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I am trying to find some ring isomorphic to $\mathbb Z[i] / \langle 5 \rangle$ . I know that

$\langle 5 \rangle=\langle (2+i)(2-i)\rangle=\langle 2+i\rangle \langle2-i\rangle$ , now if $d$ is the gcd of $2+i , 2-i$ , then

$d|2+i-(2-i)=2i$ , so $d|2$ , then $d|2+i-2=i$ , so the gcd is a unit , hence

$\langle 2+i\rangle+ \langle2-i\rangle=\mathbb Z[i]$ , so by Chinese remainder theorem ,

$\mathbb Z[i] / \langle 5 \rangle \cong \mathbb Z[i] /\langle 2+i\rangle \times \mathbb Z[i] / \langle 2-i\rangle $ . Now if gcd$(a,b)=1$ , then $\mathbb Z[i] / \langle a+ib \rangle \cong \mathbb Z_{a^2+b^2}$ , so

$\mathbb Z[i] / \langle 2-i\rangle \cong \mathbb Z_5$ and $\mathbb Z[i] /\langle 2+i\rangle \cong \mathbb Z_5$ , thus $\mathbb Z[i] / \langle 5 \rangle \cong \mathbb Z_5 \times \mathbb Z_5$ . Am I correct ? If so , then

I want to know , what is the explicit surjective ring homomorphism from $\mathbb Z[i]$ to $\mathbb Z_5 \times \mathbb Z_5$ with

kernel $\langle 5 \rangle$ ? I have already checked that the map $f(a+ib)=([a]_5 , [b]_5)$ is a group

homomorphism but not a ring homomorphism . Please help . Thanks in advance

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Yes, though I would write $\mathbb Z[i] = \mathbb Z[x]/(x^2+1)$. Then it is easier to see that $$\mathbb Z[i]/5 \cong \mathbb Z[x]/(5,x^2+1) \cong \mathbb F_5[x]/(x^2+1) \cong \mathbb F_5[x]/(x+2) \times \mathbb F_5[x]/(x-2) \cong \mathbb Z/5 \times \mathbb Z/5.$$ Following this isomorphism, you get the explicit ring homomorphism $$\mathbb Z[i] \to \mathbb Z/5 \times \mathbb Z/5,\ a+bi \mapsto (a-2b,a+2b),$$ with kernel $(5) \subseteq \mathbb Z[i]$.

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  • $\begingroup$ How did $x^2+1$ get factored into $x+1$ and $x-1$ in $\mathbb Z_5 [x] $ ? $\endgroup$ – user228168 May 6 '15 at 7:23
  • $\begingroup$ Magic!... Of course, that was very silly, sorry. I'll edit it immediately. But note that my original answer was valid for the quadratic ring $\mathbb Z[\sqrt 1] = \mathbb Z[x]/(x^2-1)$ of discriminant $1$! ;) $\endgroup$ – Thomas Poguntke May 6 '15 at 7:28
  • $\begingroup$ That's more like it , thanks $\endgroup$ – user228168 May 6 '15 at 7:33
  • $\begingroup$ Discriminant $1$ ? :-) no , I don't think so , $x^2+1=(x-1)(x+1)$ only in $\mathbb Z_2[x]$ ... isn't it ? :-) $\endgroup$ – user228168 May 6 '15 at 7:35
  • $\begingroup$ Note the sign in $x^2-1$ in my comment. $\endgroup$ – Thomas Poguntke May 6 '15 at 7:37

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