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Consider the quotient ring $\mathbb{Z}[x]/(x^2-3, 2x+4)$. Sometimes we can get identifications of rings like these with other more familiar rings if we first kill $x^2-3$ in the ring $\mathbb{Z}[x]$ and then kill $2x+4$ or vise versa. The substitution homomorphism $\mathbb{Z}[x] \rightarrow \mathbb{Z}[-2]$ doesn't seem to be relevant since its ideal is $x + 2$ and not $2x+4$. Working with $x^2 -3$ though I got that $\mathbb{Z}[x]/(x^2-3)\cong \mathbb{Z}[\sqrt{3}]$ and since the substitution homomorphism is surjective, we have that $\mathbb Z[x](x^2-3, 2x+4) \cong \mathbb{Z}[\sqrt{3}]/(2\sqrt{3} + 4)$, making use of the correspondence theorem. But my use isn't good enough since I thought the identification would be a more familiar ring.

Daniel Donnelly
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  • Hint $,\ \color{#c00}{x^2\equiv 3},,\ 2(2!+!x)\equiv 0 ,\Rightarrow, 0\equiv 2(2!+!x)(2!-!x) \equiv 2(4!-!\color{#c00}{x^2}) \equiv 2\ $ therefore $\tag*{}$ $\ (x^2!-!3,2x+4) = (x^2!-!3,2) = ((x!+!1)^2,2)\ \ $ – Bill Dubuque May 04 '15 at 15:49

1 Answers1

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Hint: Show that $2+\sqrt{3}$ is a unit in $\mathbb{Z}[\sqrt{3}]$. You could do this either by

  • finding its inverse explicitly, or
  • showing that the norm map $N:\mathbb{Z}[\sqrt{3}]\to\mathbb{Z}$ defined by $$N(a+b\sqrt{3})=a^2-3b^2$$ is multiplicative, and deducing from this that an element of $\mathbb{Z}[\sqrt{3}]$ is a unit if and only if its norm is $\pm 1$.

Since $2+\sqrt{3}$ is a unit, the ideal $(2\sqrt{3}+4)$ is the same as the ideal $(2)$. Now observe that $$\begin{align} \mathbb{Z}[\sqrt{3}]/(2)&\cong \mathbb{Z}[x]/(x^2-3,2)\\\\ &\cong \mathbb{F}_2[x]/(x^2-3)\\\\ &=\mathbb{F}_2[x]/(x^2+1)\\\\ &=\mathbb{F}_2[x]/(x+1)^2\\\\ &\cong\mathbb{F}_2[y]/(y^2) \end{align}$$ where $\mathbb{F}_2=\mathbb{Z}/2\mathbb{Z}$ is the field with two elements.

Zev Chonoles
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