Maximal ideal of $\mathbb{Z}_{n}[x]$ if $n$ is composite

Question

How should i think a maximal ideal of $\mathbb{Z}_{n}[x]$ if $n$ is composite number?

For example, is it possible to say that $\mathbb{Z}_{6}[x]/(2,x)\cong\mathbb{Z}_{6}/(2)\cong\mathbb{Z}_{2}$?

I wonder how to generalize this property.

Give some advice. Thank you!

Break up $\mathbb{Z}_n$ using the Chinese remainder theorem. Alternatively, you can characterize the maximal ideals of $\mathbb{Z}[x]$, which isn't so bad. — Qiaochu Yuan, Oct 03 '19 at 04:47

Arthur · Accepted Answer · 2019-10-03T06:22:31.020

2

Since dividing out by a maximal ideal $I$ gives a field, $\Bbb Z_n/(I\cap \Bbb Z_n)$ must be $\Bbb Z_p$ for some prime $p$ that divides $n$. So $I\cap \Bbb Z_n$ must be $(p)$. And $\Bbb Z_p$ is a principle ideal domain, so over $\Bbb Z_p$, $I$ reduces to $(f)$ for some $f\in\Bbb Z_p$. And if $I$ is maximal, that means $f$ is irreducible.

So any maximal ideal in $\Bbb Z_n[x]$ is of the form $(f,p)$ where $p$ is a prime that divides $n$, and $f$ becomes irreducible in $\Bbb Z_n[x]/(p)\cong \Bbb Z_p[x]$.

edited Oct 03 '19 at 06:22

answered Oct 03 '19 at 04:53

Arthur

199,419

@JyrkiLahtonen You're right. My form always gives maximal ideals, but it doesn't give all maximal ideals. – Arthur Oct 03 '19 at 05:56
@JyrkiLahtonen It's early morning here too, actually. And regardless of time zone, it's easy to miss things. – Arthur Oct 03 '19 at 06:05

Maximal ideal of $\mathbb{Z}_{n}[x]$ if $n$ is composite

1 Answers1