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I'm doing an elementary commutative algebra / number theory exercise, the goal is to show that $A=\mathbb{Z}[\alpha]$ ($\alpha = \frac{1+i\sqrt{19}}{2}$) is a principal ideal domain which is non euclidean.

I've done most of it but I did skip one question that I couldn't quite figure out. I am asked to show that $A/2A$ and $A/3A$ are fields. My first instinct was showing $2A$ and $3A$ were maximal ideals.

First I was able to show that $2$ and $3$ are irreducible in $A$, this tells me that $2A$ and $3A$ are maximal in the set $Princ(A)\setminus\left\lbrace A\right\rbrace$ (the set of principal ideals minus the ideal A ). But at this point, I'm not supposed to know $A$ is a PID so I can't conclude from that because I still don't know that these ideals are maximal among all ideals of A. Any clues on how to go from there ?

t_kln
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  • One idea is to see your ring as ${\mathbb Z}[x]/(x^2-x+5)$. Going further modulo $2$ gives ${\mathbb Z}[x]/(x^2-x+5,2)$ and the idea is to show that this is a field. There are quite a few questions and answers on the site that give this idea. – Magdiragdag Nov 25 '22 at 22:35
  • @Magdiragdag Damn.. the first thing I proved was indeed that my ring was isomorphic to $\mathbb{Z}[x]/(x^2-x+5)$ ... But then I forgot to ever use it... thanks – t_kln Nov 25 '22 at 22:39
  • See here and its links for more on the entire exercise. – Bill Dubuque Nov 26 '22 at 07:07
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    You can use quotient reciprocity as here. – Bill Dubuque Nov 26 '22 at 07:18

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Let $\alpha = \frac{1}{2} + \frac{i\sqrt{19}}{2}$ and $\bar{\alpha} = \frac{1}{2} - \frac{i\sqrt{19}}{2}$. Note that $\bar{\alpha} \alpha = 5$ and $\alpha + \bar{\alpha} = 1$. Therefore $\alpha$ is a root of the monic polynomial $(x - \alpha) (x - \bar{\alpha}) = x^2 - x +5$.

The kernel of the 'evaluation' ring map $\mathbb{Z}[x] \rightarrow \mathbb{Z}[\alpha]$ contains the polynomial $x^2 - x + 5$, so it factors through the ring $B = \mathbb{Z}[x]/(x^2 - x + 5)$.

Thus we get a surjection $B = \mathbb{Z}[x]/(x^2 - x + 5) \twoheadrightarrow A = \mathbb{Z}[\alpha]$. This induces surjections $B/2B \twoheadrightarrow A/2A$ and $B/3B \twoheadrightarrow A/3A$.

Even though you don't know much about $A$ yet, the quotients $B/2B$ and $B/3B$ should be easy to understand. Can you take it from here?

Badam Baplan
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  • Very helpful thanks I actually had already shown that $A\simeq \mathbb{Z}[x]/(x^2-x+5)$... I'll work out the rest thank you – t_kln Nov 25 '22 at 22:40
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    Sure thing. The point I wanted to lead you to is just that it is helpful to look at the ring $B= Z[x]/(f)$ where $f$ is the minimal polynomial of $\alpha$. Even without knowing $B =A$, it's easy to get information about the maximal ideals of $A$ from the maximal ideals of $B$, because there are no non-trivial surjections from fields. – Badam Baplan Nov 26 '22 at 00:41