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The setup is that $a \in \mathbb C$ is a root of $g(x)=x^3 - x + 1$. I'm trying to argue that the quotient $\mathbb Z[a] / (a+2)$ is isomorphic to $\mathbb Z_5$. I know that given our hypothesis, $\mathbb Z[a]$ must be a finitely generated $\mathbb Z$-module, with basis $\{1,a,a^2\}$. The strategy is to show that the map $f : \mathbb Z[a] \to \mathbb Z_5$ given by $f(r_1 + r_2a + r_3a^2) = r_1 -2r_2 + 4r_3$ is an onto homomorphism with kernel $(a+2)$. I'm having trouble showing that the kernel is contained in $(a+2)$. Any hints would be appreciated.

Bill Dubuque
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A.Z
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  • Right this is where the intuition behind the map came from. I just need to make the argument rigorous. – A.Z May 08 '24 at 23:57
  • I’m a little confused. The map I defined is from Z[a] to Z? – A.Z May 09 '24 at 00:08
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    Here is an observation that might be helpful. We know $a^3 = a - 1$. Therefore, $[a^3] = [-8]$ and $[a-1] = [-3]$. We get that $[-8] = [-3]$, it follows that $[0]$ and $[5]$ are the same class in the quotient. – Nicolas Bourbaki May 09 '24 at 00:33

3 Answers3

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Note that $x^3-x+1$ has a root in $\mathbb{Z_5}$, which is $3$. So this gives an idea to define $\varphi:\mathbb{Z}[a]\to\mathbb{Z_5}$ by $f(a)\to\bar{f}(3)$, where $\bar{f}\in\mathbb{Z_5}[x]$ is the reduction of $f$ mod $5$, i.e we take all the coefficients mod $5$. The map $\varphi$ is well defined (check this!), and is clearly a surjective ring homomorphism. We want to show that its kernel is the ideal $(a+2)$ of $\mathbb{Z}[a]$.

It is obvious that $(a+2)$ is contained in the kernel. Conversely, assume $f\in\mathbb{Z}[x]$ is a polynomial such that $f(a)\in\text{Ker}(\varphi)$. We can divide $f$ by $x+2$ with remainder in $\mathbb{Z}[x]$. (possible because $x+2$ has an invertible leading coefficient) So there are $q\in\mathbb{Z}[x]$ and $r\in\mathbb{Z}$ such that $f(x)=(x+2)q(x)+r$. Then $f(a)=(a+2)q(a)+r$, and hence:

$0=\varphi(f(a))=\varphi(a+2)\varphi(q(a))+\varphi(r)=0+\varphi(r)=r \pmod{5},$

which means that $5|r$. So it remains to show that $5\in (a+2)$, and then it will follow that $f(a)=(a+2)q(a)+5s$ (where $r=5s$) belongs to $(a+2)$ as well.

To show that $5\in (a+2)$, note first that $a^3+3=[a+2]+[a^3-a+1]=a+2\in (a+2)$. Also, $a^3+8$ is divisible by $a+2$ ($-2$ is a root of the polynomial $x^3+8$), and so belongs to $(a+2)$ as well. Thus, $5=[a^3+8]-[a^3+3]\in (a+2)$, as required.

Arturo Magidin
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Mark
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Let $\bar x$ be the image of $x$ in $\mathbb{Z}[x]/(x^3-x+1)$ and consider $$\mathbb{Z}[x]\overset{f}{\to}\mathbb{Z}[x]/(x^3-x+1)\overset{g}{\to}\frac{\mathbb{Z}[x]/(x^3-x+1)}{(\bar x +2)}.$$ The $\ker g=(\bar x +2)$ and $\ker f=(x^3-x+1)$ so $\ker gf=(x+2,x^3-x+1)=(x+2,5)$ since $x^3-x+1=-5+x^2(x+2)-2x(x+2)+3(x+2)$ and hence $$\mathbb{Z}[x]/(x+2,5)\cong\frac{\mathbb{Z}[x]/(x^3-x+1)}{(\bar x +2)}$$ now consider $$\mathbb{Z}[x]\to\mathbb{Z}[x]/(x+2,5)$$ which factors $$\mathbb{Z}_5[x]\cong\mathbb{Z}[x]/(5)\to\mathbb{Z}[x]/(x+2,5)$$ and the kernel of this map is $(x+2)$ so we get $$\mathbb{Z}_5[x]/(x+2)\cong\mathbb{Z}[x]/(x+2,5)$$ but $$\mathbb{Z}_5\cong\mathbb{Z}_5[x]/(x+2)$$ so finally $$\mathbb{Z}_5\cong\frac{\mathbb{Z}[x]/(x^3-x+1)}{(\bar x +2)}.$$

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Hint: by quotient reciprocity it is $\:\!\cong\:\! \Bbb (\color{#0a0}{\Bbb Z[x]/(x\!+\!2)})\:\!/\:\!(\underbrace{g(x)/(x\!+\!2)}_{\large g(-2)\ =\ \color{#c00}{-5}})\cong \color{#0a0}{\Bbb Z}/\color{#c00}5,\,$ by here.

Bill Dubuque
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