Any maximal ideal intersect its proper subring

Question

Let M is maximal ideal in $\mathbb{Z}[x]$ then is it true that M intersect with $\mathbb{Z}$ non trivially?

Any hint is appreciated.

Thanks

See this question. Also see your previous question. – Dietrich Burde Jun 18 '19 at 20:38 — Dietrich Burde, Jun 18 '19 at 20:38

score 3 · Answer 1 · answered Jun 18 '19 at 20:37

3

By the classification of maximal ideals of $\mathbb Z[x]$ the maximal ideals all contain a prime $p$ of $\mathbb Z$. So yes.

answered Jun 18 '19 at 20:37

rschwieb

153,510

No,, you cannot use that because I want to prove classification of maximal ideal that is why I need this. So any other process plz – Pradip Jun 18 '19 at 20:45
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@Siraj Having withheld vital context from your question, you are in no position to object to the method. – rschwieb Jun 18 '19 at 22:08

score 0 · Answer 2 · answered Jun 18 '19 at 21:05

It is shown in this reference that maximal ideals in $Z[x]$ are all of the form $\langle p, f(x) \rangle$, where $p \in \Bbb Z$ is a prime and $f(x) \in Z[x]$ is irreducible $\mod p$. Thus a maximal ideal $M \subset Z[x]$ contains a prime and hence $M \cap \Bbb Z \ne \emptyset$.

score 0 · Answer 3 · answered Jun 19 '19 at 12:49

It seems that you are trying to use this result to prove the classification of maximal ideals. To this end, note that if $M$ is a maximal ideal of $\mathbb Z[x]$, then $F:=\mathbb Z[x]/M$ is a field, and is finitely generated as a $\mathbb Z$-algebra. This cannot happen for $\mathbb Q$, so the field has positive characteristic, and hence $M$ contains some prime $p\in\mathbb Z$.

Any maximal ideal intersect its proper subring

3 Answers3