Given a field $F$ and a polynomial $P \in F[x]$ such that $P$ is irreducible over $F$. Let $L_P$ be the splitting field of $P$ and $F$. Does $\operatorname{dim}_F({L_F}) = \deg(P)$ hold?
If looking for a the minimal polynomial of $\alpha$ in a field $F$ is it sufficient to find a polynomial $P$ which is irreducible over $F$ and $P(\alpha) = 0$?
Best Answer
For the first question the answer is no. The degree of the splitting field is divisible by $n=\deg(P)$ (and it divides $n!$), but it can be greater than $n$. After forming $F'=F[x]/(P)$, the polynomial $P$ has one obvious root (the class of $x$) and maybe others, but it need not split into linear factors over $F'$. To find a splitting field on needs to decompose $P$ over $F'$, and then if any irreducible factor of degree${}>1$ remains one should adjoin a root of that factor, decompose what is left over the larger field, and so on until only linear factors remain. In the "worst" case the splitting field has dimension $n!$ over $F$.
For the second question, yes (if your polynomial is monic as well). All polynomials that annihilate $\alpha$ are multiples of its minimal polynomial, so the only monic irreducible among them is the minimal polynomial itself.