This question already has an answer here:

Consider $R=\mathbb{Z}[x]$. Also let $p$ be a prime. Then we want to find all the prime and maximal ideals of $\mathbb{Z}[x]$. The prime ideals are $(0), (p), (x)$ and $(ap + bx)$. Then we see that the last one contains all the previous ones making it a maximal ideal for $\mathbb{Z}[x]$. Is this the correct way to think about it?

Also, since $(ap + bx)$ is a maximal ideal this means that $R/$(ap + bx)$ is a field. How can I see that this quotient is indeed a field?

Finally, how could I find the same for $\mathbb{Q}$ and then for a generic $k$?


marked as duplicate by Dietrich Burde, rschwieb abstract-algebra Feb 12 '16 at 16:21

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Note that $\mathbb{Z}[x]$ is not a PID, $\endgroup$ – Dietrich Burde Feb 12 '16 at 16:21
  • $\begingroup$ No, your characterization of prime ideals is wrong. For one thing, you're saying all the maximal ideals are principal, which in a Noetherian ring would mean that all ideals are principal, but $\Bbb Z[x]$ is not a PID. There are other prime ideals like $(2, x+1)$. $\endgroup$ – rschwieb Feb 12 '16 at 16:21
  • $\begingroup$ In answer to your coat-tail question, for a field $k$, $k[x]$ is a principal ideal domain, and there the primes are exactly $\{0\}$ and $(f(x))$ where $f(x)$ is any polynomial irreducible over $k$. The nonzero prime ideals are the maximal ideals. $\endgroup$ – rschwieb Feb 12 '16 at 16:22
  • $\begingroup$ Could you make this more concrete for $\mathbb{Q}[x]$ or $\mathbb{R}[x]$. As for my characterization for prime ideals I mean it only for this case I don't see where I am wrong. It is indeed true that $(ap+bx)$ contains the others, right? $\endgroup$ – Marion Feb 12 '16 at 16:32