find minimal polynomial in ${GF(3^2)}$

Question

is this solution right ... when ${\alpha =g }$ and ${ \alpha = g^2 }$

Answer 1

Here is a computer check, using sage:

sage: R.<x> = PolynomialRing( GF(3) )
sage: R
Univariate Polynomial Ring in x over Finite Field of size 3
sage: F.<g> = GF( 3^2, modulus=x^2+2*x+2 )
sage: g.minpoly()
x^2 + 2*x + 2
sage: F
Finite Field in g of size 3^2
sage: for k in [1..8]:
....:     print( "g^%s = %s has minimal polynomial %s" 
....:            % ( k, g^k, (g^k).minpoly() ) )
....:     
g^1 = g has minimal polynomial x^2 + 2*x + 2
g^2 = g + 1 has minimal polynomial x^2 + 1
g^3 = 2*g + 1 has minimal polynomial x^2 + 2*x + 2
g^4 = 2 has minimal polynomial x + 1
g^5 = 2*g has minimal polynomial x^2 + x + 2
g^6 = 2*g + 2 has minimal polynomial x^2 + 1
g^7 = g + 2 has minimal polynomial x^2 + x + 2
g^8 = 1 has minimal polynomial x + 2
sage: 

I could not understand some lines from the picture. (For instance, $g^4=2$, the minimal polynomial is $(x-2)=x+1$...)

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Later edit: Let us get all the minimal polynomials using explicit computations. I will denote by $\Phi$ the Frobenius morphism $\Phi(h)=h^3$, $h\in F:=\Bbb F_9=\Bbb F_3[g]$.

• $g$ has minimal polynomial $f(x)=(x^2+2x+2)$, this is give. The two roots are $g$ and $\Phi g=g^3=g(g^2)=g(g+1)=g^2+g=(g+1)+g=2g+1$. Vieta check: the sum of $g,2g+1$ is $1$, and the coefficient of $x$ in $(x^2+2x+2)$ is $2=-1$, the product of $g,2g+1$ is $g(2g+1)=2g^2+g=-(g+1)+g=-1=2$.

• $g^2=g+1$, so the polynomial $f(x-1)$ annihilates it. (Here $f$ is the minimal polynomial of $g$ above.) We compute $f(x-1)= (x-1)^2+2(x-1)+2=(x^2-2x+1)+(2x-2)+2 =x^2+1$. Alternatively, we can compute explicitly $(x-g^2)(x-\Phi g^2)=(x-g^2)(x-g^6))=x^2 -(g^2+g^6)x+g^8$. Now $g$ is a generator of the unit group, so it has multiplicative order $8$, so $g^8=1$, and $g^4=-1$. This gives $g^2+g^6=g^2(1+g^4)=g^2(1+(-1))=0$.

• $g^3$ is conjugated to $g$, same minimal polynomial as computed for $g$.

• $g^5=2g$, so $f(x/2)=f(x/(-1))=f(-x)=x^2-2x+2=x^2+x+2$ is the minimal polynomial.

• $g^6$ is conjugated to $g^2$, same minimal polynomial as computed for $g^2$.

• $g^7$ is either $1/g$, so we take the reciprocal of $f$, which is $1+2x+2x^2$, adjust it to be monic, get $x^2+x+2$ as minimal polynomial for it. Or we use $g^7=g+2=g-1$, so $f(x+1)$ is this minimal polynomial, so we compute $(x+1)^2 + 2(x+1)+2$. Or we use the fact that it is conjugated to $g^5$, because $\Phi g^5=g^{15}=g^7\cdot g^8=g^7$, so the minimal polynomial coincides with the one of $g^5$.