Question

If E and F are subfield of a finite field K and E is isomorphic to F prove that $\mathit{E}\mathbf{=}\mathit{F}$.

Short answer

It is proved that $E=F$.

Step-by-step solution

Definition of splititing subfield

A splitting field of a polynomial with coefficient in a smallest field extension of that field over which the polynomial split into linear factor.

Showing that E=F

Let E and F are subfield of a finite field K . Also E is isomorphic to F that is $E\equiv F$. That is both the subfield E and F of a finite field K are equal.

Since E is isomorphic to F so isomorphic implies that the order of both the subfield are equal. This mean$\left|E\left|=\right|F\right|$.

Both have order $p^n$for some prime p. This implies:

$$\begin{gathered} \left|E\left|=\right|F\right| \\ =p^n \end{gathered}$$

Then by the given theorem:

Let Kbe an extension field of $Z_p$ and n a positive integer. Then K has order $p^n$ if and only if K is a splititing field of $x^{p^n}-x$over $Z_P$.

Each of E and F consist of the root of the polynomial of $x^{p^n}-x$in K. Therefore,

$E=Z_P\left(u_1,u_2,.....u_i\right)=F$

Here $u_i$is all the root of the polynomial of $x^{p^n}-x$ in K . Therefore the claim follows $E=F$.