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I'm trying to show that the ideal $(x^3 + 2x + 5, x^2 + 1)$ is equal to the ideal $(x^2 + 1, p)$ for some prime $p$ (ideals in $\mathbb Z[x]$). My motivation is that I believe that $(x^3 + 2x + 5, x^2 + 1)$ is a maximal ideal and hence it must have the form that I am claiming it does. I just can't seem to find the prime that would do the trick. Thanks.

J. W. Tanner
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A.Z
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    You can try to prove that $(x^{3} + 2x + 5, x^{2} + 1) = (x + 5, x^{2} + 1) = (x + 5, 26)$, so $(x^{3} + 2x + 5, x^{2} + 1)$ is not a maximal ideal. – jhzg Jun 03 '24 at 01:17

2 Answers2

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I believe your conjecture is false. By writing $I = (x^3 + 2x + 5, x^2 + 1) = (x^3 + 2x + 5 - x(x^2+1), x^2+1) = (x+5, x^2+1)$, we see that in the quotient $\mathbb{Z}[x]/I$, we have $[x] = [-5]$ and therefore $[x^2+1] = [26]$, so this quotient is isomorphic to $\mathbb{Z}/26$, which is not even an integral domain, so your ideal is not even prime.

A. Thomas Yerger
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  • That's a typo. It should be $26$. – Just a user Jun 03 '24 at 01:29
  • Thanks, @Justauser. And in your (now deleted) answer, how did you know $I$ is not maximal? – J. W. Tanner Jun 03 '24 at 01:32
  • Every maximal ideal in $\mathbb{Z}[x]$ has the form $(f(x), p)$ where $p$ is prime integer and $f(x)$ is primitive integer polynomial that is irreducible modulo $p$. $(x^{3} + 2x + 5, x^{2} + 1) = (x + 5, 26)$ and $26$ is not a prime integer. – jhzg Jun 03 '24 at 01:35
  • I see, @jhzg; thank you – J. W. Tanner Jun 03 '24 at 01:44
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    @A. Thomas Yerger Yes it looks like I was wrong. But could you explain how one can show that the quotient is indeed isomorphic to $\mathbb Z / 26$ in a more rigorous way? – A.Z Jun 03 '24 at 01:45
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    Calculations like these should be thought of as kind of like scratch work, they tell you how to define a map $\mathbb{Z}[x] \to \mathbb{Z}/26$. We send $x \mapsto -5$ and $1 \mapsto 1$. Now you prove that the kernel of this map is precisely your ideal, and conclude via the first isomorphism theorem. Alternatively, you can make a map $\mathbb{Z}[x]/I \to \mathbb{Z}/26$, and argue that it is a bijection. In my experience, both are fine, but the latter is more fiddly because of the equivalence classes. – A. Thomas Yerger Jun 03 '24 at 02:13
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$\, 3\:\!$ other ways to see the inequality $[\!$ vs. Euclidean reducing $(f,x^2\!+\!1)\,$ to $\,(\color{#c00}x\!+\!\color{#0a0}5,26)\,]$ $ \begin{align} &[\![1]\!]\ \ f=x^3\!+\!2x\!+\!5\:\!\in\:\! I=(x^2\!+\!1,p)\!\iff\! \overbrace{f\bmod x^2\!+\!1}^{\!\!\!\!\small\textstyle \color{#c00}1\:\!x+\color{#0a0}5}\:\!\in I\!\iff\,\!p\mid \color{#c00}1,\,\color{#0a0}5.\\[.6em] &[\![2]\!]\ \ \text{${\bf Or}$ they are unequal in $\,\Bbb Z[x]/(x^2\!+\!1)\cong \Bbb Z[i]\!:\ (i\!+\!5)\neq (p)\,$ by $\,p\nmid\color{#c00}i\!+\!\color{#0a0}5;$}\\[.6em] &[\![3]\!]\ \ \text{${\bf Or}$ they are unequal in $\,\Bbb Z[x]/p\!:\,\ (f,x^2\!+\!1)\,\neq\, (x^2\!+\!1)\ $ by $\!\underbrace{x^2\!+\!1\nmid f}_{\textstyle f(i)= \color{#c00}i\!+\!\color{#0a0}5\neq 0\!\!\!\!\!\!\!\!\!\!\!\!\!}\,$} \end{align}$

See quotient reciprocity for how the prior two methods are related.

Bill Dubuque
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