Let $K:F$ be an extension and $\alpha_1,\alpha_2,…,\alpha_{n+1} \in K $ algebraics over $F$.
I have to show that $[F(\alpha_1,\alpha_2,…,\alpha_{n+1}):F]\le deg(m_{\alpha_1}(x)) deg(m_{\alpha_2}(x))…deg(m_{\alpha_{n+1}}(x))$.
As $\alpha_1,\alpha_2,…,\alpha_{n+1} \in K $ are algebraics over $F$, there exist $f_i(x)\in F: f_i(\alpha_i)=0$ .
From this we can say that $m_{\alpha_i}(x)$ divides $f_i(x)$. (I don't know how to use this for the proof)
Also we have $m_{\alpha_i}(\alpha)=0$, where $m$ is monic, irreducible and has the least degree.
Also I know that $F(\alpha_1,\alpha_2,…,\alpha_{n+1})$ is the intersection of all subfields of K that contains F and $\alpha_1,\alpha_2,…,\alpha_{n+1}$.
But I still don't know how to stablish a relation with the minimal polynomial m and the degree of the extension.
Or maybe Would be better to try to prove it by contradiction?
Best Answer
In general if $L$ is a field and $\beta$ is algebraic over $L$ then $[L(\beta):L] = \deg m^L_\beta(x)$ where $m^L_\beta(x)$ is the minimal polynomial of $\beta$ over $L$.
Keeping that in mind, let $F_0 = F$ and $F_k = F_{k-1}(\alpha_k) = F(\alpha_1,\ldots,\alpha_k)$ for all $k = 1,\ldots, n$.
Since the degree of an extension is multiplicative then $$[F_n:F_0] = [F_n:F_{n-1}][F_{n-1}:F_{n-2}]\ldots [F_2:F_1][F_1:F_0] = \prod_{k=1}^n \deg m_{\alpha_k}^{F_{k-1}}(x).$$
Now $m_{\alpha_k}^{F_0}(x)$ (this is what you call $m_{\alpha_k}(x)$) is a polynomial in $F_{k-1}$ that has $\alpha_k$ as a root, to $m_{\alpha_k}^{F_{k-1}} \mid m_{\alpha_k}^{F_0}$ and therefore $\deg m_{\alpha_k}^{F_{k-1}} \leq\deg m_{\alpha_k}^{F_0}$, so
$$[F_n:F_0] = \prod_{k=1}^n \deg m_{\alpha_k}^{F_{k-1}}(x) \leq \prod_{k=1}^n \deg m_{\alpha_k}^{F_{0}}(x).$$
Quick note. Since you're studying Galois Theory you might often find that you work with minimal polynomials of a same element but over different fields. For that the notation $m_\alpha(x)$ is insufficient since it doesn't convey any information about what the field is. I used $m_\alpha^F(x)$ in this answer because it's similar to the notation you are using but I find if a bit ugly, even more so when the field or the element have substripts on their own. I suggest using something like $$\text{Irr}(\alpha,F;x)$$ This is what I used when I studied Galois Theory. It represents the irreducible (or minimal) polynomial of $\alpha$ over $F$ on the variable $x$. Using this notation, the last equation becomes
$$[F_n:F_0] = \prod_{k=1}^n \deg \text{Irr}(\alpha_k,F_{k-1};x) \leq \prod_{k=1}^n \deg \text{Irr}(\alpha_k,F_0;x).$$ Isn't it easier on the eye?