Question

If$\mathbf{[}\mathit{K}\mathbf{:}\mathit{F}\mathbf{]}\mathbf{}$is finite and u is algebraic over K, prove that $\mathbf{\left[}\mathit{F}\mathbf{\right(}\mathit{u}\mathbf{)}\mathbf{:}\mathit{F}\mathbf{]}$divides $\mathbf{\left[}\mathit{K}\mathbf{\right(}\mathit{u}\mathbf{)}\mathbf{:}\mathit{F}\mathbf{]}$.

Short answer

It is proved that $\left[F\right(u):F]$divides $\left[K\right(u):F]$.

Step-by-step solution

Describe the concept of extension field

Let K an extension field of F. then K is said to be algebraic extension of F if every element of K is algebraic over F and if it is not algebraic then it is transcendental.

Prove that [F(u):F] divides [K(u):F]

Let is algebraic over K and $[K:F]$is finite.

Claim:$\left[F\right(u):F]$divides $\left[K\right(u):F]$.

Also, as K is an algebraic extension of F. Therefore,u is algebraic over F.

Thus,$\left[F\right(u):F]$is finite.

Let$p\left(x\right)$ be the minimal polynomial ofu over F. Then $\left[F\right(u):F]=n=degp\left(x\right)$.

Now, as$[K:F]$is finite. So,$\left[K\right(u):F(u\left)\right]$is finite.

Since,$\left[K\right(u):F(u\left)\right]\le [K:F]$. Therefore,$K\left(u\right)$is algebraic extension of $F\left(u\right)$and $F\left(u\right)$ is algebraic extension of F.

This implies,

$F\subseteq F\left(u\right)\subseteq K\left(u\right)$

This gives $\left[K\right(u):F]=\left[K\right(u):F(u\left)\right]\left[F\right(u):F]$. Thus, the claim follows and$\left[F\right(u):F]$ divides $\left[K\right(u):F]$.

Hence, it is proved that$\left[F\right(u):F]$ divides$\left[K\right(u):F]$.