I guess that $\mathbb{Z}[i]/\langle 2+i\rangle$ is equal to $\mathbb{Z}_5[i]$ if it is a simple extension of $\mathbb{Z}_5$. Based on this thought, I tried to construct a surjective ring homomorphism $\varphi$ from $\mathbb{Z}[i]$ to $\mathbb{Z}_5[i]$ with a kernel $\langle 2+i\rangle$. I know that $\mathbb{Z}[i]$ is a Euclidean domain, so it makes sense for me to think $\varphi(z)$ as the "remainder" when $z$ is divided by $2+i$. But how can I express the map $\varphi$ explicitly? Thanks.
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3Note that $2^2=-1\bmod5$, so ${\bf Z}_5[i]={\bf Z}_5$. – Gerry Myerson Nov 05 '17 at 05:51
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Do you mean this: https://en.m.wikipedia.org/wiki/Primitive_element_theorem ? – lhf Nov 05 '17 at 13:57
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I guess you can write $\mathbb{Z}[i]/ \langle 2+i \rangle$ as $\mathbb{Z}[x]/ \langle x^2+1, x+2 \rangle$. As you've noted, the ideal $\langle x^2+1, x+2 \rangle$ actually contains the ideal $\langle x^2+1, 5 \rangle$ since $ (2+x)(2-x)=4-x^2=5$.
Thus, we have $\mathbb{Z}[x]/ \langle x^2+1, 2+x \rangle = \mathbb{Z}/5 \mathbb{Z}[x]/ \langle x^2+1, 2+x \rangle$. Now lets examine what the ring $ \mathbb{Z}/5 \mathbb{Z}[x]/ \langle x^2+1 \rangle$ looks like.
Modulo $5$, $x^2+1=(x+2)(x+3)$ so by Chinese remainder theorem, we have $ \mathbb{Z}/5 \mathbb{Z}[x]/ \langle x^2+1 \rangle \cong \mathbb{Z}/ 5 \mathbb{Z}[x]/\langle x+2 \rangle \times \mathbb{Z}/ 5 \mathbb{Z}[x]/ \langle x+3 \rangle \cong \mathbb{Z}/ 5 \mathbb{Z} \times \mathbb{Z}/ 5 \mathbb{Z}$.
Finally, if we quotient by $x+2$ then we see that this keeps the first factor unchanged but take the second factor to $0$. Thus, in conclusion we see that $\mathbb{Z}[x]/ \langle x^2+1, x+2 \rangle \cong \mathbb{Z}/5 \mathbb{Z}$.
Alexander
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Thanks a lot. Should I say that $\mathbb{Z}[i]/\langle 2+i\rangle$ is not a simple extension of $\mathbb{Z}_5$ or it is a simple extension of $\mathbb{Z}_5$ as $\mathbb{Z}_5[1]$? – Roy Nov 05 '17 at 17:01
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