# Explicit Field Isomorphism of Finite Fields

## Problem 233

(a) Let $f_1(x)$ and $f_2(x)$ be irreducible polynomials over a finite field $\F_p$, where $p$ is a prime number. Suppose that $f_1(x)$ and $f_2(x)$ have the same degrees. Then show that fields $\F_p[x]/(f_1(x))$ and $\F_p[x]/(f_2(x))$ are isomorphic.

(b) Show that the polynomials $x^3-x+1$ and $x^3-x-1$ are both irreducible polynomials over the finite field $\F_3$.

(c) Exhibit an explicit isomorphism between the splitting fields of $x^3-x+1$ and $x^3-x-1$ over $\F_3$.

## Proof.

### (a) Fields $\F_p[x]/(f_1(x))$ and $\F_p[x]/(f_2(x))$ are isomorphic

Let $n$ be the degree of $f_1$ and $f_2$.
Since $f_1$ is irreducible over $\F_p$, the quotient field $\F_p[x]/(f_1(x))$ is the finite field of $p^n$ elements.
Similarly, so is $\F_p[x]/(f_2(x))$.

Since a finite field of $p^n$ elements are unique up to isomorphism, these two quotient fields are isomorphic.

Here, we give an explicit isomorphism. The polynomial $f_1(x)$ splits completely in the field $F_{p^n}\cong \F_p[x]/(f_2(x))$, so let $\theta$ be a root of $f_1(x)$ in $\F_p[x]/(f_2(x))$. (Note that $\theta$ is a polynomial.)
Define a map
$\Phi: \F_p[x] \to \F_p[x]/(f_2(x))$ sending $g(x)\in \F_p[x]$ to $g(\theta)$. The map $\Phi$ is a ring homomorphism.
We want to show that the kernel $\ker(\Phi)=(f_1(x))$.

Since $\Phi(f_1(x))=f_1(\theta)=0$, we have $(f_1(x)) \subset \ker(\Phi)$.

On the other hand, if $g(x)\in \ker(\Phi)$, then we have $g(\theta)=0$.
Since $f_1(x)$ is the minimal polynomial of $\theta$, it follows that $f_1$ divides $g(x)$, and hence $g(x) \in (f_1(x))$.
Therefore we proved $\ker(\Phi)=(f_1(x))$.

By the first isomorphism theorem, we obtain an isomorphism
$\tilde\Phi: \F_p[x]/(f_1(x)) \xrightarrow{\cong} \F_p[x]/(f_2(x)),$ where $\tilde \Phi$ maps $x$ to $\theta$.

### (b) The polynomials $x^3-x+1$ and $x^3-x-1$ are irreducible over $\F_3$

Since these polynomial are of degree $3$, if they are reducible, then it has a root in $\F_3$. Evaluating these polynomials at $x=0,1,2$ shows that they have no roots in $\F_3$. Thus these two polynomial are irreducible over $\F_3$.

### (c) Explicit isomorphism between the splitting fields of $x^3-x+1$ and $x^3-x-1$ over $\F_3$

By part (a), the splitting fields
$\F_3[x]/(x^3-x+1) \text{ and } \F_3[x]/(x^3-x-1)$ are isomorphic. In the proof of part (a), we gave an explicit isomorphism.
That is, if $\theta$ is a root of $x^3-x+1$ in the field $\F_3[x]/(x^3-x-1)$, then the map sending $x\in \F_3[x]/(x^3-x+1)$ to $\theta \in \F_3[x]/(x^3-x-1)$ gives an isomorphism.

So we want to find a root $\theta$ of $f_1(x):=x^3-x+1$.
Let $\theta=a+bx+cx^2\in \F_3[x]/(x^3-x-1)$.

Then we have
\begin{align*}
&f_1(\theta)=f_1(a+bx+cx^2)\\
&=(a+bx+cx^2)^3-(a+bx+cx^2)+1\\
&=a+bx^3+cx^6-(a+bx+cx^2)+1\\
& \text{(Note that $a^3=a$ in $\F_3$ and similarly for $b$ and $c$.)}\\
&=a+b(x+1)+c(x^2+2x+1)-(a+bx+cx^2)+1\\
&\text{(Note that $x^3=x+1$ in $\F_3[x]/(x^3-x-1)$, and thus $x^6=x^2+2x+1$.)}\\
&=2cx+b+c+1\stackrel{\text{set}}{=}0.
\end{align*}

From this we deduce that $c=0$, $b=2$ gives a root $\theta$.
For example, choosing $a=0$, we have a root $\theta=2x$ of $f_1(x)$.
Therefore the explicit isomorphism is
$\Phi:\F_3[x]/(x^3-x+1) \xrightarrow{\cong} \F_3[x]/(x^3-x-1),$ which sends $x$ to $\theta=2x$.

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