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Simplify the ring $\mathbb Z[\sqrt{-13}]/(2)$. I have so far:

$$\mathbb Z[\sqrt{-13}]/(2) \cong \mathbb Z[x]/(2, x^2 + 13) \cong \mathbb Z_2[x] / (x^2 + 1)$$

Now how do I simplify it further? I know that $x^2 + 1 = (x + 1)^2$ in $\mathbb Z_2[x]$, is this useful?

shmth
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  • That last identity is useful. I believe that is all that you can say. Interestingly this means the ring has nilpotents! In particular, it's not a domain, and hence 2 is not prime in $Z[\sqrt{-13}]$. – User0112358 Feb 16 '17 at 01:25
  • I don't think the ring structure is either of those. The underlying group, maybe. The ring does not have idempotents, so it can't be a product. Also, the ring has characteristic 2, so it can't be $\mathbb{Z}_{4}$. It is neither of these rings. – User0112358 Feb 16 '17 at 01:31
  • @User0112358 I see, thanks! – shmth Feb 16 '17 at 01:31
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    The underlying group structure is $\mathbb{Z}{2} \times \mathbb{Z}{2}$. But the multiplication is not the coordinate-wise multiplication. – User0112358 Feb 16 '17 at 01:37
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    It's $\cong \Bbb Z_2[t]/(t^2),,$ the ring of dual numbers over $,\Bbb Z_2.,$ See here and here for applications (tangent spaces, product rules, infinitesimals, etc). – Bill Dubuque Feb 16 '17 at 01:47

2 Answers2

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Make the substitution $y = x + 1$ to get an isomorphism $(Z/2Z)[x] \cong (Z/2Z)[y]$. The answer is $(Z/2Z)[y]/(y^2)$.

user416819
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$x^2+1=0 \Rightarrow x^2=-1 \equiv 1(mod 2) \Rightarrow x^3=x$

$f(x)=a_0+a_1x+a_2x^2+..+a_nx^n=a_0+a_1x+a_2(-1)+a_3(x)+...=a_0+a_1x(mod 2)$

$\Rightarrow \mathbb Z_2[x]/(x^2+1) =\{I,1+I, x+I,(1+x)+I \}; I=(x^2+1)$

Mustafa
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