3
$\begingroup$

Let $\mathcal{E}$ be the ring of integers of some imaginary finite field extension of $\mathbb{Q}$. We consider only the case $\mathcal{E}$ is Euclidean. Let $\pi_1, \pi_2 $ be prime elements in $\mathcal{E}$ with equal norms. From that it follows that $\mathcal{E} /\pi_1\mathcal{E}$, $\mathcal{E} /\pi_2\mathcal{E}$ are isomorphic finite fields. Does it also follow that the coset representatives can be chosen in the same way for both quotient rings?

For example, in case $\mathcal{E} = \mathbb{Z}[w]$, $w$ is primitive cube root of unity and $\pi$ -- prime element with norm $4$ they are: $0, 1, w, w+1$, but is this a general law?

  • 3
    (if I understand your question right) It is possible, simply because of the Chinese remainder theorem2017-01-08
  • 1
    In the case of $\Bbb{Z}][w]$, $N(\pi)=4$, all those prime elements are associates, so this is kind of obvious. In general, as user8268 pointed out, you need to use CRT to show that you can find matching representative for the cosets.2017-01-08
  • 0
    @JyrkiLahtonen, is it simply smth like : $\forall x \in \mathcal {E}/\pi_1\mathcal{E}$, $\forall y \in \mathcal {E}/\pi_2\mathcal{E}$, $\exists \alpha \in \mathcal{E}$ such that $\alpha + \pi_1\mathcal{E}=x$ and$\alpha + \pi_2\mathcal{E}=y$?2017-01-10
  • 0
    which follows from $\mathcal{E}/\pi_1 \pi_2 \mathcal{E} \simeq \mathcal{E}/\pi_1 \mathcal{E} \oplus \mathcal{E}/\pi_2\mathcal{E}$, which sends coset $x$ to $(x+\pi_1\mathcal{E}) + (x+\pi_2\mathcal{E})$?2017-01-10
  • 0
    That's the idea. Of course CRT fails when the two ideals are equal, but that was the easy case :-)2017-01-10
  • 0
    @JyrkiLahtonen, :) thanks, but i really don't think that it is just an idea. In a case of euclidian ring, every prime ideal is maximal , so $\mathcal{E}/\pi_1\mathcal{E}$, $\mathcal{E}/\pi_2\mathcal{E}$ are isomorphic as a fields when $\pi_1, \pi_2$ have equal norms. And CRT garantees that for any two elements of these quotient rings the same representative from $\mathcal{E}$ can be found. And thus we are done since we can "match" proper elements of quotients. am i right?2017-01-10

0 Answers 0