Find $(1 + 2α)^{−1}$ in $F_{27}$.

Question

Given : $f=X^3+2X+1 \in \mathbb Z_3[X]$

I have that deg $f$ = $3$ and it is irreducible in $\mathbb Z_3$ as $f$ has no roots in $\mathbb Z_3$:

$f(0)=1, f(1)=1, f(2)=1$ in $\mathbb Z_3$:

My Question: Let $α = X + (f) \in \mathbb Z_3[X]/(f)$

It follows that $f(α) = 0$ and $F_{27}$ (the field with $27$ elements) is given by $F_{27} = \mathbb Z_3[X]/(f) = \mathbb Z_3(α) =\{\lambda_0 + \lambda_1α + \lambda_2α^2 : \lambda_0,\lambda_1,\lambda_2 \in \mathbb Z_3\}$.

Find $(1 + 2α)^{−1}$ in $F_{27}$. Express your answer in the form $\lambda_0 + \lambda_1α + \lambda_2α^2, \lambda_0,\lambda_1,\lambda_2 \in \mathbb Z_3$.

Let $(1 + 2α)^{−1} = \lambda_0 + \lambda_1α + \lambda_2α^2 \in \mathbb Z_3(\alpha)$. Then $1=(\lambda_0 + \lambda_1α + \lambda_2α^2)(1 + 2α)= \lambda_0 + \lambda_1α + \lambda_2α^2 +2\lambda_0α + 2\lambda_1α^2+2\lambda_2α^3$

But we have $α^3+2α+1=0$, hence $α^3=-2α-1=2α+1$ and $α^4=αα^3=α(2α+1)=(2α^2+α)$.

$F_{27}^\times$ is a group under multiplication which has $26$ elements

By Lagrange's Theorem:

$(1+2α)^{26}=1$ => $(1+2α)(1+2α)^{25}=1$

$(1+2α)^{-1}=(1+2α)^{25}$

$α^4α^4α^4α^4α^4α^3α^3=1=α^{26}=(1+2α)^2(2α^2+α)^4$

Hence: $(1+2α)^{-1}=(1+2α)(2α^2+α)^4$

Does this sufficiently Express my answer in the form $\lambda_0 + \lambda_1α + \lambda_2α^2, \lambda_0,\lambda_1,\lambda_2 \in \mathbb Z_3$ or am I making an error with this method?

Or should I multiply the entire expression out and taking the modulus of it I obtain that:

$(1+2α)^{-1}=α^2+2α$, i.e $\lambda_2 = 1$ and $\lambda_1=2$ and $\lambda_0=0$

Answer 1

$\!\bmod{\,\color{#c00}{\alpha^3\!-\alpha+1}}\!:\,\ \dfrac{\color{#c00}1}{1-\alpha}\, =\, \dfrac{\color{#c00}{\alpha-\alpha^3}}{1-\alpha}\, =\, (1+\alpha)\alpha$

Or: $\,1 = \alpha-\alpha^3 = (1\!-\!\alpha)\color{#0a0}{(1\!+\!\alpha)\alpha}\,\Rightarrow\, (1\!-\!\alpha)^{-1} = \color{#0a0}{\alpha(1+\alpha)}\ $ without fractions.

Remark $ $ Generally for coprime polynomials $\,f,g\,$ over a field we can use the extended Euclidean algorithm to obtain the Bezout identity $\,a f + b g = 1\,$ hence $\, g^{-1}\!\equiv b \pmod {\!f}.\ $ The above is simply the special case when the Euclidean algorithm terminates in a single step, i.e. when $\, f\ {\rm mod}\ g = c\,$ is constant, i.e. $\ f = qg + c,\,\ \deg c = 0,\,$ so $\!\bmod f\!:\ qg \equiv -c\,$ so $\, g^{-1} \equiv -q/c.\, $ In particular, the remainder $c$ is always constant when $g$ is linear (as here), so the extended Euclidean algorithm requires only a single step.

Answer 2

The systematic way is to use the extended Euclidean algorithm for $\gcd(1+2X,X^3+2X+1)$: $$ 1 = (4 X^2 - 2 X + 9)(1+2X) -8(X^3+2X+1) $$ So, mod 3, we get $$ (1 + 2\alpha)^{−1} = 4 \alpha^2 - 2 \alpha + 9 = \alpha^2 + \alpha $$