Question

If $\left[K:F\right]\mathbf{=}\mathit{p}$ with prime prove that there is no field E such that role="math" localid="1657879959232" $\mathit{F}\underset{\mathbf{\ne }}{\mathbf{⊑}}\mathit{E}\underset{\mathbf{\ne }}{\mathbf{⊑}}\mathit{K}$.

Short answer

$F\underset{\ne }{⊑}E\underset{\ne }{⊑}K$

Step-by-step solution

Definition of field.

A field is any set of element that satisfy the field axioms for both addition and multiplication and is commutative division algebra.

Step 2:Showing that F⊑≠E⊑≠K

Let $\left[K:F\right]=p$where it is given that p is prime.

So, suppose there exist a field E and $F\underset{\ne }{⊑}E\underset{\ne }{⊑}K$

Then by given theorem

Let F,K and L be field with$F\underset{}{⊑}E\underset{}{⊑}K\mathrm{If}\left[K:F\right],\left[L:K\right]$. If L are finite then L is finite dimensional extension of F and $\left[L:K\right]=\left[L:K\right],\left[K:F\right]$. Therefore,

$\left[K:F\right]=\left[K:E\right]\left[E:F\right]=p$

This implies

$\left[K:F\right]=1or\left[E:F\right]=1$

This means

Either $K=EorE=F$

Which is a contradicition. Hence there is no E such thatlocalid="1657880675902" $F\underset{\ne }{⊑}E\underset{\ne }{⊑}K$.