Question

Let $\mathit{K}$be a splititing of $\mathit{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$ over$\mathit{F}$ if is prime,$\mathit{u}\mathbf{\in }\mathit{K}$ is a root of $\mathit{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$ and $\mathit{u}\mathbf{\notin }\mathit{F}$ show that$\mathit{K}\mathbf{=}\mathit{F}\mathbf{\left(}\mathbf{u}\mathbf{\right)}$ .

Short answer

It is Proved that$K=F\left(u\right)$ .

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=F(u) K=F(u)

Let$K$be a splititing field of$f\left(x\right)$over$F$and$[K:F]=p$here$p$is prime number and$u\in K$is a root of the polynomial$f\left(x\right)$.

As$[K:F\left(u\right)]/[K:F]=p$

Thus$[K:F\left(u\right)]/[K:F]=1orp$

If$[K:F\left(u\right)]=p$then$[K:F]=[K:F\left(u\right)][F\left(u\right):F]$

This implies$[F\left(u\right):F]=1$

This means$\left[F\left(u\right)\right]=\left[F\right]$and$u\in F$

But this is a contradiction as$u\notin F$. Thus this gives$[K:F\left(u\right)]=1$

So, it is proved that $K=F\left(u\right)$.