I want to know why is my idea wrong in theorem

Question

In theorem : Let $~\mathbb{F}~$ be a field and $f(x)$ a nonconstant polynomial of degree $n$ in $\mathbb{F}(x)$. Then there exists a splitting field $\mathbb{K}$ of $f(x)$ over $\mathbb{F}$ s.t. $~[\mathbb{K}:\mathbb{F}] \leqslant n!~$ where $~[\mathbb{K}:\mathbb{F}]~$ is dimension of $\mathbb{K}$ over $\mathbb{F}$.

$\bullet~$ $\textbf{My question:}~$ Why it is $n!$ not $(n+1)$.

$\bullet~$ $\textbf{Idea:}~$ Let $~\mathbb{K}=\mathbb{F}(u_1,u_2,...,u_n)~$ where $~u_i~$ is root of $f(x)$ for all $i$. If the set ${1_\mathbb{F},u_1,u_2,...,u_n}$ is linear independence then $\dim(K)=n+1$, if not $\dim(K)\leqslant n+1$

Answer 1

$F(u_1,\cdots,u_n)$ is generated by $1$ and $u_1,\cdots,u_n$ as a field extension of $F$ (in the in algebraic case, as a $F$-algebra as well), but not as a $F$-vector space, and here you mean dimension as $F$-vector space. If the extension is algebraic, then $F(u_1,\cdots,u_n)$ is generated as a $F$-vector space by the monomials $u_1^{\alpha_1}u_2^{\alpha_2}\cdots u_n^{\alpha_n}$, where $(\alpha_1,\cdots,\alpha_n)\in\Bbb N^n$.

Answer 2

Consider the $F=\Bbb{Q}$ and $f(x)=x^3-2$ and find its splitting field $K$ over $F$. You will get $K=\Bbb{Q}(\sqrt[3]{2},\zeta_3)$, so $[K:F]=6>3+1$. Here you try to get where is the problem in your approach.