Question

Question:Prove that $\left[K:F\right]$is finite if and only if $\mathit{K}\mathbf{=}\mathit{F}\mathbf{(}{\mathit{u}}_{\mathbf{1}}\mathbf{,}{\mathit{u}}_{\mathbf{2}}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}{\mathit{u}}_{\mathbf{n}}\mathbf{)}$with each role="math" localid="1657950723725" ${\mathit{u}}_{\mathbf{i}}$algebraic over F .

Short answer

$\left[K:F\right]$is finite.

Step-by-step solution

Definition 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 overF and it is not algebraic then it is transcendental.

 Step 2: Showing that. [K:F] is finite.

Let K be an extension field of F.

It is enough to show that $[\mathrm{K}:\mathrm{F}]$ is finite if and only if K is generated by finite number of algebraic elements over K .

Prove the result by induction on mumber of generator of K over F.

Now, for n = 1

$K=F\left(u_1\right)$ and$u_1$ algebraic over F.

Then

$\left[K:F\right]=\left[F(u_1,u_2,....u_n):F\right]<\infty$

Since

$F\subseteq F(u_1,u_2,....u_n)$and$u_{r+1}$ is also algebraic over$F(u_1,u_2,....u_n)$ Therefore$\left[F(u_1,u_2,....u_{r+1})\right]$$:F(u_1,u_2,....u_r)<\infty$

Now by induction hypothesis,

$\left[F(u_1,u_2,....u_r):F\right]<\infty$

Thus

$\left[F(u_1,u_2,....u_r):F\right]=\left[F(u_1,u_2,....u_{r+1}):F(u_1,u_2,....u_r)\right]\times F(u_1,u_2,....u_r):F<\infty$

This implies that cliam is true for n = r + 1. So, it is true for all r

Hence$\left[K:F\right]$ is finite.

Conversely let $\left[K:F\right]$ is finite. Then K/F is algebraic extension that is every element of K is algebraic over F.

If $\left[K:F\right]=n$then there exist $\left\{v_1{v}_2,....v_n\right\}\in K$ such that$\left\{v_1{v}_2,....v_n\right\}$ forms a basis for K over F.

Therefore K is a generated by $\left\{v_1{v}_2,....v_n\right\}$over F.

$K=\left\{v_1{v}_2,....v_n\right\}$This implies that . Also each$v_i$ is algebraic overF as K/F is a algebraic extension.