Find an explicit isomorphism of rings Find and explicit isomorphism of rings $$\lambda:\mathbb{F}_5[x]/(x^2+x+2)\rightarrow\mathbb{F}_5[x]/(x^2+4x+2)$$
I know that both fields minus the zero element are isomorphic to $C_{24}$ with generator x, but how do I describe the isomorphism between the two with the zero element? 
I also know that 0 must get mapped to zero and 1 mapped to 1, so do i simply state what x gets mapped to (and map it to x) ?
 A: Denote by $t$ the image of $x$ under the canonical homomorphism $\mathbb{F}_5[x]\to\mathbb{F}_5[x]/(x^2+4x+2)$, so that $t^2+4t+2=0$.
You know that every element of $\mathbb{F}_5[x]/(x^2+4x+2)=\mathbb{F}_5[t]$ can be uniquely written as $a+bt$, with $a,b\in \mathbb{F}_5$.
If $(a+bt)^2+(a+bt)+2=0$, we get
$$
a^2+2abt+b^2(-4t-2)+a+bt+2=0
$$
that is,
$$
(a^2-2b^2+a+2)+(2ab-4b^2+b)t=0
$$
so $b=0$ (contradiction) or $2a-4b+1=0$, that is, $b=-2a-1$. Substituting in the condition $a^2-2b^2+a+2=0$, we get
$$
a^2+a=0
$$
that means $a=0$ or $a=-1$. Thus the roots of $x^2+x+2=0$ in $\mathbb{F}_5[t]$ are $-t$ and $4+t=t-1$. Thus the kernel of the homomorphism
$$
\mathbb{F}_5[x]\to\mathbb{F}_5[t]
$$
sending $x$ to $-t$ is $(x^2+x+2)$. You can also choose the homomorphism sending $x$ to $t-1$.

Of course, you could also note that the homomorphism $\mathbb{F}_5[x]$ to $\mathbb{F}_5[x]$ sending $x$ to $-x$ transforms the polynomial $x^2+x+2$ into $x^2-x+2=x^2+4x+2$.
A: Factoring the polynomials over $\mathbb F_5$, we get that the first polynomial have roots $x_1=3 \pm 3 \sqrt{3}$, and the other polynomial have roots $x_2=-3 \pm 3\sqrt{3}$. 
To find the isomorphism, note that $x_2 = x_1-6$.
