Question

If K is an extension field of F such that $\left[K:F\right]\mathbf{=}\mathbf{2}$, prove taht K is normal.

Short answer

It is proved that K is normal.

Step-by-step solution

Irreducible polynomial:

A non-constant polynomial that cannot be factored into a product of two non-constant polynomials is irreducible.

The coefficients are consideeed to express the property of irreducibility.

To show that K is normal such that [K:F]=2

Assume that $\left[K:F\right]=2$such that K is an extension field of F.

Now, to show that K is normal.

Proof:

Since, $\left[\mathrm{K}:\mathrm{F}\right]=2$therefore $\mathrm{K}:\mathrm{F}$is algebric and for u of degree $2K=F\left(u\right)$

Now, assume that$\mathrm{f}\in \mathrm{F}\left(\mathrm{x}\right)$ be an irreducible polynomial of u.

So, $\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^2+{\mathrm{a}}_1\mathrm{x}+{\mathrm{a}}_0$for ${\mathrm{a}}_{\mathrm{i}}\in \mathrm{F}$

And also, forr some $v\in K$,

$\mathrm{f}\left(\mathrm{x}\right)=\left(\mathrm{x}-\mathrm{u}\right)\left(\mathrm{x}-\mathrm{v}\right)$

So, u and v belongs to F, that is, $\mathrm{u},\mathrm{v}\in \mathrm{F}$

Which gives, $\mathrm{v}=\frac{{\mathrm{a}}_0}{\mathrm{u}}\in \mathrm{F}$

Hence, f splits in K.

Thus, K is normal.