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Our professor has given us the following isomorphism (fractions denote quotients): $$\frac{\Bbb{Z}[\sqrt{-3}]}{(2+\sqrt{-3})}\cong\frac{\frac{\Bbb{Z}[X]}{(X^2+3)}}{\frac{(2+X,\,X^2+3)}{(X^2+3)}}.$$ Alas, I am unable to find the reasoning behind it. He told us that he used that $\Bbb{Z}[\sqrt{-3}]$ is isomorphic to $\Bbb{Z}[X]/(X^2+3)$ (as with any other square-free integer). Also I'd like to point out that this is the first step of an isomorphism chain, in which the next one he used the Third isomorphism theorem and simplified the duplicated quotients, for what I believe the current ring is obtained without the use of such theorem.

Thanks in advance :)

nuuusxd
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2 Answers2

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Let $\varphi: \mathbb{Z}[X] \to \mathbb{Z}\left[\sqrt{-3}\right]$ be the unique ring homomorphism sending $X$ to $\sqrt{-3}$. It is not difficult to show that $\varphi$ is surjective and its kernel is the ideal $(X^2+3)$. By the first isomorphism theorem, $\varphi$ induces a ring isomorphism $$\overline{\varphi}: \frac{\mathbb{Z}[X]}{(X^2+3)} \longrightarrow \mathbb{Z}\left[\sqrt{-3}\right]$$ sending $X+(X^2+3)$ to $\sqrt{-3}$.

Now consider the ideal $$\frac{(2+X,X^2+3)}{(X^2+3)}=\left(2+X+(X^2+3)\right)$$ of $\frac{\mathbb{Z}[X]}{(X^2+3)}$. What does it corresponds to under $\overline{\varphi}$? We compute it: $$\overline{\varphi}\left(\frac{(2+X,X^2+3)}{(X^2+3)}\right) = \overline{\varphi}\left(2+X+(X^2+3)\right)=\left(2+\sqrt{-3}\right).$$ We conclude that $$\frac{\frac{\mathbb{Z}[X]}{(X^2+3)}}{\frac{(2+X,X^2+3)}{(X^2+3)}} \cong \frac{\overline{\varphi}\left(\frac{\mathbb{Z}[X]}{(X^2+3)}\right)}{\overline{\varphi}\left(\frac{(2+X,X^2+3)}{(X^2+3)}\right)} = \frac{\mathbb{Z}\left[\sqrt{-3}\right]}{\left(2+\sqrt{-3}\right)}.$$

Don
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  • That does makes much more sense now, thank you. One question though: how do you know that $\frac{(2+X,;X^2+3)}{(X^2+3)}=\big(2+X+(X^2+3)\big)$? Is that an ideal inside an ideal? I'm not familiar with that notation... – nuuusxd Jun 14 '23 at 20:45
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    @nuuusxd When I wrote it I was afraid it may confuse the reader... With $2+X+(X^2+3)$ I only mean the equivalent class of $2+X$ in $\mathbb{Z}[X]/(X^2+3)$. Therefore, $(2+X+(X^2+3))$ is just the principal ideal generated by $2+X+(X^2+3)$ in $\mathbb{Z}[X]/(X^2+3)$. – Don Jun 14 '23 at 20:56
  • Hello once again, still on this. Could you please elaborate how did you obtain the equality $\frac{(2+X,X^2+3)}{(X^2+3)}=\left(2+X+(X^2+3)\right)$? It confuses me altogether with the fact that we're taking the quotient of an ideal, and not from a ring. – nuuusxd Dec 07 '23 at 19:55
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    @nuuusxd Hi! Consider the ideal $\big(2+X+(X^2+3)\big)$ of $\mathbb{Z}[X]/(X^2+3)$ generated by $2+X+(X^2+3)$. Recall that by the Correspondence Theorem, the ideals of $\mathbb{Z}[X]/(X^2+3)$ are in correspondence with the ideals of $\mathbb{Z}[X]$ containing $(X^2+3)$. Namely, the correspondence is given by $I \mapsto \pi^{-1}(I)$, with inverse $J \mapsto J/(X^2+3) := \pi(J)$ (here $\pi: \mathbb{Z}[X] \to \mathbb{Z}[X]/(X^2+3)$ is the natural homomorphism). In our case, we have the equality $\pi^{-1}\big(2+X+(X^2+3)\big) = (2+X,X^2+3)$, so $(2+X,X^2+3)/(X^2+3) = \big(2+X+(X^2+3)\big)$. – Don Dec 08 '23 at 09:31
  • One last question: regarding this expression, could it be that writing $\left(\mathbb{Z}[X]/(X^2+3)\right)/(2+X)$ was just a simplified way of typing $\left(\mathbb{Z}[X]/(X^2+3)\right)/\left(2+X+(X^2+3)\right)$, i.e, avoiding indicating the class? Thank you so much for your answers btw :) – nuuusxd Dec 11 '23 at 11:18
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    @nuuusxd Strictly speaking, writing $\big(\mathbb{Z}[X]/(X^2+3)\big)/(2+X)$ doesn't make sense, since $2+X$ is not an element of $\mathbb{Z}[X]/(X^2+3)$ but of $\mathbb{Z}[X]$. On the other hand, the element $2+X+(X^2+3)$ is an element of $\mathbb{Z}[X]/(X^2+3)$, and thus writing $\big(\mathbb{Z}[X]/(X^2+3)\big)/\big(2+X+(X^2+3)\big)$ makes sense. This being said, if you want to make the abuse of notation of writing $2+X$ instead of $2+X+(X^2+3) that's also fine, but then the convention should be stated explicitly. – Don Dec 11 '23 at 17:48
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Show that in this ring $X$ plays the role of $\sqrt{3}$ and compare dimensions by finding basis. The isomorphism you are looking for is not very hard to express with this piece of info.

julio_es_sui_glace
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