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I know this question is so easy that should be a first year algebra, but It has been a long time since I last studied this argument.

I have to find the inverse of $x+2$ in the ring $\mathbb{Q}[x]/\left<x^6 + x^4 + x^2 + 1\right>$.

I tried to divide $(x^6 + x^4 + x^2 + 1)+1$ for $x + 2$ but I obtained a reminder of $86$. Then I tried to solve $$(x + 2)(ax^5+bx^4+cx^3+dx^2+ex+f)=q(x^6 + x^4 + x^2 + 1)+1$$ and obtained that q must be a complex number. So now I'm wondering if there's a better way to proceed. Can anybody give me an hint? Thanks

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Use Horner's scheme to divide by $x+2$: $$\begin{matrix}\\\\\times -2\end{matrix}\quad\begin{matrix} 1&0&1&0&1&0&1\\ &-2&4&-10&20&-42&84 \\ \hline 1&-2&5&-10&21&-42&\!\!\vert\; 85 \end{matrix}$$ Thus $\;x^6+x^4+x^2+1=(x^5-2x^4+5x^3-10x^2+21x-42)(x+2)+85$, so that \begin{align}-\frac1{85}(x^5-2x^4+5x^3-10x^2+21x-42)(x+2)&=1-(x^6+x^4+x^2+1)\\&\equiv 1\mod (x^6+x^4+x^2+1). \end{align}

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Equate co-efficients of each power of x and you get the following equations:

$a=q$

$2a+b=0$

$2b+c=q$

$2c+d=0$

$2d+e=q$

$2e+f=0$

$2f=q+1$

Then solve the equations to find a,b,c,d,e,f and q.

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  • $\begingroup$ This is what I've done $\endgroup$ – Dac0 Apr 20 '17 at 11:06
  • $\begingroup$ Can also be that I just made an error in calculations $\endgroup$ – Dac0 Apr 20 '17 at 11:06
  • $\begingroup$ Indeed... thanks, was just an error in the calculations $\endgroup$ – Dac0 Apr 20 '17 at 11:07
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You can write $x^6+x^4+x^2+1=(x^5-2x^4+5x^3-10x^2+21x-42)(x+2)+85$

So, from here, $85=(x^6+x^4+x^2+1)-(x^5-2x^4+5x^3-10x^2+21x-42)(x+2)$, i.e.,

$85\equiv -(x^5-2x^4+5x^3-10x^2+21x-42)(x+2) \pmod{x^6+x^4+x^2+1}$

So, $(x+2)^{-1}=-\frac1{85}(x^5-2x^4+5x^3-10x^2+21x-42)$ in the quotient ring.

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