Find the multiplicative inverse of $\phantom{\rule{thinmathspace}{0ex}}{x}^{2}+({x}^{3}-x+2)$ in the quotient $\phantom{\rule{thinmathspace}{0ex}}{F}_{3}[x]/({x}^{3}-x+2)$

Zachariah Norris
2022-09-18
Answered

Find the multiplicative inverse of $\phantom{\rule{thinmathspace}{0ex}}{x}^{2}+({x}^{3}-x+2)$ in the quotient $\phantom{\rule{thinmathspace}{0ex}}{F}_{3}[x]/({x}^{3}-x+2)$

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zmikavtmz

Answered 2022-09-19
Author has **2** answers

You wrote you can find f,g such that ... and that's what you need to do. This is just th e(extended) Euclidean algorithm as known from integers, but requiring polynomial divisions. Thius

$\begin{array}{rl}{x}^{3}-x+2& =x\cdot {x}^{2}-(x-2)\\ {x}^{2}& =(x-2)\cdot x+2x\\ x-2& =2x\cdot \frac{1}{2}-2=2x\cdot 2+1\end{array}$

(where only the last step is aware of us working in ${\mathbb{F}}_{3}$). From this with $p(x)={x}^{3}-x+2,q(x)={x}^{2}$ we obtain step by step:

$\begin{array}{rl}x-2& =xq-p\\ 2x& =q-x(xq-p)\\ 1& =(xq-p)-2(q-x(xq-p))\end{array}$

So

$(x+2)\cdot p(x)+(2{x}^{2}+x+1)\cdot q(x)=1$

$\begin{array}{rl}{x}^{3}-x+2& =x\cdot {x}^{2}-(x-2)\\ {x}^{2}& =(x-2)\cdot x+2x\\ x-2& =2x\cdot \frac{1}{2}-2=2x\cdot 2+1\end{array}$

(where only the last step is aware of us working in ${\mathbb{F}}_{3}$). From this with $p(x)={x}^{3}-x+2,q(x)={x}^{2}$ we obtain step by step:

$\begin{array}{rl}x-2& =xq-p\\ 2x& =q-x(xq-p)\\ 1& =(xq-p)-2(q-x(xq-p))\end{array}$

So

$(x+2)\cdot p(x)+(2{x}^{2}+x+1)\cdot q(x)=1$

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