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I understand geometrically why the Gaussian integers modulo $i+1$ is $\mathbb{Z}_{2}$, using lattices. Is there an algebraic isomorphism construction, however?

boldbrandywine
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3 Answers3

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Consider $a+bi$: we can write $a-b=2q+r$, where $q$ is an integer and $r$ is $0$ or $1$. Then $$a+bi=(b+2q+r)+bi=r+b(1+i)+q(1-i)(1+i)\equiv r\ .$$

David
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You could use $\mathbb{Z}[i] \cong \mathbb{Z}[x]/(x^2+1)$. Then $$ \mathbb{Z}[i]/(i+1) \cong \mathbb{Z}[x]/(x^2+1, x+1) \cong \mathbb{Z}[x]/(-x+1,x+1) \cong \mathbb{Z}[x]/(2,x+1) \cong \mathbb{Z}/2\mathbb{Z}[x]/(x+1)\cong \mathbb{Z}/2\mathbb{Z}. $$

Youngsu
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  • Could you explain the isomorphism $\mathbb{Z}[x]/(x^2+1, x + 1) \cong \mathbb{Z}[x]/(-x + 1, x + 1)$? I think I understand the other isomorphisms (hopefully). – boldbrandywine Feb 04 '16 at 02:37
  • @boldbrandywine: It suffices to see that the ideals $(x^2+1, x+1)$ and $(-x+1,x+1)$ are equal in $\mathbb{Z}[x]$. Observe that $x^2 + 1 = x(x+1) + (-x+1)$. Equivalently, one has $-x + 1 = (x^2 + 1) - x(x+1)$. This shows the equality. – Youngsu Feb 04 '16 at 02:47
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Since $i^2=-1$, if we pass to a quotient where $i+1=0$, we must have $-1=(-1)^2=1$, so such a quotient contains $\mathbb{Z}/2$.

In fact, such a quotient is exactly $\mathbb{Z}/2$, because $1\mapsto 1$ and $i\mapsto 1$, so any $a+bi$ is mapped to $a+b\in\mathbb{Z}/2$.

Andrew Dudzik
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