Find inverse element of $1+2\alpha$ in $\mathbb{F}_9$

Question

Let $$\mathbb{F}_9 = \frac{\mathbb{F}_3[x]}{(x^2+1)}$$ and consider $\alpha = \bar{x}$. Compute $(1+2 \alpha)^{-1}$

I think I should use the extended Euclidean algorithm: so I divide $x^2 +1 $ by $(1+2x)$:

$$x^2 + 1 = (1+2x)(2x+2)+2$$

$$(2x+2)(1+2x) + 2(x^2+1) = 1$$

Therefore, considering $\text{mod}(x^2+1) $, I have $$(2x+2)(1+2x) = 1\text{mod}(x^2+1)$$

and so $2x+2 = (1+2x)^{-1}$

Is it okay, or did I misunderstood something?

Answer 1

$(2x+2)(2x+1)=2(x+1)(2x+1)=2(2x^2+3x+1).$ Since we are in $\mathbb{F}_3,$ then $3x=0$ and $4x^2=x^2$ so $(2x+2)(2x+1)=2(2x^2+1)=4x^2+2=x^2+2=(x^2+1)+1,$ reducing modulo $(x^2+1)$ we get $1,$ so yeah, you are correct.

Answer 2

Your answer is correct.

Note that $\alpha^2=-1$, and the conjugate of $\alpha$ is $-\alpha$,

so $\dfrac1{1+2\alpha}=\dfrac1{1+2\alpha}\dfrac{1-2\alpha}{1-2\alpha}=\dfrac{1-2\alpha}{1-4\alpha^2}=\dfrac{1-2\alpha}{1-\alpha^2}=\dfrac{1-2\alpha}2=2^{-1}-\alpha=2+2\alpha$

in $\mathbb F_9$.