ and applying the operator $\Delta$ to both sides:
$\Delta {e}_{0}$=-$\left(x\otimes 1+1\otimes x-1\right)$$\left(x\otimes 1+1\otimes x-2\right)\cdot\cdot\cdot(x\otimes 1+1\otimes x-(p-1))$=- ${x}^{p-1}\otimes 1+$ terms which haven’t a pure number as second tensor factor.
On the right side of $\Delta { e }_{ 0 }=\Sigma_{0}^{p-1} { e }_{ k }\otimes { e }_{ p-k }$ we have:
${ A }_{ 0 }(x-1)(x-2)\cdot\cdot(x-(p-1)\otimes(x-1)(x-2)\cdot\cdot(x-(p-1))+x(\cdot\cdot\cdot)\otimes x(\cdot\cdot\cdot)+\cdot\cdot\cdot+x(\cdot\cdot\cdot)\otimes x(\cdot\cdot\cdot)$
the presence of x in these secondary terms forbides a contribution of them to something like (…)$\otimes 1$ and the first term can contribute to something like ${x}^{p-1}\otimes 1$ only with
${A}_{0}(x)(x)$$(x)\otimes (-1)(-2)$$(-(p-1))$ = ${A}_{0}{x}^{p-1}{(-1)}^{p-1}((p-1)!)$. But for Wilson $(p-1)!= -1$ mod p and comparing the coefficients of the two sides we obtain ${A}_{0}{(-1)}^{p-1}(-1)= -1$, that is ${A}_{0}{(-1)}^{p-1}= 1$. If char(p)=2, ${A}_{0}=1,$ if char(p) >2 (p-1) is even and so ${A}_{0}=1$ too.
To proceed with the case k=1, we rework $\Delta {e}_{0}$:
$\prod_{1}^{p-1}{( x\otimes 1+ 1\otimes x-k)}=\prod_{1}^{p-1}{(( x-1)\otimes 1+ 1\otimes (x-1)-k+2)}$ = ${(x-1)}^{p-1}\otimes 1+$…that is again ${(x-1)}^{p-1}\otimes 1$ is obtained exclusively multiplying the initial terms of each parenthesis! On the right side of $\Delta { e }_{ 0 }=\Sigma_{0}^{p-1}{A}_{k} { e }_{ k }\otimes { e }_{ p-k }$ only $\cfrac {{x}^{p}}{x-(p-1)}\otimes \cfrac{ {x}^{p}-x}{x-1}$ is not in the form $(\cdot\cdot)\otimes (x-1)(\cdot\cdot)$. So, with the same reasoning of case $k=0$, we have ${A}_{p-1}{(-1)}^{p-1}((p-1)!) = -1$, hence ${ A }_{ p-1 }= 1$.

Similarly we obtain ${A}_{k}=1$ and finally $\Delta { e }_{ 0 }=\Sigma_{r=0}^{p-1} { e }_{ k }\otimes { e }_{ p-k },$ and, by translation, all other $\Delta {e}_{k}.$
Step 8)

Let’s verify the $\varepsilon$ rule:
$\varepsilon ({e}_{0})= \varepsilon(-(x-1)(x-2)\cdot\cdot\cdot (x-(p-1))) = -(0-1)(0-2)\cdot\cdot\cdot (0-(p-1)) = -{(-1)}^{p-1}((p-1)!) = 1$
$\varepsilon ({e}_{k})$=$\varepsilon (-x(\cdot\cdot )) = (-0(\cdot\cdot ))= 0,$ when $k\neq 0$.
The theorem is established.

References

W.C. Waterhouse: Introduction to affine Group schemes,
Springer Verlag
Demazure, M. Gabriel, P. : Groupes Algèbriques I
(Amsterdam 1970)
Demazure, M. Grothendieck, A., et al.: Séminaire de Geométrie Algébrique,
Schemas en Groupes (SGA 3),
Lecture Notes in Math.
(New York: Springer, 1970)
Borel, A.: Linear Algebraic Groups
(New York: Benjiamin 1969)
Milne J.S.: Basic theory of affine group schemes
http://www.jmilne.org/math/CourseNotes/
Schoof, R.: Introduction to finite group schemes
https://math.dartmouth.edu/

Pink. R: Finite group schemes, ETH Zurich

https://people.math.ethz.ch/~pinkri/ftp/FGS/CompleteNotes.pdf

Vitali Alessio, Masserano Biella Italy , 28/03/2016
alvital88@yahoo.it

Keywords: Isomorphism Algebraic geometry
Finite affine group schemes Zp
Abstract

Group scheme Zp, two isomorphic forms