Question

Prove that K has exactly one basis over F if and only if$\mathbf{K}\mathbf{=}\mathbf{F}\mathbf{\cong }{\mathbf{Z}}_{\mathbf{2}}$

Short answer

It is proved that K has exactly one basis over F if$\mathrm{K}=\mathrm{F}\cong {\mathrm{Z}}_2$.

Step-by-step solution

The vector space

If the vectors are linearly independent and every vector in vector space is a linear combination of the given set, then every set of elements in vector space V is called a basis.

A linearly independent spanning set is a basis.

To show that K has exactly one basis over F if and only if K=F≅Z2

Assume that the extension of F be K.

Now, to show that K has exactly one basis over F if and only if $\mathrm{K}=\mathrm{F}\cong {\mathrm{Z}}_2$

Proof:

Assume that$\mathrm{K}=\mathrm{F}\cong {\mathrm{Z}}_2$.

Since $\mathrm{K}=\mathrm{F}$.

This shows that $\left[\mathrm{K}:\mathrm{F}\right]=1$.

Thus K has dimension 1 over F.

So, the basis of K contains exactly 1 basis over F.

Conversely, assume that K has exactly one basis over F.

As,$\left[\mathrm{K}:\mathrm{F}\right]=1$

So $\left[\mathrm{K}:\mathrm{F}\right]=1$means that K has dimension one over F

Which further implies that $\mathrm{K}=\mathrm{F}$

Now, let the basis of K over F be $\mathrm{\beta }=\left\{0,{\mathrm{v}}_1,{\mathrm{v}}_2\right\}$

Which implies that $\left\{{\mathrm{v}}_1\right\},\left\{{\mathrm{v}}_2\right\}$are also basis of K over F.

Now, ${\mathrm{Z}}_2=\left\{0,1\right\}$and there are two elements in the basis.

Therefore there are two elements of Z₂ in basis.

Thus, $\mathrm{K}=\mathrm{F}\cong {\mathrm{Z}}_2$

Hence, it is proved that K has exactly one basis over F if$\mathrm{K}=\mathrm{F}\cong {\mathrm{Z}}_2$.