Elementary tensorial equations

In this article we give examples of elementary tensorial equations:

Let's\quad solve\quad the\quad tensorial\quad equation\quad (x\otimes 1+1\otimes x)y=0\quad in\quad k[x],\quad k\quad a\quad field.\\ Let\quad y\in k[x]\otimes k[x]\quad be\quad y=\sum _{ i,j=0 }^{ n }{ { a }_{ i,j }{ x }^{ i }\otimes { x }^{ j }.\quad We\quad write\quad the\quad quation\quad in\quad the\quad form: } \\ (x\otimes 1+1\otimes x)\begin{pmatrix} { a }_{ 0,0 }1\otimes 1+{ a }_{ 0,1 }1\otimes x+\cdot\cdot\cdot{ a }_{ 0,n }1\otimes { x }^{ n } \\ { a }_{ 1,0 }x\otimes 1+{ a }_{ 1,1 }x\otimes x+\cdot\cdot\cdot{ a }_{ 1,n }x\otimes { x }^{ n } \\\cdot\cdot\cdot\\ { a }_{ n,0 }{ x }^{ n }\otimes 1+{ a }_{ n,1 }{ x }^{ n }\otimes { x }^{ 1 }+\cdot\cdot\cdot{ { a }_{ n,n } }{ x }^{ n }\otimes { x }^{ n } \end{pmatrix}=0 \quad where\quad the\quad second\quad factor\quad is\quad not\quad a\quad true\quad matrix\quad but\quad a\quad tensorial\quad polynomial\\written\quad in\quad this\quad columned\quad way!\quad Let's\quad begin\quad with\quad n=2:\\ (x\otimes 1+1\otimes x)\begin{pmatrix} { a }_{ 0,0 }1\otimes 1+{ a }_{ 0,1 }1\otimes x+{ a }_{ 0,2 }1\otimes { x }^{ 2 } \\ { a }_{ 1,0 }x\otimes 1+{ a }_{ 1,1 }x\otimes x+{ a }_{ 1,n }x\otimes { x }^{ 2 } \\ { a }_{ 2,0 }{ x }^{ 2 }\otimes 1+{ a }_{ 2,1 }{ x }^{ 2 }\otimes { x }^{ 1 }+{ { a }_{ 2,2 } }{ x }^{ 2 }\otimes { x }^{ 2 } \end{pmatrix}=0\\ \begin{pmatrix} { a }_{ 0,0 }x\otimes 1+{ a }_{ 0,1 }x\otimes x+{ a }_{ 0,2 }x\otimes { x }^{ 2 } \\ { a }_{ 1,0 }{ x }^{ 2 }\otimes 1+{ a }_{ 1,1 }{ x }^{ 2 }\otimes x+{ a }_{ 1,n }{ x }^{ 2 }\otimes { x }^{ 2 } \\ { a }_{ 2,0 }{ x }^{ 3 }\otimes 1+{ a }_{ 2,1 }{ x }^{ 3 }\otimes { x }+{ { a }_{ 2,2 } }{ x }^{ 3 }\otimes { x }^{ 2 } \end{pmatrix}+\begin{pmatrix} { a }_{ 0,0 }1\otimes x+{ a }_{ 0,1 }1\otimes { x }^{ 2 }+{ a }_{ 0,2 }1\otimes { x }^{ 3 } \\ { a }_{ 1,0 }x\otimes x+{ a }_{ 1,1 }x\otimes { x }^{ 2 }+{ a }_{ 1,n }x\otimes { x }^{ 3 } \\ { a }_{ 2,0 }{ x }^{ 2 }\otimes x+{ a }_{ 2,1 }{ x }^{ 2 }\otimes { { x }^{ 2 } }+{ { a }_{ 2,2 } }{ x }^{ 2 }\otimes { x }^{ 3 } \end{pmatrix}=0\\ and\quad we\quad readily\quad obtain\quad { a }_{ i,j }=0\quad for\quad all\quad i\quad and\quad j.\quad It\quad is\quad easy\quad to\quad see\quad that\quad this\\ generalizes\quad to\quad all\quad n.\quad So\quad the\quad only\quad solution\quad in\quad k[x]\quad is\quad o. Case\quad \frac { k[x] }{ { (x }^{ 3 }-1) } :\quad the\quad ``matrixes''\quad are\quad the\quad same\quad of\quad above\quad where\quad we\quad change { x }^{ 3 }\quad with\quad 1:\\ \begin{pmatrix} { a }_{ 0,0 }x\otimes 1+{ a }_{ 0,1 }x\otimes x+{ a }_{ 0,2 }x\otimes { x }^{ 2 } \\ { a }_{ 1,0 }{ x }^{ 2 }\otimes 1+{ a }_{ 1,1 }{ x }^{ 2 }\otimes x+{ a }_{ 1,2 }{ x }^{ 2 }\otimes { x }^{ 2 } \\ { a }_{ 2,0 }1\otimes 1+{ a }_{ 2,1 }1\otimes { x }+{ { a }_{ 2,2 } }1\otimes { x }^{ 2 } \end{pmatrix}+\begin{pmatrix} { a }_{ 0,0 }1\otimes x+{ a }_{ 0,1 }1\otimes { x }^{ 2 }+{ a }_{ 0,2 }1\otimes 1 \\ { a }_{ 1,0 }x\otimes x+{ a }_{ 1,1 }x\otimes { x }^{ 2 }+{ a }_{ 1,2 }x\otimes 1 \\ { a }_{ 2,0 }{ x }^{ 2 }\otimes x+{ a }_{ 2,1 }{ x }^{ 2 }\otimes { { x }^{ 2 } }+{ { a }_{ 2,2 } }{ x }^{ 2 }\otimes 1 \end{pmatrix}=0\\we\quad have\quad the\quad following\quad relations\quad subdivided\quad in\quad ``cycles'':\begin{pmatrix} { a }_{ 0,0 }+ { a }_{ 1,2 } \\ \quad\quad\quad\quad\quad { a }_{ 1,2 }+ { a }_{ 2,1 } \\ \quad\quad { a }_{ 0,0 }+ \quad\quad\quad { a }_{ 2,1 } \end{pmatrix}=\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}\quad cycle\quad 1\\ \begin{pmatrix} { a }_{ 2,2 }+ { a }_{ 1,0 } \\ \quad\quad\quad\quad\quad { a }_{ 1,0 }+ { a }_{ 0,1 } \\ \quad\quad { a }_{ 2,2 }+ \quad\quad\quad { a }_{ 0,1 } \end{pmatrix}=\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}\quad cycle\quad 2\\ \begin{pmatrix} { a }_{ 2,0 }+ { a }_{ 0,2 } \\ \quad\quad\quad\quad\quad { a }_{ 0,2 }+ { a }_{ 1,1 } \\ \quad\quad{ a }_{ 2,0 }+ \quad\quad\quad { a }_{ 1,1 } \end{pmatrix}=\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}\quad cycle\quad 3\\that\quad correspond\quad to\quad the\quad coefficients\quad in\quad diagonals:\\ \begin{pmatrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 3 & 1 & 2 \end{pmatrix}\\Every\quad cycle\quad gives\quad one\quad independent\quad coefficients,\quad so\quad in\quad this\\ case\quad dimker(x\otimes 1+1\otimes x)=3.\\ Case\quad \frac { k[x] }{ { (x }^{ n }-1) } :\quad the\quad pattern\quad is\quad similar.\quad We\quad have\quad n\quad cycles\quad \\ corresponding\quad to\quad the\quad n\quad ``diagonals''\quad each\quad giving\quad one\quad independent\\ solution\quad and\quad so\quad \quad dimker(x\otimes 1+1\otimes x)=n. Case\quad \frac { k[x] }{ { x }^{ 2 } } :\quad (x\otimes 1+1\otimes x)\begin{pmatrix} { a }_{ 0,0 }1\otimes 1+{ a }_{ 0,1 }1\otimes x \\ { a }_{ 1,0 }x\otimes 1+{ a }_{ 1,1 }x\otimes x \end{pmatrix}=0\\ \begin{pmatrix} { a }_{ 0,0 }1\otimes 1+{ a }_{ 0,1 }1\otimes x \\ { a }_{ 1,0 }0\otimes 1+{ a }_{ 1,1 }0\otimes x \end{pmatrix}+\begin{pmatrix} { a }_{ 0,0 }1\otimes 1+{ a }_{ 0,1 }1\otimes 0 \\ { a }_{ 1,0 }x\otimes 1+{ a }_{ 1,1 }x\otimes 0 \end{pmatrix}=0\\ that\quad give:\begin{cases} { a }_{ 0,0 }=0; \\ { a }_{ 0,1 }+{ a }_{ 1,0 }=0; \\ { a }_{ 1,1 }\quad every\quad value; \end{cases}\\ \Rightarrow ker(x\otimes 1+1\otimes x)=\{ a(x\otimes 1-1\otimes x)+bx\otimes x\} \\ \Rightarrow \frac { (\frac { k[x] }{ { x }^{ 2 } } \otimes \frac { k[x] }{ { x }^{ 2 } } ) }{ ker(x\otimes 1+1\otimes x) } = \{ c1\otimes 1+d1\otimes x\}=\{ 1\otimes (c+dx)\} = \frac { k[x] }{ { x }^{ 2 } } .

 

References

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Milne J.S.: Basic theory of affine group schemes
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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