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 $P(\alpha) = 0$?
On
No. You have $x^3 -2$ that is degree 3 but the splitting field has degree 6 over $\Bbb{Q}$ (the splitting field is $\Bbb{Q}(\sqrt[3]{2}.e^{2\pi i/3})$.
For your second question, you need $P$ to be monic: The minimal polynomial by definition is the monic polynomial of lowest degree with $\alpha$ as a root, making it irreducible by definition.
On
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.
On
The answer is no, enough examples have been given above.
Basically the point is that often it is not enough to adjoin one root of a polynomial to get the splitting field. In these cases the degree of the extension will not be the degree of the polynomial but will be bigger.
In the cases where you can just adjoin one root your claim is true.
No, it does not hold in general, for example take $P=x^{3}-2\in\mathbb{Q}[x]$ : $[L_{P}:\mathbb{Q}(\sqrt[3]{2})]>1$ since $L_{P}$ have elements that are not in $\mathbb{R}$, but $[\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q}]=3$ hence $[L_{P}:\mathbb{Q}]>\operatorname{deg}(p)=3$ .
For your second question the answer is almost: If $g(\alpha)=0$ then $m_{\alpha,F}(x)\mid g(x)$ but since $g$ is irreducible then you have that $m_{\alpha,F}(x)=cg(x)$ for some constant $c\in F$ (otherwise you would've had a decomposition of $g$ ).
The minimal polynomial is defined so it is monic i.e s.t that the leading coefficient is $1$ so $g$ divided by its leading coefficient is monic and equals to $m_{\alpha,F}(x)$.