How to show the isomorphism I'm not sure how to show this
$$\mathbb{Z}_3[X]/(x^3 -x +1)\cong\mathbb{Z}_3[X]/(x^3 -x^2+x +1).$$
Any help or hint would be helpful.
 A: Since the two polynomials are irreducible (no roots, small degree) you simply need to invoke the uniqueness of finite fields of a given order:  Finite field extensions are uniquely characterized by their order. Since both of them are finite fields of order $3^3=27$ they are isomorphic, and in fact are exactly the set of all solutions to $x^{27}-x=0$.

If you're not aware of this, then we can prove it:
Let $F$ be a field with $27$ elements. Then since $F^\times$ is a cyclic group of order $26$, we have, by LaGrange's theorem, that $x^{26}-1$ is satisfied by all non-zero elements of $F$, and so $x^{27}-x$ is satisfied by all elements of $F$.
Hence $F$ is isomorphic to the splitting field for $x^{27}-x$ over $\Bbb F_3$, the field with three elements. Since isomorphism is an equivalence relation, any two fields of this order are isomorphic to one another.
A: First, observe that each of the polynomials $x^3-x+1$ and $x^3-x^2+x+1$ are irreducible over $\mathbb Z_3$. To prove this, we use the following fact.
If $\deg p\leq 3$ and $p$ has no roots, then $p$ is irreducible
Proof:
If $p$ was reducible, there would be non-constant polynomials $f,g$ such that $p=fg$. Then by degree considerations, we have either $\deg f\leq 1$ or $\deg g\leq 1$. Consequently we have found a root of $p$, which is a contradiction. $\square$
Now by plugging in 0,1,2 into the polynomials, we see that neither has a root over $\mathbb Z_3$. Thus each defines a degree 3 extension of $\mathbb Z_3$. Recall that $\mathbb Z_3$ is the field of order 3; therefore each extension defines the field of order 27. Hence they are isomorphic.
A: Absolutely nothing wrong with the other answers. But if you want a concrete isomorphism you can find one (in this case) with the following ad hoc trick.
Let $\beta$ be a zero of the polynomial $p(x)=x^3-x^2+x+1$. Then
$$
(\beta+1)^3-(\beta+1)^2=(\beta^3+1)-(\beta^2+2\beta+1)=p(\beta)-1=-1.
$$
So
$$
(\beta+1)^3-(\beta+1)^2+1=0.
$$
Dividing this equation by $(\beta+1)^3$ gives
$$
1-\frac1{\beta+1}+\frac1{(\beta+1)^3}=0.
$$
This means that $\alpha=1/(\beta+1)$ is a zero of the polynomial $q(x)=x^3-x+1$.
Your former field $\Bbb{Z}_3[x]/\langle q(x)\rangle$ is isomorphic to $\Bbb{Z}_3[\alpha]$ by the standard argument. Similar your second field
$\Bbb{Z}_3[x]/\langle p(x)\rangle$ is isomorphic to $\Bbb{Z}_3[\beta]$. An isomorphism
is thus given by extending $\alpha\mapsto 1/(\beta+1)$.

How did I find this? In characteristic three you cannot make the quadratic term of a cubic vanish of $p(x)$ by a linear substitution. Therefore I wanted to make the linear term vanish instead. Then I "went reciprocal" and instead of looking for $q(x)$ I looked for its reciprocal polynomial $x^3q(1/x)$. This time I was lucky. I could have also tried replacing $\alpha$ or $\beta$ with its negative. I don't know, if it's worth your while to learn all these tricks, but they do make an appearance every now and then.
