Constructing Isomorphism between finite field Consider $\mathbb{F}_3(\alpha)$ where $\alpha^3 - \alpha +1 = 0$ and $\mathbb{F}_3(\beta)$ where $\beta^3 - \beta^2 +1 =0$.
I know these two fields are isomorphic but I have difficulty buliding an isomorphism between them.
I know I have to determine where $\alpha$ is mapped to under the isomorphism map but I can't figure it out. 
Any help is much appreciated.
 A: One way to do it is just to slug through it:
An isomorphism $\mathbb{F}_3(\alpha)\to\mathbb{F}_3(\beta)$ is completely determined by the image of $\alpha$; and, as you note in comments, the image of $\alpha$ must be of the form $a\beta^2 + b\beta + c$, with $a,b,c\in\mathbb{F}_3$. Moreover, the image of $\alpha$ must satisfy $\alpha^3 = \alpha - 1$; that is, you want to find $a,b,c$ such that
$$(a\beta^2 + b\beta + c)^3 = a\beta^2 + b\beta + c-1.$$
So you can just expand the left hand side, using $\beta^3 = \beta^2-1$, and figure out the coefficients. We have:
$$\begin{align*}
\beta^3 &= \beta^2 - 1\\
\beta^4 &= \beta^3-\beta\\
&= \beta^2 - \beta - 1\\
\beta^6 &= (\beta^3)^2
= \beta^4 - 2\beta^2 + 1\\
&= \beta^2 - \beta - 1 -2\beta^2 + 1\\
&= -\beta^2 - \beta.
\end{align*}$$
And so, since we are in characteristic $3$,
$$\begin{align*}
(a\beta^2 + b\beta + c)^3 &= a^3\beta^6 + b^3\beta^3 + c^3\\
&= a^3(-\beta^2-\beta) + b^3(\beta^2-1) + c^3\\
&= (b^3-a^3)\beta^2 + (-a^3)\beta + c^3-b^3\\
&= a\beta^2 + b\beta + c - 1.
\end{align*}$$
So we need to solve the equations
$$\begin{align*}
b^3-a^3 &= a\\
-a^3 &= b\\
c^3 -b^3&= c-1.
\end{align*}$$
Since $a,b,c\in\mathbb{F}_3$, where $x^3=x$ for all $x$, we get
$$\begin{align*}
b-a &= a\\
-a&=b\\
c-b&= c-1
\end{align*}$$
The first two equations both give $b=-a$; the last equation gives $b=1$. So $a=-1$, $b=1$, and $c$ is free (this gives the three roots of the polynomial). That is, $f(\alpha)$ can be any of $-\beta^2+\beta$, $-\beta^2+\beta+1$, $-\beta^2+\beta-1$. You can verify they all work.
A: From $\beta^3 - \beta^2 + 1 = 0$, we get $1 - \beta^{-1} + \beta^{-3} = 0$, and
since $\alpha^3 - \alpha + 1 = 0$, we can map $\alpha$ to $\beta^{-1}$ to get 
the desired isomorphism.  Of course, from $\beta^3 - \beta^2 + 1 = 0$ we get 
$\beta^2 - \beta + \beta^{-1} = 0$, that is, $\beta^{-1} = -\beta^2 + \beta$.
In other words, we can take $f(\alpha) = -\beta^2 + \beta$ exactly as Arturo
Magidin found.
Edit: Added note We could instead have chosen to map $\alpha$ to the conjugates $\beta^{-3}$ or $\beta^{-3^2}$ of $\beta^{-1}$ where we
have that
$$\beta^{-3} = \beta^{-1} - 1 = -\beta^2 + \beta - 1,$$ and 
$$\beta^{-3^2} = (\beta^{-1} - 1)^3 = \beta^{-3}-1 = -\beta^2 + \beta + 1$$which are the other two possible images of $\alpha$ given in Arturo's answer. 
As in implied in Jyrki Lahtonen's comment, a simpler derivation is 
possible in this case because $\alpha$ and $\beta$ are roots of reciprocal
polynomials.
