# Thread: Eisenstein integers

1. ## 11

aa

2. Originally Posted by aaa0909
a) the rational prime 2 is an Eisenstein prime
The proof for (b) will do this one.

b) a rational prime of the form 3k+2, where k is a positive integer, is an Eisenstein prime
Say that $p = \alpha \beta$ where $\alpha , \beta$ are non-units. It follows that $N(\alpha),N(\beta) > 1$ and so $p^2 = N(\alpha)N(\beta)$ thus $p = N(\alpha)$. Let $\alpha = a+b\omega$ then $p = a^2 - ab + b^2$. Thus, $4p = (2a-b)^2 + 3b^2$ which means $p\equiv (2a-b)^2 (\bmod 3)$. If $3\not | p$ then $p\equiv 1(\bmod 3)$ but $1$ is the only square mod 3. Thus, rational primes which are of form $3k+2$ are Eisenstein primes.

c) a rational prime of the form 3k+1, where k is a positive integers, factors into the product of two primes that are not associates of one another
Say that $p$ has form $3k+1$. By using quadradic reciprocity it follows that $(-3/p) = 1$. Thus, there is an integer $n$ so that $n^2 \equiv -3(\bmod p)$ thus $p|(n^2+3)$. Thus, $p|(n+i\sqrt{3})(n-i\sqrt{3}) = (n+1+2\omega)(n-1-2\omega)$. If were prime then it would divide both those factors but that is impossible since then $p|[(n+1+2\omega)+(n-1-2\omega)] \implies p|2n$, since $p\not = 2$ it means $p|n$. This is impossible and so $p$ is reducible. Thus, $p = \alpha \beta$. We have $p^2 = N(\alpha)N(\beta)$ and so $N(\alpha) = N(\beta) = p$. Since the norm is a rational prime it means $\alpha,\beta$ are prime. Note, $p = N(\alpha) = \alpha \bar \alpha$. And so $p$ factors into two prime $\alpha , \bar \alpha$ which are not associate.