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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!

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    $\begingroup$ Are you asking for the multiplicative inverse of $\alpha$? $\endgroup$ May 7 '19 at 3:08
  • $\begingroup$ 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 $a_2$ $\endgroup$ May 7 '19 at 3:14
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Just note that $1=-2=\alpha^3+\alpha^2=\alpha(\alpha^2+\alpha)$ and so the inverse of $\alpha$ is $\alpha^2+\alpha$.

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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 $\alpha+\alpha^2$.

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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 $y=x^{-1}$.

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

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