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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 $\displaystyle p = \alpha \beta$ where $\displaystyle \alpha , \beta$ are non-units. It follows that $\displaystyle N(\alpha),N(\beta) > 1$ and so $\displaystyle p^2 = N(\alpha)N(\beta)$ thus $\displaystyle p = N(\alpha)$. Let $\displaystyle \alpha = a+b\omega$ then $\displaystyle p = a^2 - ab + b^2$. Thus, $\displaystyle 4p = (2a-b)^2 + 3b^2$ which means $\displaystyle p\equiv (2a-b)^2 (\bmod 3)$. If $\displaystyle 3\not | p$ then $\displaystyle p\equiv 1(\bmod 3)$ but $\displaystyle 1$ is the only square mod 3. Thus, rational primes which are of form $\displaystyle 3k+2$ 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 $\displaystyle p$ has form $\displaystyle 3k+1$. By using quadradic reciprocity it follows that $\displaystyle (-3/p) = 1$. Thus, there is an integer $\displaystyle n$ so that $\displaystyle n^2 \equiv -3(\bmod p)$ thus $\displaystyle p|(n^2+3)$. Thus, $\displaystyle p|(n+i\sqrt{3})(n-i\sqrt{3}) = (n+1+2\omega)(n-1-2\omega)$. If were prime then it would divide both those factors but that is impossible since then $\displaystyle p|[(n+1+2\omega)+(n-1-2\omega)] \implies p|2n$, since $\displaystyle p\not = 2$ it means $\displaystyle p|n$. This is impossible and so $\displaystyle p$ is reducible. Thus, $\displaystyle p = \alpha \beta$. We have $\displaystyle p^2 = N(\alpha)N(\beta)$ and so $\displaystyle N(\alpha) = N(\beta) = p$. Since the norm is a rational prime it means $\displaystyle \alpha,\beta$ are prime. Note, $\displaystyle p = N(\alpha) = \alpha \bar \alpha$. And so $\displaystyle p$ factors into two prime $\displaystyle \alpha , \bar \alpha$ which are not associate.