If $F\in k[{x}_{0},...,{x}_{n}]$ is homogeneous and $F={x}_{0}^{\alpha}G$ (where ${x}_{0}$ does not divide G ) then $\beta \circ \alpha (F)=G$

rialsv
2022-10-02
Answered

If $F\in k[{x}_{0},...,{x}_{n}]$ is homogeneous and $F={x}_{0}^{\alpha}G$ (where ${x}_{0}$ does not divide G ) then $\beta \circ \alpha (F)=G$

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trutdelamodej0

Answered 2022-10-03
Author has **11** answers

Let's write $G=\sum _{|\alpha |=d(G)}{a}_{\alpha}{x}^{\alpha}$, we can do so as G is homogeneous. Now we have

$\begin{array}{rl}{x}_{0}^{d(G)}G(1,\frac{{x}_{1}}{{x}_{0}},\dots ,\frac{{x}_{n}}{{x}_{0}})& ={x}_{0}^{d(G)}\cdot \sum _{|\alpha |=d(G)}{a}_{\alpha}{1}^{{\alpha}_{0}}\cdot \frac{{x}_{1}^{{\alpha}_{1}}}{{x}_{0}^{{\alpha}_{1}}}\cdots \frac{{x}_{n}^{{\alpha}_{n}}}{{x}_{0}^{{\alpha}_{n}}}\\ \text{}& ={x}_{0}^{d(G)}\cdot \sum _{|\alpha |=d(G)}{a}_{\alpha}{1}^{{\alpha}_{0}}\cdot \frac{{x}_{1}^{{\alpha}_{1}}\cdots {x}_{n}^{{\alpha}_{n}}}{{x}_{0}^{{\alpha}_{1}+\cdots +{\alpha}_{n}}}\\ \text{}& ={x}_{0}^{d(G)}\cdot \sum _{|\alpha |=d(G)}{a}_{\alpha}\frac{{x}_{1}^{{\alpha}_{1}}\cdots {x}_{n}^{{\alpha}_{n}}}{{x}_{0}^{d(G)-{\alpha}_{0}}}\\ \text{}& =\sum _{|\alpha |=d(G)}{a}_{\alpha}{x}_{0}^{{\alpha}_{0}}\cdot {x}_{1}^{{\alpha}_{1}}\cdots {x}_{n}^{{\alpha}_{n}}\\ \text{}& =\sum _{|\alpha |=d(G)}{a}_{\alpha}{x}^{\alpha}\\ \text{}& =G({x}_{0},\dots ,{x}_{n}).\end{array}$

$\begin{array}{rl}{x}_{0}^{d(G)}G(1,\frac{{x}_{1}}{{x}_{0}},\dots ,\frac{{x}_{n}}{{x}_{0}})& ={x}_{0}^{d(G)}\cdot \sum _{|\alpha |=d(G)}{a}_{\alpha}{1}^{{\alpha}_{0}}\cdot \frac{{x}_{1}^{{\alpha}_{1}}}{{x}_{0}^{{\alpha}_{1}}}\cdots \frac{{x}_{n}^{{\alpha}_{n}}}{{x}_{0}^{{\alpha}_{n}}}\\ \text{}& ={x}_{0}^{d(G)}\cdot \sum _{|\alpha |=d(G)}{a}_{\alpha}{1}^{{\alpha}_{0}}\cdot \frac{{x}_{1}^{{\alpha}_{1}}\cdots {x}_{n}^{{\alpha}_{n}}}{{x}_{0}^{{\alpha}_{1}+\cdots +{\alpha}_{n}}}\\ \text{}& ={x}_{0}^{d(G)}\cdot \sum _{|\alpha |=d(G)}{a}_{\alpha}\frac{{x}_{1}^{{\alpha}_{1}}\cdots {x}_{n}^{{\alpha}_{n}}}{{x}_{0}^{d(G)-{\alpha}_{0}}}\\ \text{}& =\sum _{|\alpha |=d(G)}{a}_{\alpha}{x}_{0}^{{\alpha}_{0}}\cdot {x}_{1}^{{\alpha}_{1}}\cdots {x}_{n}^{{\alpha}_{n}}\\ \text{}& =\sum _{|\alpha |=d(G)}{a}_{\alpha}{x}^{\alpha}\\ \text{}& =G({x}_{0},\dots ,{x}_{n}).\end{array}$

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