0

What is the order of $ \Bbb{Z}[ \sqrt{-1}]/(2)$ ? I think this ring is isom to $\Bbb{F}_2[ \sqrt{-1}]$, so the order is $2×2＝4$, but on the other hand, I know $ {\Bbb{Z}[ \sqrt{-1}]/(2)}^×$ is isom to $GL_2(\Bbb{Z}/2 \Bbb{Z})$, so the order must be larger than $6$.

Where am I missing and what is the correct order of $ \Bbb{Z}[ \sqrt{-1}]/(2)$? Thank you for your help.

Pont
  • 5,909
  • In $\mathbb{F}_2$, $-1=1$ already has a square root, so $\mathbb{F}_2[\sqrt{-1}]=\mathbb{F}_2$. Also, as $2= (1+i)(1-i)$ is not prime, $(2)$ is not prime, hence not maximal, so the quotient cannot be a field. – Arturo Magidin Aug 01 '22 at 14:27
  • And the multiplication semigroup is commutative, so no subgroup can be isomorphic to $S_3$. – Arturo Magidin Aug 01 '22 at 14:28

0 Answers0