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Why is $(2, 1+\sqrt{-5})$ not principal?

\begin{align*}\mathbf Z[\sqrt{-5}]/(2, 1+\sqrt{-5})&\simeq \mathbf Z[x]/(2,x+1,x^2+5)\simeq \mathbf Z_2[x]/(x+1,x^2+1)\\ &=\mathbf Z_2[x]/\bigl(x+1,(x+1)^2\bigr)=\mathbf Z_2[x]/(x+1)\simeq\mathbf Z_2. \end{align*}

It is said to be used that $(R/I)/(J/I)\simeq R/J$, but can you explain in more detail?

And I still don't understand about the other homomorphisms and equations (especially $\mathbf Z_2[x]/(x+1,x^2+1)=\mathbf Z_2[x]/\bigl(x+1,(x+1)^2)$

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1 Answers1

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This is why: $\;\mathbf Z[\sqrt{-5}]\simeq\mathbf Z[x]/(x^2+5)$, in which $x$ corresponds to $\sqrt{-5}$, so \begin{align*}\mathbf Z[\sqrt{-5}]/(2,1+\sqrt{-5})&\simeq\mathbf Z[x]/(x^2+5)\textbf{/}(2,1+x)\mathbf Z[x]/(x^2+5)\\&=\mathbf Z[x]/(x^2+5)\textbf{/}(2,1+x, x^2+5)/(x^2+5) \end{align*} Note. – The general situation is this:

Suppose you have a ring $R$ and two ideals $I,J$ in $R$. Then $I\cdot R/J=(I+J)/J$, so that $$ (R/J)\textbf{/}(I\cdot R/J)\simeq R/(I+J).$$

For your second question, it comes from the fact that, in characteristic $2$, $\;(x+1)^2=x^2+1^2=x^2+1$.

Bernard
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  • Random question, but: Can the fact that $\Bbb Z[\sqrt{-5}]/(2,1+\sqrt{-5})\simeq \Bbb Z_2$ be used to conclude that $\Bbb Z[\sqrt{-5}]$ is not a PID? – MyNameIs Jul 26 '16 at 14:54
  • I don't see how one might deduce non factoriality. All rings of integers have finite residual fields, whether they be principal or not. – Bernard Jul 26 '16 at 14:58
  • Where does the $1+x$ come from? – Addem Apr 08 '18 at 04:42