I am studying Galois theory from Malik,Mordeson,Sen.There is a theorem which is as follows:
$K/F$ be a finite extension and $\alpha\in K$,then there exists a polynomial $g(x)$ and a field $E\supset K$ such that $g(\alpha)=0$ and $E$ is the splitting field of $g(x)$.
I always try to produce my own proof.So,I thought as follows:
$K/F$ is finite,hence algebraic,so $\alpha\in K$ is algebraic element over $F$.Take its minimal polynomial and call it $g(x)$ and take $E$ to be a splitting field of $g(x)$ .
But I think the problem with this argument is that it does not guarantee that $E\supset K$.That is why I think the book has done in this way.They have chosen a basis $\{v_1,...,v_n\}$ and taken $g(x)$ to be the product of the minimal polynomials of $\alpha,v_1,...,v_n$ and taken $E$ to be the splitting field of $g(x)$.Then we have $E\supset F(\alpha,v_1,...,v_n)\supset K$.
Am I logically correct?
The problem with your method is that let $K=\Bbb{Q}(\sqrt{3},i)$ and let $\alpha=\sqrt{3}$. Then the splitting field of $m_{\alpha}(x)=x^{2}-3$ is $\Bbb{Q}(\sqrt{3})\subsetneq\Bbb{Q}(\sqrt{3},i)$ .
Whereas if you choose $g(x)=m_{\sqrt{3}}(x)\cdot m_{i}(x)=(x^{2}-3)(x^{2}+1)$ then you get the splitting field to be $\Bbb{Q}(\sqrt{3},i)$
This is precisely what the book is trying to do.
as $K$ is a finite extension let $[K:F]<\infty$ and let $\{v_{1},...,v_{n}\}$ be an $F$ basis for $K$.
Then $K=F(v_{1},...,v_{n})$ and hence if $\displaystyle g(x)=(\prod_{i=1}^{n}m_{v_{i}}(x))\cdot m_{\alpha}(x)$ . Then the splitting field of $g(x)$ contains $v_{1},...v_{n}$ and hence $K\subset E$ and $g(\alpha)=0$
Here $m_{\beta}(x)\in F[x]$ denotes the minimal polynomial of $\beta$ for $\beta$ in some algebraic closure of $F$.
Bonus:- You can actually get a Galois Extension $E$ if you remove the repeated factors in $g(x)$ . In that case $g(x)$ becomes separable and hence $E$ is splitting field of a separable polynomial $g(x)$ and hence Galois.