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Prove that if $\alpha \in \mathbb{E}$ is an Eisenstein integer and $\pi$ is an Eisenstein prime, than $\pi \mid \alpha^{N(\pi)}-\alpha$.

$\mathbb{E} = \mathbb{Z}[\varepsilon] = \{ a+\varepsilon b \mid a, b \in \mathbb{Z} \}$ is the ring of the Eisenstein integers (where $\varepsilon = -\frac{1}{2}+i\frac{\sqrt{3}}{2}$). An Eisenstein prime is a prime in this ring. $N(z) = z\overline{z} = a^2 -ab + b^2 \ \ \ \ \ \forall z = a+\varepsilon b \in \mathbb{E}$

Hi! I saw this problem in an algebra textbook, but I think it has more things to do with number theory. The statement strongly reminds me of Fermat's Little Theorem, but I haven't been able to "translate" its proof to ring theory yet.

I know the characterization of the Eisenstein primes:

  • ordinary primes congruent to $2$ modulo $3$
  • $\lambda = i\sqrt{3} = 1 + 2\varepsilon \in \mathbb{E}$
  • any ordinary prime congruent to $1$ modulo $3$ factors as $\pi\pi^*$ where each of $\pi$ and $\pi^*$ are primes in $\mathbb{E}$

I thought it could be a good idea to prove the proposition in these 3 cases, but I don't know how to proceed.

Any help will be appreciated!

Bill Dubuque
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1 Answers1

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Fermat's little theorem says that $a^{p-1}=1$ in the cyclic group $\left(\mathbb{Z}/p\mathbb{Z}\right)^*\cong \mathbb{Z}/(p-1)\mathbb{Z}$ (the order of an element divides the order of the group). For Eisenstein integers, the analogous statement would be $\left(\mathbb{Z}[\varepsilon]/\pi\mathbb{Z}[\varepsilon]\right)^*\cong \mathbb{Z}/(N(\pi)-1)\mathbb{Z},$ since, in that case, you would have $\alpha^{N(\pi)-1}=1$. For example, if $p\in\mathbb{Z}$ is prime in $\mathbb{Z}[\varepsilon]$, then $\mathbb{Z}[\varepsilon]/(p)\cong \frac{\mathbb{Z}/p\mathbb{Z}[x]}{(x^2+x+1)}$ is a field with $p^2=N(p)$ elements. Hence, the group of units is cyclic of order $p^2-1.$

PQH
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