# X^p+Y^(p+1)+Z^(p+1)/Charakteristik p/Matrix zu Hauptteilen/Beispiel

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${\displaystyle {}F=X^{p}+Y^{p+1}+Z^{p+1}\,}$

in Charakteristik ${\displaystyle {}p}$. Die relevanten Taylor-Ableitungen sind

${\displaystyle {}{\frac {1}{p!}}{\left({\frac {\partial }{\partial X}}\right)}^{p}(F)=1\,,}$
${\displaystyle {}{\left({\frac {\partial }{\partial Y}}\right)}(F)=Y^{p}\,,}$
${\displaystyle {}{\frac {1}{2}}{\left({\frac {\partial }{\partial Y}}\right)}^{2}(F)={\frac {p}{2}}Y^{p-1}=0\,,...,}$
${\displaystyle {}{\frac {1}{(p-1)!}}{\left({\frac {\partial }{\partial Y}}\right)}^{p-1}(F)=0\,,}$
${\displaystyle {}{\frac {1}{p!}}{\left({\frac {\partial }{\partial Y}}\right)}^{p}(F)=Y\,,}$
${\displaystyle {}{\frac {1}{(p+1)!}}{\left({\frac {\partial }{\partial Y}}\right)}^{p+1}(F)=1\,,}$

( ${\displaystyle {}p\geq 3}$)

${\displaystyle {\begin{pmatrix}&1&A&B&C\\1&0&0&0&0\\A&0&0&0&0\\B&Y^{p}&0&0&0\\C&Z^{p}&0&0&0\\A^{2}&0&0&0&0\\AB&0&Y^{p}&0&0\\AC&0&Z^{p}&0&0\\B^{2}&0&0&Y^{p}&0\\BC&0&0&Z^{p}&Y^{p}\\C^{2}&0&0&0&Z^{p}\end{pmatrix}}}$
${\displaystyle {\begin{pmatrix}&1&A&B&C&A^{2}&AB&AC&B^{2}&BC&C^{2}\\1&0&0&0&0&0&0&0&0&0&0\\A&0&0&0&0&0&0&0&0&0&0\\B&Y^{p}&0&0&0&0&0&0&0&0&0\\C&Z^{p}&0&0&0&0&0&0&0&0&0\\A^{2}&0&0&0&0&0&0&0&0&0&0\\AB&0&Y^{p}&0&0&0&0&0&0&0&0\\AC&0&Z^{p}&0&0&0&0&0&0&0&0\\B^{2}&0&0&Y^{p}&0&0&0&0&0&0&0\\BC&0&0&Z^{p}&Y^{p}&0&0&0&0&0&0\\C^{2}&0&0&0&Z^{p}&0&0&0&0&0&0\\A^{3}&0&0&0&0&0&0&0&0&0&0\\A^{2}B&0&0&0&0&Y^{p}&0&0&0&0&0\\A^{2}C&0&0&0&0&Z^{p}&0&0&0&0&0\\AB^{2}&0&0&0&0&0&Y^{p}&0&0&0&0\\ABC&0&0&0&0&0&Z^{p}&Y^{p}&0&0&0\\AC^{2}&0&0&0&0&0&0&Z^{p}&0&0&0\\B^{3}&0&0&0&0&0&0&0&Y^{p}&0&0\\B^{2}C&0&0&0&0&0&0&0&Z^{p}&Y^{p}&0\\BC^{2}&0&0&0&0&0&0&0&0&Z^{p}&Y^{p}\\C^{3}&0&0&0&0&0&0&0&0&0&Z^{p}\end{pmatrix}}.}$