Prove that $\mathbb{Z}[x]$ doesn't have principal maximal ideals.
Please, I need help with this problem. Thanks!
Asked
Active
Viewed 897 times
3
-
Given any principal ideal, you can easily find a larger ideal. Take the two cases of principal ideals which are generated by constants and by non-constants. – Thomas Andrews May 23 '16 at 16:30
-
The maximal ideals of $\mathbb{Z[x]}$ are of the form $(p, f(x))$ where $f(x)$ is an irreducible polynomial modulo $p$. Now assume that this ideal is generated by a signle polynomial and you would derive a contradiction. – May 23 '16 at 17:26
-
Duplicate of http://math.stackexchange.com/questions/1591558/nonconstant-polynomials-do-not-generate-maximal-ideals-in-mathbb-zx/1591594#1591594 – user26857 Aug 26 '16 at 13:29
1 Answers
6
Here is a detailed outline:
Let $I$ be a principal ideal in $\mathbb{Z}[x]$. Now there is some $f(x)\in \mathbb{Z}[x]$ such that $\langle f(x)\rangle=I$.
If $\deg(f(x))=0$, then $f(x)$ is a constant $c$. If $c=0$ or $c=1$, then $I$ cannot be maximal. (why?) Now assume $c\neq 0,1$ and show that $\langle c\rangle \subsetneq \langle c, x \rangle \subsetneq \mathbb{Z}[x]$. Thus $I$ is not maximal.
If $\deg(f(x))>0$, then $f(x)=a_0+a_1x+\cdots + a_nx^n$ for some $n>0$. Now show that $\langle f(x)\rangle \subsetneq \langle f(x), q \rangle \subsetneq \mathbb{Z}[x]$, where $q$ is a prime that does not divide $a_n$. Thus $I$ is not maximal.
M47145
- 4,106
-
Could you kindly explain why $\langle f(x), q \rangle \subsetneq \mathbb{Z}[x]$ ? – Sameer Kulkarni May 23 '16 at 19:14
-
@SameerKulkarni Note that $\mathbb{Z}[x]/\langle f(x),q\rangle = \mathbb{Z}_q[x]/\langle f(x) \rangle $ which isn't the trivial ring. – M47145 May 23 '16 at 20:06
-
See also this answer. Let me mention that this proof works for $R[X]$, where $R$ is a UFD having infinitely many primes. (+1) – user26857 Aug 26 '16 at 13:28