Algebraic Number Theory - Lemma for Fermat's Equation with $n=3$ I have to prove the following, in my notes it is lemma before Fermat's Equation, case $n=3$. I was able to prove everything up to the last two points:
Let $\zeta=e^{(\frac{2\pi i}{3})}$. Consider $A:=\mathbb{Z}[\zeta]=\{a+\zeta b \quad|\quad a,b\in \mathbb{Z}\}$. Then


*

*$\zeta$ is a root of the irreducible poly. $X^2+X+1$.

*The field of fractions of $A$ is $\mathbb{Q}(\sqrt{-3})$

*The norm map $N:\mathbb{Q}(\sqrt{-3})\rightarrow \mathbb{Q},$ given by $a+\sqrt{-3}b \mapsto a^2+3b^2$ is multiplicative and sends every element in $A$ to an element in $\mathbb{Z}$. In particular, $u\in A$ is a unit iff $N(u)\in\{-1,1\}$. Moreover, if $N(a)=\pm$ prime number, then $a$ is irreducible.

*The unit group $A^x$ is cyclic of order $6$. ($A^x=\{\pm 1, \pm\zeta, \pm\zeta^2\}$)

*The ring $A$ is Euclidean with respect to the norm $N$ and hence a unique factorisation domain.

*The element $\lambda=1-\zeta$ is a prime element in $A$ and $3=-\zeta^2\lambda^2$.

*The quotient $A$ / $(\lambda)$ is isomorphic to $\mathbb{F}_3$.

*The image of the set $A^3=\{a^3|a\in A\}$ under $\pi: A \rightarrow A / (\lambda^4)=A / (9)$ is equal to $\{0+(\lambda^4),\pm 1+(\lambda^4),\pm \lambda^3+(\lambda^4)\}$


I was not able to prove 7 and 8. For 7 I do not even know which isomorphism, I guess it should be an isomorphism of rings?
I hope anybody knows what to do or has at least some hints,
Thanks in advance for your help!

It is field isomorphism.
We are trying to compute $F = \mathbb Z[\sqrt{-3}]/(1-\zeta)$ and $(1-\zeta)$ is a prime ideal.
In $F$ we have $3=0$ because $\zeta = \frac{-1 + \sqrt{-3}}{2}$ and the norm is $(1-\frac{-1 + \sqrt{-3}}{2})(1-\frac{-1 - \sqrt{-3}}{2}) = 3$. We also have $\zeta = 1$ in this field (when you take a quotient $F/(p)$ think of multiples of $p$ being equal to zero), so $F=\mathbb F_3$
