I was having some trouble finding an explicit isomorphism between $\mathbb{Z}[i]/(7)$ and $\mathbb{Z}[\sqrt{-2}]/(7)$.

$\textbf{What I have noticed is}:$

  • 7 is a prime element in $\mathbb{Z}[i]$ so $(7)$ is a maximal ideal in $\mathbb{Z}[i]$ and $\mathbb{Z}[i]/(7)$ is a field.
  • 7 is also a prime element in $\mathbb{Z}[\sqrt{-2}]$

$\textbf{What I have been trying to do is this}$

  • Find a surjective homomorphism between $\mathbb{Z}[i]/(7) \rightarrow \mathbb{Z}[\sqrt{-2}]/(7)$. Since $\mathbb{Z}[i]/(7)$ was a field, the kernel of this homomorphism will be either the whole ring or just $0$. In the latter case it would be a isomorphism.

I am having trouble finding this surjective homomorphism:

I have noticed that $\bar{{i}}^{2}=-1$ so the image of $\bar{{i}}$ must be sent to something whose square is $-1$. Any help would be appreciated. I may be missing some obvious insight.

  • 4
    $\begingroup$ $-8 \equiv -1 \pmod{7}$ $\endgroup$ – Daniel Fischer Nov 6 '14 at 16:22
  • $\begingroup$ Thanks, so $-8=\sqrt{-2}^{6}$ so we map $i$ to $\sqrt{-2}^{6}$? $\endgroup$ – user135520 Nov 6 '14 at 16:29
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    $\begingroup$ Not $\sqrt{-2}^6$, $\pm (\sqrt{-2})^3$. $\endgroup$ – Daniel Fischer Nov 6 '14 at 16:33
  • $\begingroup$ Yes, that's right. Thanks a lot. I would have been stuck on that for some time. I didn't occur to me to think of congruence classes of -1. Thanks again. $\endgroup$ – user135520 Nov 6 '14 at 16:36
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    $\begingroup$ Well, it isn't so bad anyway. There are only $49$ residue classes modulo $7$, so it's a finite undertaking. Then $(a+b\sqrt{-2})^2 = a^2 + 2b^2 + 2ab\sqrt{-2}$ shows that one of $a$ and $b$ must be $\equiv 0 \pmod{7}$. Since $-1$ is not a square modulo $7$, it must be $a$. $\endgroup$ – Daniel Fischer Nov 6 '14 at 16:40

Consider the unique homomorphism $f\colon\mathbb{Z}[X]\to\mathbb{Z}[\sqrt{-2}]/(7)$ sending $X\mapsto 2\sqrt{-2}+(7)$. Note that $f(4X)=8\sqrt{-2}+(7)=\sqrt{-2}+(7)$, so $f$ is surjective. Its kernel contains $X^2+1$, as $$ f(X^2+1)=(2\sqrt{-2})^2+1+(7)=-8+1+(7)=(7) $$ so $f$ induces a (surjective) homomorphism $g\colon\mathbb{Z}[i]\to\mathbb{Z}[\sqrt{-2}]/(7)$.

What's the kernel of $g$?

  • $\begingroup$ Readers may find it illuminating to ponder my hint before reading this (or similar proofs). $\endgroup$ – Bill Dubuque Jul 3 '17 at 22:00

Hint $\ $ We get an image of $\,\Bbb Z[\sqrt{\color{#c00}{-9}}]\,$ in $\,\Bbb Z[\sqrt{-1}]\,$ by mapping $\,\sqrt{-9}\,\mapsto\, 3\sqrt{-1}$

$\!\bmod 7\!:\ \color{#c00}{{-}9\equiv -2}\,$ and the map is onto, since $3$ is a unit so $\,3^{-1}\sqrt{-9}\,\mapsto \sqrt{-1}$

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    $\begingroup$ The essence of the matter is that the radicands differ by a multiple of a unit square factor. $\endgroup$ – Bill Dubuque Jul 3 '17 at 21:57
  • $\begingroup$ Sorry I cannot understand your hint... $\endgroup$ – Aolong Li Jul 4 '17 at 16:56
  • $\begingroup$ @AolongLi The idea is: we obtain an injective image of $\Bbb Z[\sqrt{c^2 d}]$ in $\Bbb Z[\sqrt{d}]$ by essentially pulling $c$ out of the radical. To do that ring theoretically, show that the map $\,x\mapsto c\sqrt d\,$ from $\,\Bbb Z[x]\to \Bbb Z[\sqrt d]$ has kernel $\,(x^2-c^2d),\,$ hence its image $\cong \Bbb Z[x]/(x^2\!-\!c^2d)\cong \Bbb Z[\sqrt{c^2d}],\,$ assuming that $\sqrt d$ is irrational. Furthermore, if $\,c\,$ is a unit then the map is onto, so an isomorphism. $\endgroup$ – Bill Dubuque Jul 4 '17 at 17:34

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