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!
 A: Here is a direct bash to show (7).  We have
$$\begin{eqnarray*} \Bbb{Z}[\zeta]/(1 - \zeta) &\cong& \left[\frac{\Bbb{Z}[x]}{(x^2 + x + 1)}\right]\bigg/ \left[\frac{\left(1-x ,x^2 + x + 1\right)}{(x^2 + x + 1)}\right] \\
&\cong& \Bbb{Z}[x]\Big/\left(x^2 + x + 1,1-x\right) \hspace{1in} (\text{Third Isomorphism Theorem}) \\
&\cong& \Bbb{Z}[x]/(1-x,3) \\
&\cong& \Bbb{F}_3.\end{eqnarray*}$$
If at any point you don't understand how I got those isomorphisms please tell me and I will elaborate. 
Edit: For the sake of the OP let me elaborate how I got from the third last line to the second last line. I claim that in fact we have an equality of ideals
$$(1-x,3) = (x^2 + x + 1,1-x).$$
How do we show this equality? We just need to show that both generators on the left are in the right, and both generators on the right are in the left one. Now by long division from school I get
$$(x^2 + x + 1) = (1-x)(-x-2) + 3.$$
This shows that $x^2 +x +1 \in (1-x,3)$ and $3 \in (x^2 + x+1,1-x)$ and so we are done!
A: For 7), note that 6) tells you that $3 \in (\lambda)$, and since by 6) $\lambda$ is prime, $A \ne (\lambda)$. Moreover $a + \zeta b = a + (1-\lambda) b \equiv a + b \pmod{\lambda}$. So if you want an explicit isomorphism, it is $a + \zeta b + (\lambda) \mapsto a+ b \pmod{3}$.
A: 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 $F=\mathbb F_3$
A: 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.  
