Question about minimal polynomial and extension degree I've been studying for my Galois Theory exam and I realised I had been using something I've been unable to find in my course notes. I've been naturally using it since I intuitively assumed it's true, but then I hesitated and decided to ask here if it's true or not.
My question is, is this correct?

Given $F$ a field extension of $L$, let's say $F=L(\alpha)$, then it's
verified that $[F:L]$ equals the degree of the minimal polynomial of
$\alpha$ over $L$.

Is this always true? I assumed from the start this was true without seeing it in my course notes, but my hesitation came from thinking that maybe the minimal polynomials of elements of the general form $t_1+t_2\alpha$ with $t_1,t_2\in L$ may have higher degree, thus giving a higher $[F:L]$.
If this is true in the end, then it would mean that given $F=L(\alpha,\beta)$, verifying that $\beta\notin L(\alpha)$ and $\alpha\notin L(\beta)$, then $[F:L]$ would be the product of the degrees of the minimal polynomials of $\alpha$ and $\beta$, right?
Is all I mentioned correct? If not, what could be a counterexample? Any help will be appreciated, thanks in advance.
 A: Let $\mu_\alpha$ be the minimal polynomial of $\alpha$ over $L$. It is irreducible over $L$, because otherwise one of its proper factors would be a polynomial of smaller degree over $L$ vanishing on $\alpha$.
Notice that $L(\alpha)=L[\alpha]$. Namely, $L[\alpha]$ is already a field, containing $L$ and contained in $L(\alpha)$, so it must coincide with $L(\alpha)$.
Proof: let $p(\alpha)\in L[\alpha]\setminus\{0\}$. As $p(\alpha)\ne 0$, it cannot be divisible by $\mu_\alpha$. Therefore, it is coprime to $\mu_\alpha$ (recall $L[x]$ supports Euclidean algorithm and so we can use our intuition about prime factorization here). Therefore, again by Euclidean algorithm, $1=up+v\mu_\alpha$ for some polynomials $u, v\in L[x]$. Now that means (substitute $x=\alpha$) that $1=u(\alpha)p(\alpha)+v(\alpha)\mu_\alpha(\alpha)=u(\alpha)p(\alpha)$, i.e. $p(\alpha)$ has an inverse $u(\alpha)$ already in $L[\alpha]$.$\blacksquare$
Now, the degree $[L(\alpha):L]=[L[\alpha]:L]$ is the same as the degree of $\mu_\alpha$, because, taking $m=\deg(\mu_\alpha)$, the elements $1, \alpha, \alpha^2,\ldots,\alpha^{m-1}$ are the base of $L[\alpha]$ over $L$ (as a vector space). This is again because of Euclidean division: take arbitrary $p(\alpha)\in L[\alpha]$, divide $p$ by $\mu_\alpha$ (with quotient $q$ and remainder $r$) and notice that $p=q\mu_\alpha+r$, so $p(\alpha)=q(\alpha)\mu_\alpha(\alpha)+r(\alpha)=r(\alpha)=r_0+r_1\alpha+r_2\alpha^2+\ldots+r_{m-1}\alpha^{m-1}$ because $\deg(r)<\deg(\mu_\alpha)=m$.

As for your second question, the logic you are using is incorrect: $[L(\alpha, \beta):L]=[L(\alpha, \beta):L(\alpha)][L(\alpha):L]$ and, for this to be equal to $[L(\beta):L][L(\alpha):L]$ you would need to prove that the minimal polynomial of $\beta$ over $L(\alpha)$ is the same as the minimal polynomial of $\beta$ over $L$. This need not be the case: the same polynomial $\mu_\beta$, which is irreducible over $L$ may suddenly become reducible over $L(\alpha)$, so it's not minimal over $L(\alpha)$ anymore. The fact that $\beta\not\in L(\alpha)$ just helps prove that it won't have a linear factor, but it may e.g. factorize into quadratic factors, or factors of higher degree.
For an example, try this out: $L=\mathbb Q$, $\alpha=\sqrt[4]{2}$, $\beta=e^{i\pi/4}=\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2}$. The minimal polynomial of $\alpha$ over $L$ is $x^4-2$ (irreducible by Eisenstein for prime $p=2$), while the minimal polynomial of $\beta$ is $x^4+1$ (also easily proven to be irreducible). However, the minimal polynomial of $\beta$ over $L(\alpha)$ is $(x-\beta)(x-\overline{\beta})=x^2-\sqrt{2}x+1=x^2-\alpha^2x+1$ and is of degree $2$. Thus, $[L(\alpha, \beta):L]=8$ and not $16$, as you would expect.
