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I am checking if the ideals $$I=(5,x^3+2x+3), \qquad J=(4,x^2+x+1,x^2+x-1) \trianglelefteq \mathbb{Z}[x]$$ are either maximal or prime.

For the ideal $I$, by factorizing $x^3+2x+3=(x+1)(x^2-x+3)$ I discovered that $I \subseteq (5,x+1)$. Moreover I proved successfully that $I$ is indeed not maximal, since $1 \notin (5,x+1)$ and $x+1 \notin I$, which makes $(5,x+1)$ become a proper intermediate ideal between $I$ and $\mathbb{Z}[x]$. I guess that $I$ is prime because $x^2-x+3 \notin I$ but $(x+1)(x^2-x+3) \in I$. To prove this, I intend to show that there exists no $p(x),q(x) \in \mathbb{Z}[x]$ such that $$x^2-x+3=5p(x)+(x+1)(x^2-x+3)q(x) \qquad (*)$$ From ($*$) we have $(x^2-x+3) \mid 5p(x) \color{red}{\Rightarrow} (x^2-x+3) \mid p(x) \Rightarrow p(x)=(x^2-x+3)r(x)$. Substituting $p(x)$ in (*) we have $$1=5r(x)+(x+1)q(x),$$ which I have proved that it is false (there are no such $r(x)$ and $q(x)).

For the ideal $J$, by calculation I have $$\begin{array}{rcll} J &=& (x^2+x+1,x^2+x-1) & \text{(since } 4=2(x^2+x+1)-2(x^2+x-1) \text{)} \\ &=& (x^2+x+1,2) & \text{(since } 2=(x^2+x+1)-(x^2+x-1) \text{)} \end{array}$$

However, I have no idea to continue my effort.

Questions

  1. Is the red rightarrow "$(x^2-x+3) \mid 5p(x) \color{red}{\Rightarrow} (x^2-x+3) \mid p(x)$" valid?

  2. Are there some other ways to check that $I$ is maximal or prime? I think that $I$ is not maximal since we can factorize $x^3+2x+3$, and $J$ is not since $x^2+x+1$ is irreducible. I don't know if this idea can be generalized for arbitrary ideals in $\mathbb{Z}[x]$.

Thanks for your help.

Mystery girl
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  • Modding out first by $(2)$ gives you an ideal in $\mathbb{Z}_2[x]$ generated by a quadratic; it is easy to check if it is prime/maximal, since you have a Euclidean domain there. – Arturo Magidin Jul 04 '22 at 01:45
  • Do you mean modding out $x^2+x+1$ in $\mathbb{Z}_2[x]$? Ah, and could you tell me which question you are answering to me, since I have 2 questions? – Mystery girl Jul 04 '22 at 01:48
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    Having more than one question typically requires more than one question post to get good answers on this site. For your first question, does $x=2\to x^2-x+3=5$ matter for that implication? – abiessu Jul 04 '22 at 01:54
  • @abiessu Nope. | is the divisibility in $\mathbb{Z}[x]$. – Mystery girl Jul 04 '22 at 02:01
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    For $J$: check the quotient to see if the ideal is prime/maximal. And to mod out by $J=(2,x^2+x+1)$, you can first mod out by $(2)$, then mod out the result by the image of $(x^2+x+1)$; since moding out by $(2)$ drops you to $\mathbb{Z}_2[x]$, which is a Euclidean domain and your polynomials is degree $2$, checking whether $(x^2+x+1)$ is maximal/prime there is a trivial matter. – Arturo Magidin Jul 04 '22 at 02:03
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    Related threads 1, 2, 3. I think we have a classification of prime/maximal ideals of $\Bbb{Z}[x]$ here somewhere, but I don't have the time to search for it right now. – Jyrki Lahtonen Jul 06 '22 at 06:45
  • @JyrkiLahtonen Ah, thanks a lot :D Actually I have done the exercise, but I am still thinking about some general results about maximal and prime ideals in $\mathbb{Z}[x]$. Your sources may help a lot, I will read them more. :D – Mystery girl Jul 06 '22 at 07:01
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    A more definitive source, I think. – Jyrki Lahtonen Jul 06 '22 at 08:51

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