Understanding a congruence relation in $\mathbb{Z}[\omega]$ I'm having some difficulty understanding the relation between two different congruences I've been dealing with. These come from Exercise 25 of Chapter 3 in Ireland and Rosen's Number Theory. 

Let $\lambda=1-\omega\in\mathbb{Z}[\omega]$, where $\omega$ is the cube root of unity. If $\alpha\equiv 1\pmod{\lambda}$, then $\alpha^3\equiv 1\pmod{9}$. 

Some algebra shows that $3=-\omega^2\lambda^2$. If $9|\alpha^3-1$, then I want to show that $\omega^4\lambda^4=\omega\lambda^4|\alpha^3-1$. Factoring, I see $\alpha^3-1=(\alpha-1)(\alpha-\omega)(\alpha-\omega^2)$. Now it's given $\lambda|\alpha-1$, also,
$$
\alpha-\omega\equiv 1-\omega\equiv 0\pmod{\lambda},
$$
and
$$
\alpha-\omega^2\equiv 1-\omega^2=(1-\omega)(1+\omega)\equiv 0\pmod{\lambda}.
$$
From this I see $\lambda^3|\alpha^3-1$. But how can I show there is a fourth factor $\lambda$ and one of $\omega$ ? Thanks for any help.
 A: A useful way to solve these sorts of problems is as follows:
Write $\alpha = 1 + x \lambda,$ for some $x \in \mathbb Z[\omega]$.  (This is possible by the assumption that $\alpha \equiv 1 \bmod \lambda$.)
Cubing both sides gives
$$\alpha^3 = 1 + 3 x \lambda + 3 x^2 \lambda^2 + x^3 \lambda^3
= 1 +(- \omega^2 x + x^3)\lambda^3 - \omega^2 x^2 \lambda^4$$
(the last equality following from $3 = - \omega^2 \lambda^2$).
The point is that I have written $\alpha^3$ as a sum of multiples of increasing powers of $\lambda$.
What you want to show is that $\alpha^3 \equiv 1 \bmod 9$, or equivalently $\alpha^3 \equiv 1 \bmod \lambda^4$, and so looking at the preceding expression,
we see that this is the same as proving
$-\omega^2 x + x^3 \equiv 0 \bmod \lambda,$ for all $x$ (because any choice of $x$ in the original formula for $\alpha$ gives a choice of $\alpha$ which is $1 \bmod \lambda$).
Since $\omega \equiv 1 \bmod \lambda,$ we see that the congruence we have to prove can be simplified to
$$x^3 - x \equiv 0 \bmod \lambda$$
for all $x \in \mathbb Z[\omega]$.
Why is this true? Because $Z[\omega]/ \lambda Z[\omega]$ is a field of order $3$ (i.e. a copy of $\mathbb F_3$), and in that field every element is its own $3$rd power.  (In general in a field of order $p$, we always have $x^p = x$.)  So we're done. 
The nice thing about this approach (expanding in powers of $\lambda$) is that you can follow your nose in the proof: you turn the congruence you have to prove into a simpler congruence.  (Our first congruence was something subtle about $\alpha$ that are $1 \bmod \lambda$, but we converted it into a congruence that had to be true for all $x$, and if  a congruence holds for all $x$, it will normally have to hold for some fairly obvious reason.)
One last remark: in congruences involving powers, the binomial theorem 
is very often helpful (as it was here).
A: [Edit: rewritten as the original version had errors.] The point is (as it must) that in your factorization of $\alpha^3-1$ one of the factors is divisible by $\lambda^2$. It turns out that exactly one of them is divisible by $\lambda^2$. 
Let $\mathfrak{P}$ be the ideal of $\mathbf{Z}[\omega]$ generated by $\lambda$. Your identity $3=-\omega^2\lambda^2$ shows that $3\in\mathfrak{P}^2$, and as $-\omega^2$ is a unit, it shows that $\mathfrak{P}^2=3\mathbf{Z}[\omega]$. Therefore all the cosets of $\mathfrak{P}^2$ have a unique representative of the form $a+b\omega$ with $a,b\in\{0,1,-1\}$. As $$a+b\omega=a+b(1-\lambda)=(a+b)-b\lambda,$$ we see that because $\alpha\equiv 1\pmod{\lambda}$, then $\alpha$ will be in a coset $a+b\omega+\mathfrak{P}^2$ such that $a+b\equiv1 \pmod 3$. So $\alpha$ will be either in the coset $1+\mathfrak{P}^2$ (corresponding to $a=1,b=0$), or in the coset $\omega+\mathfrak{P}^2$ (corresponding to $a=0,b=1$) or in the coset $-1-\omega+\mathfrak{P}^2$ (corresponding to $a=b=-1$). But $1+\omega+\omega^2=0$, so the last coset is $-1-\omega+\mathfrak{P}^2=\omega^2+\mathfrak{P}^2$.
Depending on which of these cosets modulo $\mathfrak{P}^2$ your element $\alpha$ happens to fall into, you see that either $\alpha-1$, $\alpha-\omega$ or $\alpha-\omega^2$ will be divisible by $\lambda^2$.
A: The inference is easily verified by a few minutes of rote congruence arithmetic mod $3$ & $9$, viz.
$\rm\  \lambda^2 =\: 3\:\lambda -3\:\ \Rightarrow\:\ \alpha^3-1\: =\: (1+x\:\lambda)^3-1\:\equiv\: \:3\:\color{Brown}{(x-x^3)\:\lambda}\:\equiv\: 0 \pmod{9}\:,\:$ by $\rm\ x = m + n\:\lambda$   
$\rm\: \Rightarrow\ \color{brown}{(x - x^3)\:\lambda} \equiv (m-m^3)\:\lambda\equiv 0 \pmod{3}\ $ via  $\rm\ \lambda^2 \equiv 0,\ m^3\equiv m\pmod{3}\:,\ \forall\ m\in \mathbb Z\:.$
