Explicitation of an isomorphism between two rings [closed]

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, hardmathNov 19 '16 at 2:51

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• I think you mean $\mathbb Z[\sqrt{-5}]/\langle 2+\sqrt{-5} \rangle$. And this is isomorphic to $\mathbb Z/9\mathbb Z$. – MooS Nov 18 '16 at 21:19
• @MooS I believeit is $\mathbb{Z}_3$. – J. Hartmann Nov 18 '16 at 22:02
• @MooS I made some changes in the question. There was a (killer) typo. – J. Hartmann Nov 18 '16 at 22:23
• There is still another typo... – 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?
• $a + b \sqrt{-5} + \langle 2 + \sqrt{-5} \rangle \to a + b$? – J. Hartmann Nov 18 '16 at 21:22
• 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.. – MooS Nov 18 '16 at 21:22
• @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. – Arthur Nov 18 '16 at 21:27