Determining number of cosets in a quotient ring

Asked Feb 19 '19 at 01:36

Active Feb 19 '19 at 01:36

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It was proved in class that $\mathbb{Z}[i]/<a+bi>$ has $a^2+b^2$ elements which I did not understand.

$a+bi$ is equivalent to $0$ in $\mathbb{Z}[i]/<a+bi>$. $a+bi=0 \implies i=\frac{-a}{b}\implies a^2+b^2=0$. Which means that $\mathbb{Z}[i]/<a+bi>$ has $a^2+b^2$ distinct cosets. So I didn't understand the last step, why does that imply $a^2+b^2$ cosets? Thanks and appreciate a hint.

asked Feb 19 '19 at 01:36

trickymaverick

3,372

See here for a complete simple proof. – Bill Dubuque Feb 19 '19 at 01:49
A short, geometric answer is that the number of lattice points inside the fundamental domain of a sublattice is the area of that fundamental domain, which in this case is $c^2=a^2+b^2$, by the pythagorean theorem. – jgon Feb 19 '19 at 03:03
That is neat, thanks. – trickymaverick Feb 19 '19 at 13:10
@manifolded See also this thread – Bill Dubuque Feb 19 '19 at 16:11

Determining number of cosets in a quotient ring

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