5

Let $a+bi\in\mathbb{Z}[i]$ with $\gcd(a,b)=1$.

I know that $\mathbb{Z}[i]/\langle a+bi\rangle\cong\mathbb{Z}_{a^{2}+b^{2}}$ by a ring homomorphism $\phi:\mathbb{Z}[i]\to\mathbb{Z}_{a^{2}+b^{2}}$, which maps $x+yi$ to $x-(ab^{-1})y\pmod{a^{2}+b^{2}}$ with $\ker\phi=\langle a+bi\rangle$.

My question is:

Is there a ring homomorphism as above $\phi$ if $\gcd(a,b)=d>1$?

I found the relation $\mathbb{Z}[i]/\langle a+bi\rangle\cong\mathbb{Z}_{d}\times\mathbb{Z}_{(a^{2}+b^{2})/d}$ if $\gcd(a,b)=d>1$.

But, i couldn't find the ring homomorphism between $\mathbb{Z}[i]$ and $\mathbb{Z}_{d}\times\mathbb{Z}_{(a^{2}+b^{2})/d}$ with kernel $\langle a+bi\rangle$

Give some advice! Thank you!

AnonyMath
  • 1,354
  • If $\mathbb{Z}[i]/\langle a+bi\rangle\cong\mathbb{Z}{d}\times\mathbb{Z}{(a^{2}+b^{2})/d}$, then couldn't you compose that isomorphism with the canonical quotient homomorphism $\Bbb Z[i]\to \Bbb Z[i]/\langle a + bi\rangle$ to get what you want? Or have I missed something? – Arthur Aug 14 '18 at 16:12
  • @Arthur They are asking how to prove the first isomorphism in your comment. One method to do so would be do find a surjective homomorphism $\mathbb Z[i]\to \mathbb Z_d \times \mathbb Z_{(a^2+b^2)/d}$ whose kernel is $\langle a+bi\rangle$. – Mike Earnest Aug 14 '18 at 16:16
  • Chinese remainder theorem? – Steve D Aug 14 '18 at 16:29
  • @MikeEarnest For example, I've tried a map $\mathbb{Z}[i]\to\mathbb{Z}{d}\times\mathbb{Z}{(a^{2}+b^{2})/d}$ defined $x+yi\mapsto (\square,x-\tfrac{a}{d}\left(\tfrac{b}{d}\right)^{-1}y)$. But, i couldn't compose the suitable $\square$. – AnonyMath Aug 15 '18 at 06:33

0 Answers0