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Let $ a+b\sqrt{-1} $ be a non-zero element of the ring of Gaussian integers $ \mathbb Z[\sqrt{-1}]. $ Show that $ | \mathbb Z[\sqrt{-1}]/(a+b\sqrt{-1}) |=a^2+b^2 $. [Jacobson, Basic Algebra, P202.3]

My attempt:

Since $ \mathbb{Z}[\sqrt{-1}] $ is a p.i.d. and we can factor $ a+b\sqrt{-1} $ into primes: $ a+b\sqrt{-1}=p_1^{j_1}p_2^{j_2}...p_n^{j_n} $, $\ p_i $ are primes in $ \mathbb Z[\sqrt {-1}] $, then we only have to prove the case for the primary quotient submodule in the following sense:

$$ | \mathbb Z[\sqrt{-1}]/(p_i^{j_i}) |=a^2+b^2\, ,\text{where $ a+bi=p_i^{j_i} $}. $$

From here I am stuck......

Bach
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1 Answers1

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The ideal $(a+b\sqrt{-1})$ forms a sublattice of $\Bbb Z[\sqrt{-1}]$ in the plane. The side length of a mesh square is clearly $\sqrt{a^2+b^2}$, so the area of a mesh is $a^2+b^2$ and hence contains essentially $a^2+b^2$ mesh squares of the $\Bbb Z[\sqrt{-1}]$ lattice.

cqfd
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Hagen von Eitzen
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