Question

Let K be a finite field of characteristic p. Prove that the map $\mathbf{f}\mathbf{:}\mathbf{K}\mathbf{\to }\mathbf{K}$given by $\mathbf{f}\left(a\right)\mathbf{=}{\mathbf{a}}^{\mathbf{p}}$ is an isomorphism. Conclude that every element of K has a pth root in K .

Short answer

$\mathrm{f}\left(\mathrm{a}\right)={\mathrm{a}}^{\mathrm{p}}$is isomorphism.

Step-by-step solution

Definition of isomorphism

A function is said to be isomorphism if it satisfies the homomorphism and bijective properties. That is a function is said to be isomorphism if it is one one onto and also is preserves operations.

Showing that f(a)=ap is isomorphism

Let K be a finite field characteristic. Define a map $f:K\to K$given by $f\left(a\right)=a^p$.

Consider:

$\begin{array}{rcl}\mathrm{f}\left(\mathrm{ab}\right)& =& {\left(\mathrm{ab}\right)}^{\mathrm{p}}\\ & =& {\mathrm{a}}^{\mathrm{p}}{\mathrm{b}}^{\mathrm{p}}\\ & =& \mathrm{f}\left(\mathrm{a}\right)\mathrm{f}\left(\mathrm{b}\right)\end{array}$

And:

$\begin{array}{rcl}\mathrm{f}\left(\mathrm{a}+\mathrm{b}\right)& =& {\left(\mathrm{a}+\mathrm{b}\right)}^{\mathrm{p}}\\ & =& {\mathrm{a}}^{\mathrm{p}}+\left(\begin{array}{c}\mathrm{p}\\ 1\end{array}\right){\mathrm{a}}^{\mathrm{p}-1}{\left(\mathrm{b}\right)}^1+.......+\left(\begin{array}{c}\mathrm{p}\\ \mathrm{r}\end{array}\right){\mathrm{a}}^{\mathrm{p}-\mathrm{r}}{\left(\mathrm{b}\right)}^{\mathrm{r}}\end{array}$

Since each of the middle coefficients $\left(\begin{array}{c}\mathrm{p}\\ \mathrm{r}\end{array}\right)=\frac{\mathrm{p}!}{\mathrm{r}!\left(\mathrm{p}-\mathrm{r}\right)!}$is an integer.

As every term in the denominator is strictly less than the prime p and the factor of p in the numerator does not cancel therefore $\left(\begin{array}{c}p\\ r\end{array}\right)$is divisible by p say $\left(\begin{array}{c}\mathrm{p}\\ \mathrm{r}\end{array}\right)=\mathrm{mp}$. Therefore all the term between a^p and b^p are zero being multiple of p. Thus:

$\begin{array}{rcl}\mathrm{f}\left(\mathrm{a}+\mathrm{b}\right)& =& {\left(\mathrm{a}+\mathrm{b}\right)}^{\mathrm{p}}\\ & =& {\left(\mathrm{a}\right)}^{\mathrm{p}}+{\left(\mathrm{b}\right)}^{\mathrm{p}}\\ & =& \mathrm{f}\left(\mathrm{a}\right)+\mathrm{f}\left(\mathrm{b}\right)\end{array}$

Therefore f is a homomorphism. As every field homomorphism is injective.

This means f is an injective function. Since K is a finite field characteristic p and the function f from K to K is injective. This implies f is onto also.

Now as the function f is bijective and is a homomorphism.

Therefore the claim follows and f is an isomorphism.