The inverse of $2x^2+2$ in $\mathbb{Z}_3[x]/( x^3+2x^2+2)$

Question

What is the independent coefficient in the inverse of $2x^2+2$ in $\mathbb{Z}_3[x]/(x^3+2x^2+2)$ ?

I have been calculating some combinations, but I don't know how I can calculate the inverse.

Answer 1

Hint $\, {\rm mod}\ 3\!:\ g = x^3\!+\!2x\!+\!2 \equiv (x\!+\!1)(x^2\!+\!x\!-\!1).\, $ Compute $\, f = \dfrac{1}{2x^2\!+\!2}\,$ by CRT:

${\rm mod}\,\ \color{#0a0}{x^2\!+\!x\!-\!1}\!:\ \color{brown}{x^2}\equiv 1\!-\!x\,\Rightarrow\,f = \dfrac{1}{2\color{brown}{x^2}\!+\!2} \equiv \dfrac{1}{x+1} \equiv \color{#c00}x,\ $ by $\ x(x\!+\!1) \equiv 1$

${\rm mod}\,\ x\!+\!1\!:\ x\equiv -1\,\Rightarrow\,1\equiv f\equiv \color{#c00}x+(\color{#0a0}{x^2\!+\!x\!-\!1})c \equiv -1 -c\,\,\Rightarrow\, \color{#c0f}{c \equiv 1}$

Therefore, we conclude that $\ f \equiv x+(x^2\!+\!x\!-\!1)\color{#c0f}{(1)}\, \pmod{g} $

Answer 2

It must take the form $ax^2+bx+c$.
Multiply this by $2x^2+2$, and you get a degree-four polynomial.
Since $x^3+2x+2=0$, it follows that $x^3=x+1$ and so $x^4=x(x^3)=x^2+x$.
So your degree-four polynomial becomes a quadratic. This quadratic equals 1, or $0x^2+0x+1$. Equate the three coefficients, and get three linear equations in $a,b,c$.

Answer 3

$$ (ax^2 + bx + c)(2x^2 + 2) + (dx + e)(x^3 + 2x + 2) = 1\\ (2a + d)x^4 + (2b + e)x^3 + (2a + 2c + 2d)x^2 + (2b + 2d + 2e)x + (2c + 2e) = 1\\ $$

$$\left[ \begin{array}{ccccc} 2 & 0 & 0 & 1 & 0 \\ 0 & 2 & 0 & 0 & 1 \\ 2 & 0 & 2 & 2 & 0 \\ 0 & 2 & 0 & 2 & 2 \\ 0 & 0 & 2 & 0 & 2 \\ \end{array} \right] \left[ \begin{array}{c} a \\ b \\ c \\ d \\ e \\ \end{array} \right] = \left[ \begin{array}{c} 0 \\ 0 \\ 0 \\ 0 \\ 1 \\ \end{array} \right] $$

Solve for $a$, $b$, and $c$ and you find the inverse.

Answer 4

Set $f(x) = 2x^2+2$, $g(x) = x^3+2x^2+2$. Since $f(x),g(x)$ are coprime, find

$$p(x)f(x)+q(x)g(x) = 1$$ using Euclidean division.

Then $p(x)$ is the inverse of $f(x)$ in $F[x]/\langle g(x)\rangle$.

Answer 5

This is in no way a standard solution, but I just want to include it for fun.

Note that $x^3=-(2x^2+2)$, we will find $x^{-3}$. Now $$x^3-x^2=-2=1$$ implies $$x(x^2-x)=x^2(x-1)=1.$$ Thus $$x^{-1}=x(x-1)\text{ and }x^{-2}=x-1.$$ Multiplying the two we get $$x^{-3} = x(x-1)^2=x(x^2+x+1)=x^3+x^2+x.$$ Using $x^3=x^2+1$ on the RHS, we see that $$x^{-3}=2x^2+x+1.$$ Thus $$(2x^2+2)^{-1} = -x^{-3} = -2x^2-x-1 = x^2+2x+2.$$