MHB The splitting field of f over F is also the splitting field of f also over E

[mathmari]

Hey!

Let $F\subseteq E\subseteq K$ be consecutive field extensions and $f\in F[x]$ be non-constant.
I want to show that if $K$ is the splitting field of $f$ over $F$, then $K$ is the splitting field of $f$ also over $E$.

Since $K$ is the splitting field of $f$ over $F$, we have that $f(x) = c\prod (x-a_i)^{m_i}$, where $m_i$ are non-negative integers and $(x-a_i)\in K[x]$.
Since $F\subseteq E$ we have that $f\in E[x]$. Therefore, $K$ must be also the splitting field of $f$ also over $E$.

Is this correct? (Wondering)

 

[mathmari]

I changed something... Since $K$ is the splitting field of $f$ over $F$, we have that $f(x) = c(x-a_1)\cdots (x-a_n)$, where $c\in F, a_1, \dots a_n\in K$ and $K=F[a_1, \dots , a_n]$.

Since $F\subseteq E$ we have that $f(x) = c\prod (x-a_i)^{m_i}\in E[x]$ and $F[a_1, \dots , a_n]\subseteq E[a_1, \dots , a_n] \Rightarrow K\subseteq E[a_1, \dots , a_n]$.

Since $E\subseteq K$ and $a_i\in K$ we have that $E[a_1, \dots , a_n]\subseteq K$.

Therefore, $K=E[a_1, \dots , a_n]$.

So, $K$ is the splitting field of $f$ also over $E$.
Is this correct? Could I improve something? (Wondering)