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!

For (7), notice first that, since $\lambda\mid 3$, the characteristic of the quotient ring $A/(\lambda)$ is $3$. Now $N(\lambda)=3$, so by (3), $\lambda$ is irreducible. This implies that the quotient ring is in fact a field. Again since $N(\lambda)=3$, we find that the inertia degree of $\lambda$ over $\mathbb Q$ is $1$, i.e. $A/(\lambda)\cong F_3$.
For (8), since $(a+b\zeta)^3=a^3+b^3-3ab^2+\zeta(3a^2b-3ab^2)$, we find that $(a+b\zeta)^3$ is sent to $0$ if and only if both $9$ divides $a^3+b^3-3ab^2$ and $3$ divides $ab(a-b)$. The latter occurs if and only if $3$ divides $a$, $b$, or $a\equiv b\pmod 3$. In each of the cases, we conclude that $3$ divides both $a$ and $b$. So, to find the image of $\mathbb A^3$ under $\pi$, one just checks each of $9$ cases, and noting that $\zeta\equiv1\pmod{\lambda}$ and $\lambda^2=-3\zeta$.
Hope this helps.  
