<s>[INST] Find inverse element of $1+2\alpha$ in $\mathbb{F}_9$ 
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?
 [/INST] $(2x+2)(2x+1)=2(x+1)(2x+1)=2(2x^2+3x+1).$
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.
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