# Inverse of $α∈Z_3(α)$ where $α^3+α^2+2=0$ [closed]

Find the inverse of the element in the given field. The field is a finite extension F(α). Express your answer in the form $$a_0 + a_1 α + \cdots + a_{n−1} α^{n−1}$$, where $$a_i ∈ F$$ and $$[F(α):F]=n$$.

$$α \in GF(27) =Z_3(α),\text{ where }α^3 +α^2 +2=0$$.

I'm a bit confused on how to start this problem. I understand the terminology and notation but I don't know how to find the inverse of what is asked. Any help would be great, thank you in advance!

• Are you asking for the multiplicative inverse of $\alpha$? May 7 '19 at 3:08
• Note that elements of $GF(27)$ are $a_0+a_1\alpha+a_2\alpha^2$ with $a_0,a_1,a_2\in GF(3)$; assume that times $\alpha$ is $1$ and solve for $a_0,a_1,$ and $a_2$ May 7 '19 at 3:14

Just note that $$1=-2=\alpha^3+\alpha^2=\alpha(\alpha^2+\alpha)$$ and so the inverse of $$\alpha$$ is $$\alpha^2+\alpha$$.

Elements of $$GF(27)$$ are $$a_0+a_1\alpha+a_2\alpha^2$$ with $$a_0,a_1,a_2\in GF(3)$$.

If $$(a_0+a_1\alpha+a_2\alpha^2)\alpha=1$$

then $$a_0\alpha+a_1\alpha^2+a_2\alpha^3=1$$

so $$a_0\alpha+a_1\alpha^2+a_2(-2-\alpha^2)=1$$

so $$-2a_2+a_0\alpha+(a_1-a_2)\alpha^2=1$$

so $$a_2=\frac1{-2}=1, a_0=0,$$ and $$a_1=a_2=1$$.

Thus the inverse of $$\alpha$$ in $$GF(27)$$ is $$\alpha+\alpha^2$$.

For any element $$x\in \mathrm{GF}(3^3)$$, there exist $$a_i\in \mathbb{Z}_3$$, $$i=0,1,2$$, such that $$x=\sum_{i=0}^{2}a_i\alpha^i$$. Suppose $$y=\sum_{i=0}^{2}b_i\alpha^i$$ such that $$xy=1$$, where $$b_i\in \mathbb{Z}_3$$. Then $$(\sum_{i=0}^{2}a_i\alpha^i)(\sum_{i=0}^{2}b_i\alpha^i)=1+0\alpha+0\alpha^2.$$

Note that $$\alpha^3=2\alpha^2+1$$ and $$\alpha^4=2\alpha^3+\alpha=\alpha^2+\alpha+2$$. You can deuce the degrees of $$\alpha$$ to less than 3. Then compare the coeffiences of $$\alpha^i$$ in both sides of the equation. By solving the systems of $$3$$ linear equations of $$b_i$$, where $$a_i$$ are treated as constants, you can obtain $$y=x^{-1}$$.

• Did you mean $\alpha^\color{red}2+\alpha+2$ ? May 7 '19 at 14:32