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I am looking for a bit of help on part of this problem.

Let $R = \mathbb{Z}\lbrack X \rbrack$. For each prime $p \in \mathbb{Z}$, let $I_p$ denote the ideal of $R$ generated by $p$ and $X^2+1$.

(a) Is $I_2$ a prime ideal of $R$? What are the ideals of $R$ containing $I_2$?

(b) Is $I_3$ a prime ideal of $R$? What are the ideals of $R$ containing $I_3$?

For (b):

Consider the quotient ring $R/I_3$. This ring is isomorphic to $\mathbb{Z}_3 \lbrack X\rbrack / (X^2+1)$. For all values $m \in \mathbb{Z}_3$, $(m^2 + 1)$ is not congruent to 0 mod 3, hence $X^2 + 1$ is irreducible in $\mathbb{Z}_3 \lbrack X \rbrack$. Since $\mathbb{Z}_3$ is a field, $\mathbb{Z}_3\lbrack x \rbrack$ is a PID, $(X^2+1)$ is a maximal ideal, hence it is also prime. Since it is maximal, $\mathbb{Z}_3/(X^2+1)$ is a field, so $(3,X^2+1)$ is a maximal ideal of $R$, and hence it is prime. Since $I_2$ is maximal, there are no other proper ideals which contain it.

For (a), I am not sure what to do. In $\mathbb{Z}_2$, $(X^2+1) = (X+1)(X+1)$, both of which aren't in the ideal $(2,X^2+1)$, so the ideal is not prime. However, I'm not sure how to find an ideal which contains $I_2$. I'm also not sure how to rigorously show that $(X+1) \notin I_3$. I would appreciate any help.

Important_man74
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  • Maybe this can help: https://math.stackexchange.com/questions/174595/classification-of-prime-ideals-of-mathbbzx – Temoi Jan 03 '24 at 23:35
  • b) should be ok. For a): Take quotients $\mathbb{Z}[X]/(2,X^2+1) \cong (\mathbb{Z}[X]/(2))/((2,X^2+1)/(2))\cong \mathbb{Z}_2[X]/(X^2+1)=\mathbb{Z}_2[X]/(X+1)^2$ since $(X+1)^2 = X^2+1$ in $\mathbb{Z}_2$. So $\mathbb{Z}_2[X]/(X^2+1)$ is not an integral domain and therefore $(2,X^2+1)$ not prime in $\mathbb{Z}[X]$. – psl2Z Jan 03 '24 at 23:52

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(b) looks good. For (a), good start! All of the answers come from your factorisation. For a hint, use this to get a principal ideal containing $I_2$.

As for a rigorous demonstration of $x+1\not\in I_2$: suppose $x+1\in I_2$. Then $x+1=2f+(x^2+1)g$ for $f,g\in \mathbb{Z}[x]$. Let's see what is wrong here. Well, reducing modulo $2$, $x+1=(x^2+1)\overline{g}\bmod 2$ which is not possible due to the degrees (I'm writing a line on top, as in $\overline{g}$, to mean reduce all coefficients mod $2$). This is a contradiction, so $x+1\not\in I_2$ in fact!

Joshua Tilley
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  • Well the ideal $(X+1)$ in $\mathbb{Z}_2\lbrack X \rbrack$ definitely contains $(X^2+1)$. I think I am familiar with a result that says that at polynomial of degree 1 in a polynomial ring over a PID is irreducible, so the ideal $(X+1)$ would be maximal and contain $(X^2+1)$? – Important_man74 Jan 04 '24 at 03:43
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    Yes, that's right. Now extract the ideal of $R$ containing $I_2$ that corresponds to this one and you have an example. Are there more? – Joshua Tilley Jan 04 '24 at 11:36
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    So in $R$ it would be the ideal $(2, X+1)$ right? I don't think there are any others, since they would need to contain $2$ and $X+1$ but $(2, X+1)$ is maximal. – Important_man74 Jan 04 '24 at 21:34
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    Well, there are two more but they are trivial. Namely, $R$ itself and $I_2$ itself. You can reason as follows: the ideals in question correspond to the ideals of $\mathbb{F}_2[x]$ containing $x^2+1$. As this ring is a PID, all its ideals are principal, so this is equivalent to looking for factors of $x^2+1$ which are $1$, $x+1$, and $x^2+1$ leading to the three ideals mentioned. – Joshua Tilley Jan 05 '24 at 11:39
  • Oh that makes sense. Thanks a lot this has really helped me understand this! – Important_man74 Jan 05 '24 at 19:15