Question

Let $\mathit{F}\mathbf{,}\mathit{E}\mathbf{,}\mathit{K}$ be field such that $\mathit{F}\mathbf{\subseteq }\mathit{E}\mathbf{\subseteq }\mathit{K}$and $\mathit{E}\mathbf{=}\mathit{F}\mathbf{(}{\mathbf{u}}_{\mathbf{1}}\mathbf{,}\mathbf{.}\mathbf{.....}{\mathbf{u}}_{\mathbf{t}}\mathbf{)}$ where the${\mathit{u}}_{\mathbf{t}}$ are some of the root of $\mathit{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{\in }\mathit{F}\mathbf{\left[}\mathbf{x}\mathbf{\right]}$. Prove that $\mathit{K}$ is a splititing field of $\mathit{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$ over$\mathit{F}$ if and only if $\mathit{K}$ is a splititing field of $\mathit{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$ over$\mathit{E}$ .

Short answer

It is proved that $K$ is the splititing field of $f\left(x\right)$ over$F$ if and only if$K$ is a splititing field of $f\left(x\right)$over$E$ .

Step-by-step solution

Definition of splititing field 

A field $K$ is a splititing field of $f\left(x\right)$ if $f\left(x\right)$ completely split into linear factor in $K$.

Showing thatK   is the splititing field off(x)  

Let$F,E,K$are field such that$F\subseteq E\subseteq K$and$E=F(u_1,\mathrm{......}{u}_t)$.

Here the$u_t$are some of the root of$f\left(x\right)\in F\left[x\right]$. Let that$K$is a splititing field of$f\left(x\right)$over$F$. Now to prove that$K$is a splititing field of$f\left(x\right)$over$E$.

Since$K$is a splititing field of$f\left(x\right)$over$F$and$F\subseteq E$that mean that$K$is a splititing field of$f\left(x\right)$over$E$.

Conversely suppose that$K$is splititing field of$f\left(x\right)$over$E$.

This implies that

$K=E(v_1,\mathrm{......}{v}_q)$

Here$v_q$are some of the root of$f\left(x\right)\in F\left[x\right]$. But also

$E=F(u_1,\mathrm{......}{u}_t)$

This gives:

$K=\left(F(u_1,\mathrm{......}{u}_t)(v_1,\mathrm{......}{v}_q)\right)$

This show that$K$is the splititing field of$f\left(x\right)$over$F$if and only if$K$is a splititing field of$f\left(x\right)$over$E$.