Let $a$ be a root of irreducible polynomial of degree 3 over $\mathbb{Z}_3$. What will be the minimal polynomial of $a^2$?

Question

So I figured out this much that this $a$ lies in some splitting field of $f(x)$.But I have no clue about this $a$.How can I guess the minimal polynomial?

Answer 1

The degree of $K=Z_3[X]/f(x)$ is $3$, $a^2$ generates a subfield $L$ of $K$. We have $[K:L][L:Z_3]=3$, since $a^2$ is not an element of $Z_3$ (otherwise the degree of $a$ would have been inferior to $2$) we deduce that the degree of $[L:Z_3]=3$.

Write $f(x)=u+vx+wx^2+x^3$, we have $f(a)=0$ implies that $u+wa^2=-(va+a^3)$. We deduce that $(u+wa^2)^2=(va+a^3)^2$ and $u^2+w^2a^4+2uwa^2=v^2a^2+2va^4+a^6$.

Let $g(x)=u^2+w^2x^2+2uwx-vx-2vx^2-x^3$; $g(a^2)=0$, since the degree of $g$ is $3$, $-g$ is the minimal polynomial of $a^2$.

Answer 2

Write the polynomial in the form $$u+va^2=a(r+sa^2)$$ for suitable coefficients $u,v,r,s.$

Square both sides of this equation and you will have a polynomial satisfied by $a^2$. Finally check whether or not this new polynomial factorises.