Question

If $\mathit{u}$is algebraic over$\mathit{F}$ and is a normal extension of role="math" localid="1658901983256" $\mathit{F}$ prove that $\mathit{K}$ is a splitting field over data-custom-editor="chemistry" $\mathit{F}$ of the minimal polynomial of $\mathit{u}$.

Short answer

It is proved that $K$ is a splititing field over $F$ of the minimal polynomial of $u$.

Step-by-step solution

Definition of the 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 a splitting field

Let$u$is algebraic over$F$andis a normal extension of$F$. Since$u$is algebraic over$F$therefore there exist a polynomial$f\left(x\right)\in F$such that$f\left(u\right)=0$.

As$K=F\left(u\right)$is a normal extension of$F$. By definition of normal extension,

If a root of an irreducible polynomial belongs to nomal extension then all the root of the polynomial are contained in normal extension.

So, let the minimal polynomial of$u$be$p\left(x\right)$. Then$p\left(x\right)$is irreducible in$F\left(x\right)$and one root in$K$.

Therefore all the root are contained in$K$ . Thus $K$is the splititing field over$F$ of the minimal polynomial of $u$.