Question

Prove that the following condition on a field K are equivalent.

(i)Every nonconstant polynomial in $K\left[x\right]$has a root in K .

(ii)Every nonconstant polynomial in $K\left[x\right]$splits over K .

(iii)Every irreducible polynomial in$K\left[x\right]$has degree 1.

(iv)There is no algebraic extension field of K except K itself.

Short answer

(i)Every nonconstant polynomial in$K\left[x\right]$has a root in K.

(ii)Every nonconstant polynomial in$K\left[x\right]$splits over K .

(iii)Every irreducible polynomial in$K\left[x\right]$has degree 1.

(iv)There is no algebraic extension field of K except K itself.

Step-by-step solution

Definition of irreducible polynomial 

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

Showing that Kx  has a root in  K 

Let$f\in K\left[x\right]$be a non constant polynomial and has a one root in K . Since$f\in K\left[x\right]$has at least one root. Then$f=\left(X-{\alpha }_1\right)g$for g some polynomial g in$K\left[x\right]$.

Ifis a constant then the claim follows. So f splits over K . If g is non constant the assume it has one root and$g=\left(X-{\alpha }_1\right)h$for some h .

Repeat the process inductively. Thus $f\in K\left[x\right]$ be non constant polynomial splits over .

Showing that  Kx has degree 1 

Let $f\in K\left[x\right]$be a non constant polynomial splits over K and $f\in K\left[x\right]$ be an irreducible polynomial thus non constant. By assumption it is a product of linear factors but $f\in K\left[x\right]$ is irreducible polynomial so there can be only one such factors.

Therefore every irreducible polynomial in $K\left[x\right]$ has degree 1 or is linear.

Showing that there is no algebraic extension of K  except K

Let every irreducible polynomial in $K\left[x\right]$ has degree 1 or is linear. 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 f is irreducible and assume is of the form $X-\alpha \in K\left[x\right]$ thus $\alpha \in K\left[x\right]$and$K=E$.

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