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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.

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up vote 7 down vote accepted

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).

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Thanks Matt, this looks really helpful and what the authors intended, especially since one of the exercises was to prove $x^p\equiv x\pmod{p}$. Also, the previous exercise was to show $a+b\omega$ is always congruent to $-1,1,$ or $0$ modulo $1-\omega$, and I take it that's why $Z[\omega]/ \lambda Z[\omega]\cong\mathbb{F}_3$. – yunone Aug 3 '11 at 14:42

[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 $\lambda^2$.

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Hmm, this is very interesting. Thanks Jyrki, I really appreciate your help, and I understand this version better now. – yunone Aug 3 '11 at 10:28
@yunone: I rewrote my answer. Thanks for pointing out that the original version didn't really make any sense. The upshot is that your factors $\alpha-1$, $\alpha-\omega$ and $\alpha-\omega^2$ are all in the ideal generated by $\lambda$, but because they are pairwise noncongruent modulo $\lambda^2$, exactly one of them will be divisible by $\lambda^2$, because the ideal generated by $\lambda^2$ has only three cosets within the prime ideal generated by $\lambda$. – Jyrki Lahtonen Aug 3 '11 at 10:28

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 $\rm\ \lambda^2 \equiv 0,\ m^3\equiv m\pmod{3}\:,\ \forall\ m\in \mathbb Z\:.$

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Thanks Bill, I always appreciate your alternative methods. – yunone Aug 3 '11 at 23:27

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