I want to show that $$\mathbb{Z}_3[\sqrt{-5}] /\langle 3, 2 + \sqrt{-5} \rangle \cong \mathbb{Z}_3$$ What bijective map should I build to show the isomorphism relation?

Important Edit : I made a mistake. I meant $\langle 3, 2 + \sqrt{-5} \rangle$ not $\langle 2 + \sqrt{-5} \rangle$


closed as off-topic by MooS, Adam Hughes, Leucippus, астон вілла олоф мэллбэрг, hardmath Nov 19 '16 at 2:51

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – MooS, Adam Hughes, Leucippus, астон вілла олоф мэллбэрг, hardmath
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ I think you mean $\mathbb Z[\sqrt{-5}]/\langle 2+\sqrt{-5} \rangle$. And this is isomorphic to $\mathbb Z/9\mathbb Z$. $\endgroup$ – MooS Nov 18 '16 at 21:19
  • $\begingroup$ @MooS I believeit is $\mathbb{Z}_3$. $\endgroup$ – J. Hartmann Nov 18 '16 at 22:02
  • $\begingroup$ @MooS I made some changes in the question. There was a (killer) typo. $\endgroup$ – J. Hartmann Nov 18 '16 at 22:23
  • 1
    $\begingroup$ There is still another typo... $\endgroup$ – MooS Nov 19 '16 at 0:22

Dividing out by $(2+\sqrt{-5})$ means we "set" $2+\sqrt{-5}=0$, or in other words $\sqrt{-5}=1$. What do you think the isomorphism will be?

  • $\begingroup$ Nitpick: I think you want to say "modding out by" rather than "dividing out by." $\endgroup$ – Adam Hughes Nov 18 '16 at 21:21
  • 1
    $\begingroup$ $a + b \sqrt{-5} + \langle 2 + \sqrt{-5} \rangle \to a + b $? $\endgroup$ – J. Hartmann Nov 18 '16 at 21:22
  • $\begingroup$ I highly doubt that it should be $\mathbb Z_3$ on the left hand side, since $-5=1$ in that ring and we can't even pick a square root of $1$ in a canonical way...It could be $1$ or $2$. If we pick $2$, the ideal is the unit ideal, hence the quotient is zero. This make no sense at all.. $\endgroup$ – MooS Nov 18 '16 at 21:22
  • $\begingroup$ @AdamHughes It's called a "quotient ring", which validates terminology from fractions. So does the notation. I stand by my choice of words, although yours is just as good. $\endgroup$ – Arthur Nov 18 '16 at 21:23
  • $\begingroup$ @MooS If you interpret $\Bbb Z_3[\sqrt{-5}]$ as $\Bbb Z_3[X]/(X^2+5)=\Bbb Z[X]/(X^2-1)$, with $\sqrt{-5}=X+(X^2-1)$, there is no problem at all. $\endgroup$ – Arthur Nov 18 '16 at 21:27

Not the answer you're looking for? Browse other questions tagged or ask your own question.