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let, $R = \mathbb{Z}[x]$, and then we let $I = (x)$, the ideal generated by $x$. Let, $ab \in (x)$, then $\deg(a) + \deg(b) \ge 1$, which implies $\deg(a)$ or $\deg(b)$ are greater than or equal to $1$; hence, $a \in I$ or $b \in I$. We have that, $I \subset (2,x),$ since $I$ does not contain $2$.

Does my proof look good? Any issues? Thanks.

Tobias Rieper
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  • Maybe writing $I\subsetneq (2,x)$ is better notation. – Kenta S Jul 19 '22 at 17:04
  • In addition to the perceptive answer given by Lost in Space, to complete your argument, it would also be necessary to show that $(2,X) \neq R$. Otherwise, $I \subset (2,X)$ does not necessarily imply non-maximality. – legionwhale Jul 19 '22 at 18:24
  • $1\notin (x, 2) $ . It's trivial – Sourav Ghosh Jul 20 '22 at 01:37
  • @LostinSpace 1) it's not super trivial, 2) even if it was, it is worth mentioning. – Kenta S Jul 21 '22 at 19:17
  • $(x, 2) $ contains all polynomial with even constant terms. Not all polynomials have this feature. It's trivial. – Sourav Ghosh Jul 21 '22 at 19:20
  • @LostinSpace But still worth mentioning, even if it is stated without justification. – legionwhale Jul 22 '22 at 18:03
  • I assumed this as trivial because op already mentioned in the post about the ideal $(x, 2) $. So i belive op can prove it's a proper ideal. Otherwise it's a good exercise to make it trivial. – Sourav Ghosh Jul 22 '22 at 18:09
  • @legionwhale Feel free to edit my answer to provide more details if you want. – Sourav Ghosh Jul 22 '22 at 18:11
  • @LostinSpace My comment was never a criticism of your answer. Your answer is concise and provides all of the relevant information. I was only reminding OP that it would be better to remark on this fact when writing a proof (as that's what they said they were doing). Apologies for any misunderstanding. – legionwhale Jul 22 '22 at 18:12
  • @legionwhale You have made no mistake. Don't need to feel sorry. If you wish, you can edit my answer to look better. I know "trivial" is a very dangerous word. But it will be a good exercise for the readers. Thanks. – Sourav Ghosh Jul 22 '22 at 18:17

1 Answers1

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$p\in (x) $ implies the constant term of $p$ is zero.

By proving $\deg(a) \ge 1$ is not sufficient to conclude that $a\in (x) $

$pq\in (x) $ implies $p(x) q(x) =xh(x) $ for some $h(x) \in\Bbb{Z}[x]$ .

Then the constant term of $p(x) q(x) $ is zero implies at least one polynomial $p$ or $q$ has constant term $0$ . Hence $p$ or $q$ one of them must be in $(x) $

Alt: $I$ is an ideal of a commutative ring $R$ with $1$. Then

  1. $I$ is prime iff ${R}/{I}$ is an integral domain.

  2. $I$ is maximal iff ${R}/{I}$ is a field.

$\Bbb{Z}[x]/(x) \cong \Bbb{Z}$

Hence $(x) $ is prime but not maximal.

Sourav Ghosh
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