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Consider the quotient ring $\mathbb{Z}[i]/I$ where $I$ is the ideal $\{m+in : m,n \in \mathbb{Z} \text{ with same parity}\}$. To which commutative ring is it isomorphic?

I thought of finding an appropriate ring homomorphism for which $I$ is the kernel, but somehow I did not succeed. Any help would be appreciated.

user26857
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TheGeekGreek
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2 Answers2

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Consider the (surjective) map

$$\begin{cases}\Bbb Z[i]\to\Bbb Z/2 \\ m+ni\mapsto m+n\mod 2\end{cases}$$

which clearly has as kernel the ideal $I$. By the first isomorphism $\Bbb Z[i]/I\cong\Bbb Z/2$.

Adam Hughes
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    Does the downvoter care to comment what he thinks is incorrect? – Adam Hughes Dec 12 '16 at 17:14
  • Sorry to ask you again, but how would I prove the property $\varphi(rs) = \varphi(r)\varphi(s)$? – TheGeekGreek Dec 12 '16 at 19:18
  • @TheGeekGreek you can just write out the definitions and verifty $(a+bi)(c+di) = (ac-bd)+i(ad+bc)$ so you get $a(c+d)+b(c-d)$ on the LHS and $(a+b)(c+d) = a(c+d)+b(c+d)$ on the RHS. But since $1=-1\mod 2$, you get that the images match. – Adam Hughes Dec 12 '16 at 19:30
  • The part with $1 = -1 \mod 2$ was crucial and I was not entirely sure. Thanks. – TheGeekGreek Dec 12 '16 at 19:48
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Here's a slightly different way to do it.

Claim: $I = (1+i)$. The containment $\supseteq$ is clear. Note that $2 = (1-i)(1+i) \in (1+i)$. Given $m+in \in I$, if $m,n$ are both even, then $m+in \in (2) \subseteq (1+i)$. If $m,n$ are both odd, then $m-1,n-1$ are both even, so $m+in - (1+i) = (m - 1) + i (n - 1) \in (2) \subseteq (1+i)$. Thus $m+in \in (1+i)$.

By the Third Isomorphism Theorem, then \begin{align*} \frac{\mathbb{Z}[i]}{I} = \frac{\mathbb{Z}[i]}{(1+i)} \cong \frac{\mathbb{Z}[x]}{(1+x, x^2+1)} \cong \frac{\mathbb{Z}}{((-1)^2 + 1)} = \frac{\mathbb{Z}}{(2)} \, . \end{align*}

Viktor Vaughn
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