Splitting field and subextension Definition: Let $K/F$ be a field extension and let $p(x)\in F[x]$,
we say that $K$ is splitting field of $p$ over $F$ if $p$ splits
in $K$ and $K$ is generated by $p$'s roots; i.e. if $a_{0},...,a_{n}\in K$
are the roots of $p$ then $K=F(a_{0},...a_{n})$.
What I am trying to understand is this: in my lecture notes it is
written that if $K/E$,$E/F$ are field extensions then $K$ is splitting
field of $p$ over $F$ iff $K$ is splitting field of $p$ over $E$.
If I assume $K$ is splitting field of $p$ over $F$ then 
$$\begin{align*}K=F(a_{0,}...,a_{n})\subset E(a_{0,}...,a_{n})\subset K &\implies F(a_{0,}...,a_{n})=E(a_{0,}...,a_{n})\\
&\implies K=E(a_{0,}...,a_{n}).
\end{align*}$$
Can someone please help with the other direction ? help is appreciated! 

This is false. Let $F=\mathbb{Q}$, let $E=\mathbb{Q}(\sqrt{2})$, let $K=\mathbb{Q}(\sqrt{2},\sqrt{3})$, and let $p=x^2-3\in F[x]$. Then the splitting field for $p$ over $E$ is $K$, but the splitting field for $p$ over $F$ is $\mathbb{Q}(\sqrt{3})\subsetneq K$.
Let's say that all fields under discussion live in an algebraically closed field $L$. Letting $M$ be the unique splitting field for $p$ over $F$ inside $L$, then the splitting field for $p$ over $E$ inside $L$ is equal to $M$ if and only if $ME=M$, which is the case if and only if $E\subseteq M$. 
In other words, you'll get the same splitting field for $p$ over $F$ and over $E$ if and only if $E$ were already isomorphic to a subfield of the splitting field for $p$ over $F$. When $E$ does not have that property, there is "extra stuff" in $E$ (for example, $\sqrt{2}\in E$ in the example) that will need to also be contained in the splitting field for $p$ over $E$.
