Is it true that the order of any quotient ring $\mathbb Z[i]/\langle a+ib \rangle $ is $a^2+b^2$ ? ( I know it is atmost finite ) Please help . Thanks in advance .
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Are you sure it's finite? – JC574 Apr 16 '15 at 14:50
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@JC574 : Yes , if not both $a , b$ are zero ... – Apr 16 '15 at 14:54
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are we talking order as in the size of the set? – JC574 Apr 16 '15 at 14:54
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@JC574 : Do you know any other kind of "order " :-) ? Yes , we are talking about the size of the quotient ring .... – Apr 16 '15 at 14:55
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yes sorry, I was confused for a sec – JC574 Apr 16 '15 at 14:56
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1@JC574 the ideal generates an integer, making the quotient ring a finite extension of $\mathbb Z/n\mathbb Z$ for some $n$. – Dustan Levenstein Apr 16 '15 at 14:57
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thanks yeah, i don't know why i got confused – JC574 Apr 16 '15 at 14:58
2 Answers
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Elementary answer. The ring $\mathbb Z[i]$ is the lattice generated by the vectors $1$ and $i$. The ideal $\mathfrak a = (a+bi)$ is the subgroup generated by the vectors $a+bi$ and $ia -b$, or by the basis $A = \tiny \begin{pmatrix} a&-b\\b&a\end{pmatrix}$. Therefore, (for instance by the elementary divisor theorem), the order of the group $\mathbb Z[i]/\mathfrak a$ is $\det(A)$, which is conveniently $a^2+b^2$.
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Hint: Leaving off the multiplicative structures, $\mathbb Z[i]$ becomes a free $\mathbb Z$ module with basis $\{1,i\}$, and the ideal $\langle a+ib \rangle$ becomes a submodule of it generated by $(a,b)$ and $(-b,a)$.
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