$R = \Bbb{Z}[\sqrt{-3}]$, what is the quickest route to write the elements of $R/(2)$?

Asked Aug 10 '17 at 19:31

Active Aug 10 '17 at 19:44

Viewed 49 times

My intuition tells me that:

$R/(2) = \{a + b\sqrt{-3} : a, b \in \{0, 1\}\}$ is a complete set of coset representatives, but not sure how to prove that.

~~Would it suffice to show that $R/(2) \approx \Bbb{Z}/(2) \oplus \Bbb{Z}/(2)$?~~

Would it be best to use the opposite construction $R = \Bbb{Z}[X] /(X^2 + 3)$?

edited Aug 10 '17 at 19:44

asked Aug 10 '17 at 19:31

Daniel Donnelly

2

Yes, use that $R = \Bbb{Z}[X] /(X^2 + 3)$ and conclude $R/(2)\simeq \Bbb{Z/(2)}[X] /((X+1)^2)$. Btw, this isn't isomorphic to the direct sum you mentioned. – user26857 Aug 10 '17 at 19:38
@user26857 how did you get $(X+1)^2$? – Daniel Donnelly Aug 10 '17 at 19:43
1

Since $3=1$ and $X^2+1=(X+1)^2$. – user26857 Aug 10 '17 at 19:44
@user26857 is $\Bbb{Z}[X]/(X^2 + 3)$ represented by ${ a X + b : a,b \in \Bbb{Z}}$? – Daniel Donnelly Aug 10 '17 at 19:47
I'd say so.${}$ – user26857 Aug 10 '17 at 19:48
Reason as in the first line of the proof I gave here, and see also here. – Bill Dubuque Aug 10 '17 at 21:40

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