Ring isomorphism (polynomials in one variable) Find an explicit isomorphism of the map $h: \Bbb F_5[x]/(x^2+x+2) \to \Bbb F_5[x]/(x^2+4x+2)$
I take the question to mean showing the 2 rings are isomorphic by finding an explicit mapping of each element. 
I think $x^2+x+2$ and $x^2+4x+2$ are irreducible over $\Bbb F_5$, so these are both fields.
Also $x^2=4x+3$ in the first case and $x^2=x+3$ in the second.
But beyond these basic facts I'm not sure how I should proceed. 
 A: Think about quotients by irreducible polynomials as essentially adjoining an imaginary root, like how you get $\mathbb C$ from $\mathbb R$ by "pretending" there's an $i$ such that $i^2=-1$. In more precise language, $i$ is simply the name we give to $(x^2+1)$, the element of the quotient $\mathbb R[X]/(x^2+1)$.
So in the first structure we're working in $F_5$ but adjoint $i$ satisfying $i^2+i=-2$. All elements of the resulting structure are going to have a unique expression of the form $a+bi$ where $a,b\in F_5$, as you should be able to show.
Now, in the second structure we're adjoining a $j$ satisfying $j^2+4j=-2$. Now imagine that in the first structure, you can find an element $x+yi$ satisfying that relation. Then you could write all elements of the first structure as $a+b(x+yi)$, and in fact it's easy to show that this expression would be unique. But now the map
$$a+bj\to a+b(x+yi)$$
From the second structure to the first one is an isomorphism, because $x+yi$ "acts exactly like" $j$ does, so all calculations involving expressions of the form $a+b(x+yi)$ will go exactly the same way as the equivalent calculations involving $j$.
A: Hint $\ $ They are $\,(-x+2)^2-2\,$ and $\,(X+2)^2-2\ $ in $\,\Bbb F_5[x]$
A: Note that $\Bbb F_5[x]$ is a PID and that the ideals $I=\langle x^2+x+2\rangle$ and $J=\langle x^2+4\,x+2\rangle$ are prime ideals. It follows that $I$ and $J$ are maximal in $\Bbb F_5[x]$ so that $\Bbb F_5[x]/I$ and $\Bbb F_5[x]/J$ are fields. How many elements do $\Bbb F_5[x]/I$ and $\Bbb F_5[x]/J$ have? What do we know about fields with finitely many elements?
Followup: The fact that $\Bbb F_5[x]$ is a PID and that the ideals $I=\langle x^2+x+2\rangle$ and $J=\langle x^2+4\,x+2\rangle$ are prime ideals implies that $\Bbb F_5[x]/I$ and $\Bbb F_5[x]/J$ are fields. In particular, $\Bbb F_5[x]/I$ and $\Bbb F_5[x]/J$ are finite fields each with $25$ elements and any two finite fields with the same number of elements are isomorphic. This makes constructing an isomorphism $\Bbb F_5[x]/I\to \Bbb F_5[x]/J$ quite easy.
