# Thread: How to use Euclidian Algorithm to find multiplicative inverse of a polynomial??

1. ## How to use Euclidian Algorithm to find multiplicative inverse of a polynomial??

For example, finding the multiplicative inverse of$\displaystyle [x] in Z_5[x]/<x^2+x+1>$

or even something like

$\displaystyle [a+bx] in R[x]/<x^2+1>$

I cant seem to figure out this concept.

Thank you!

2. Originally Posted by elninio
For example, finding the multiplicative inverse of$\displaystyle [x] in Z_5[x]/<x^2+x+1>$

or even something like

$\displaystyle [a+bx] in R[x]/<x^2+1>$

I cant seem to figure out this concept.

Thank you!

Divide $\displaystyle x^2+1$ by $\displaystyle bx+a$ with residue:

$\displaystyle x^2+1=(bx+a)(dx+c)+r$ , with $\displaystyle r=0\,\,\,or\,\,\,\deg(r)<1\Longrightarrow r\in\mathbb{R}\setminus{0}$.

As $\displaystyle x^2+1$ has no real roots, it can't be $\displaystyle r=0\Longrightarrow 0\ne r\in\mathbb{R}$ , and thus $\displaystyle (bx+a)\left(\frac{d}{r}x+\frac{c}{r}\right)=1\!\!\ !\pmod{x^2+1}\Longrightarrow (bx+a)^{-1}=\frac{d}{r}x+\frac{c}{r}\,\,\,in\,\,\,\mathbb{R }[x]\slash<x^2+1>$

Tonio

3. Tonio, I think, the inverse of bx+a should be -(dx+c)/r.

4. Originally Posted by Shanks
Tonio, I think, the inverse of bx+a should be -(dx+c)/r.

Yes, of course: forgot the sign. Thanx

Tonio

5. Can we not write it in terms of a and b like

$\displaystyle (bx+a)^{-1}=\frac{a}{a^2+b^2}-\frac{b}{a^2+b^2}x$?