Question

If K be a algebraic extension field of F such that every polynomial in $F\left(x\right)$ split over K prove that K is an algebraic closure of F .

Short answer

It is proved that K is an algebraic closure of F.

Step-by-step solution

Step-1: Definition of irreducible polynomial

Irreducible polynomial is a non constant polynomial that cannot be factored into the product of two non constant polynomial.

Showing that  K is an algebraic closure of F 

Let K be an algebraic extension field of F such that every polynomial in $F\left(x\right)$ splits over K . It is enough to show that K is algebraic closed that is K has no proper algebraic extension.

Every irreducible polynomial in $K\left(x\right)$has degree 1 or is linear if and only if there is no algebraic extension field of k except itself K .

Suppose on contrary E be an algebraic extension field of K other than K .

Let $\alpha \in E$with minimal polynomial f over K. Then is irreducible and assume is of the form $X-\alpha \in K\left[x\right]$. Thus $\alpha \in K\left[x\right],K=E$

Therefore there is no algebraic extension field of K except itself.

Conversely let $P\left(x\right)$ be an irreducible polynomial over $K\left[x\right]$ and u be the root of $p\left(x\right)$. Therefore $\left[K\left(u\right):K\right]=\mathrm{deg}p\left(x\right)$. Thus $k\left(u\right)/F$ is also algebraic extension.

This implies $K\left(u\right)/F$ is an algebraic extension.

Now let $q\left(x\right)$ be the minimal polynomial of u over F . Therefore $q\left(x\right)\in F\left[x\right]$ and every polynomial in $F\left[x\right]$ splits over k . So, $q\left(x\right)$split over k.

As $F\subseteq K$ so $q\left(x\right)$ can be considered as a polynomial over k and $q\left(u\right)=0$

This implies $P\left(x\right)/q\left(x\right)$ over$K\left[x\right]$ and $q\left(x\right)$ splits completely over $K\left[x\right]$.

Also $p\left(x\right)$ is an irreducible polynomial. So, $p\left(x\right)$ must be of degree 1.

Thus this follows and K is algebraic closure of F .