Question

Let$\mathit{K}$ be a splititing field of $\mathit{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$over $\mathit{F}$. If$\mathit{E}$ is a field such that $\mathit{F}\mathbf{\subseteq }\mathit{E}\mathbf{\subseteq }\mathit{K}$ show that$\mathit{K}$ is a splititing field of $\mathit{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$overlocalid="1658901166243" $\mathit{E}$.

Short answer

$K$is splitting field of $f\left(x\right)$over $E$.

Step-by-step solution

Definition of splitting field.

A splitting field of a polynomial in a field is a smallest field extension of that field over which the polynomial splitting or split into linear factor.

Showing that K is splitting field f(x)

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

$K$is a splititing field of$f\left(x\right)$implies that$K$is a field contains all the roots of the polynomial$f\left(x\right)$over$F$.

That is$f\left(u\right)=0$for all$u\in E$. Since$K$is a splititing field of the polynomial$f\left(x\right)$over$F$. As$F$is contained in$E$that is$F\subseteq E$.

Therefore the polynomial$f\left(x\right)$over$F$serve as the same polynomial over$E$also.$E$is field such that both$E$and$F$are contained in$K$.

Therefore$K$contain all roots of the polynomial$f\left(x\right)$over$E$. This implies$f\left(u\right)=0$for all$u\in E$. So,

$F\subseteq E\subseteq K$

Therefore$K$can be considered as a splititing field over$E$also.

Thus $K$ is a splititing field of the polynomial $f\left(x\right)$ over$E$ .