Let $K$ be a splitting field of $x^3-1$ over $F_{11}$. How many roots has $(x^2-3)(x^3-3)$ in $K$? My solution: If $x^3-1$ splits then $F_{11^n}$ has an element of order $3$. Thus $3|11^n-1$ the smallest such $n$ is 2. Thus $K=F_{11^2}$. Now $x^2-3$ already has a solutions $5,6$. $x^3-3$ also has one solution $9$. Now let $\alpha,\beta$ be the other two roots of $x^3-1$ then $9\alpha,9\beta$ are roots of $x^3-3$. Hence the polynomial splits. Is this correct? If $x^3-3$ has no roots in $F_{11}$ what would be the procedure to find its roots in $F_{11^2}$?
 A: This is a detailed explanation of my comment above.

Let $F$ be a finite field with $q$ elements, and write $F_n/F$ for the unique extension of $F$ of degree $n$. If $P(x) \in F[x]$ is a polynomial of degree $3$, then there are only two possibilities:


*

*If $P$ has a root in $F$, then it factorizes as $P(x) = (x - \alpha)Q(x)$ with $Q(x)\in F[x]$ having degree $2$. Since every polynomial over $F$ of degree $2$ has two roots in $F_2$, we see that all three roots of $P$ are in $F_2$.

*If $P$ has no root in $F$, then it is an irreducible polynomial in $F[x]$ (note that this is only true for polynomials of degree $\leq 3$). It follows that the ideal generated by $P$ is a maximal ideal of $F[x]$, and hence the quotient ring $F[x]/(P)$ is a field, which has degree $3$ over $F$, thus isomorphic to $F_3$. And this field obviously contains a root of $P$, namely the image of $x$.
Therefore $P$ has a root $\alpha$ in $F_3$. But then $\alpha^q, \alpha^{q^2}$ are also roots of $P$, and the three roots are all different, because e.g. $\alpha = \alpha^q$ would imply $\alpha \in F$. Hence all three roots of $P$ are in $F_3$.
This shows that $P$ has no root in $F_2$, because $F_2 \cap F_3 = F$ (intersection taken in some algebraic closure of $F$).
