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I have to prove that

$$\langle x^2-x-1 \rangle \text{ is not a maximal ideal of } \mathbb{Z}[x].$$

My attempt: I am not that much familiar with field theory but i know that an ideal $I$ is maximal over a commutative ring $R$ iff $\displaystyle\frac{R}{I}$ is a field ,but here $\mathbb{Z[x]}$ is a UFD and given ideal is irreduicble over $\mathbb{Z[x]}$ so it is prime and thus $\displaystyle\frac{\mathbb{Z[x]}}{\langle x^2-x-1 \rangle }$ is an integral domain ,but don't know about is it field or not?

user26857
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TheStudent
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    Can you find a larger proper ideal, for example what do you get when quotienting by $2$ ? Otherwise what is the characteristic of $\Bbb{Z}[x]/(x^2-x-1)$ and does it contain the smallest field of that characteristic ? – reuns Dec 14 '19 at 04:58
  • Quotienting by 2 mean $\frac{\mathbb{Z[x]}}{<2>}$ which is $\mathbb{Z_{2}[x]}$ – TheStudent Dec 14 '19 at 05:02
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    Quotienting the quotient not $\Bbb{Z}[x]$ – reuns Dec 14 '19 at 05:03
  • is itl $\frac{\mathbb{Z_{2}[x]}}{<x^2-x-1>}$ ? i just guessed it . – TheStudent Dec 14 '19 at 05:05
  • Alternatively you could define or construct it by $\mathbb Z[x]/\langle 2,x^2-x-1 \rangle$. – hardmath Dec 14 '19 at 05:08
  • Is $\mathbb{Z_{2}[x]}/(x^2-x-1)$ the zero ring or is it a true ring, if so what does it means about the ideal $(2,x^2-x-1)$ – reuns Dec 14 '19 at 05:17
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    ok here this ideal is contained in $(2,(x^2-x-1))$ ideal so the given ideal is not maximal (as per the defination of maximal ideal) – TheStudent Dec 14 '19 at 05:21

1 Answers1

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The ideal is not maximal, because $ (2, x^2-x-1) $ is a proper ideal properly containing it.

To see it is a proper ideal, note that there do not exist $f,g\in\Bbb Z[x]$ with $f(x)\cdot 2+g(x)\cdot(x^2-x+1)= 1$.

To see that the containment is proper note that $2\not\in(x^2-x+1)$.

  • what can we say about ideal $(3,x^2-x-1)$ is this ideal also contain the given ideal? – TheStudent Dec 14 '19 at 05:23
  • I believe it will also work, as $(3, x^2-x+1)\ne(1)$. –  Dec 14 '19 at 05:31
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    That is, of course it contains the given ideal. And it also possesses the other features necessary to show the given ideal is not maximal. –  Dec 14 '19 at 05:40
  • @JyrkiLahtonen thanks. Foiled again. May as well just go with $1$. –  Dec 14 '19 at 10:00