Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I want to prove the following result:

Let $R$ be a ring and $M$ a maximal ideal in $R$. If $P$ is a prime ideal in $R[x]$ that (strictly) contains $M[x]$, then $P$ is a maximal ideal in $R[x]$.

I have an idea how to prove that, but I'm not quite sure if the argument is totally valid; maybe someone has a cleaner proof. My argument is like that:

  1. $R[x]/M[x]$ is isomorphic to $(R/M)[x]$, that is a PID (since $R/M$ is a field). So, any prime ideal in this ring is maximal;
  2. If $P$ contains $M[x]$, then $P$ corresponds to a (prime??) ideal in $R[x]/M[x]$ (this needs more clarification);
  3. The prime ideal (that corresponds to $P$) in $R[x]/M[x]$ is then maximal, and this guarantees that $P$ is maximal in $R[x]$.

I'll appreciate any comment.

share|cite|improve this question
In 2, your argument is valid because if $P[x]$ contains $M[x]$, then $P[x] / M[x]$, as an ideal of $R[x] / M[x]$, must be prime because $(R/M)[x]$ is a PID. The other two parts are already clear. – Tunococ Aug 14 '12 at 2:29
Yeah man this makes sense to me. Let $A$ be a ring, $I$ an ideal. Then ideals of $A/I$ correspond to ideals of $A$ containing $I$ under pullback under $q: A \rightarrow A/I$. I think of 'correspond' pretty much as 'are': primes go to primes, maximals go to maximals, and inclusion relations are respected! So your argument is baller! – uncookedfalcon Aug 14 '12 at 2:31
You should define $P$ before talking about $P[x]$. – Georges Elencwajg Aug 14 '12 at 6:24
@uncookedfalcon. Yeah, I was worried about this: primes go to primes? I think you're right; under pullback yes. Thanks. – fmoura2005 Aug 14 '12 at 15:37
up vote 7 down vote accepted

In (2), your argument is valid because of the following more general fact:

Let $f : A \to B$ be a surjective ring homomorphism. If $P$ is a prime ideal of $A$ that contains $\ker f$, then $f(P)$ is a prime ideal in $B$.

Proof: We have $\ker f \subseteq P \subseteq A$. So by the third isomorphism theorem, we have that $P/\ker f$ is an ideal of $A/\ker(f)$. Furthermore, we have that

$$(A/\ker f)/(P/\ker f) \cong A/P.$$

The latter is an integral domain because $P$ is a prime ideal, this proves that $P/\ker f$ is a prime ideal in $A/\ker f$. Furthermore by the first isomorphism theorem you know that because $f$ is surjective,

$$A/\ker f \cong B.$$

It follows that because $P/\ker f$ is a prime ideal in $A/\ker f$ that $f(P)$ is a prime ideal in $B$.

$\hspace{6in} \square$

Once you know this, you just need to apply it with $A = R[x]$, $B = R[x]/M[X]$, $f = \pi$ where $\pi$ is the canonical projection of $R[x]$ onto the quotient $R[X]/M[x]$ and its pre-image $\pi^{-1}$ define bijective correspondences between prime ideals in $R[x]$ that contain $M[x]$ and prime ideals in $B$. The correspondence is as follows:

$$\{P' \in \textrm{Spec}(B) \implies \pi^{-1}(P) \text{ a prime ideal in $A$ that contains} \hspace{2mm} M[x]\}$$ $$\updownarrow$$ $$\{P \hspace{2mm} \text{a prime ideal in $A$ that contains $M[x]$} \implies \pi(P) \in \textrm{Spec}(B) \}$$

share|cite|improve this answer
Thanks. That's exactly what I needed. – fmoura2005 Aug 14 '12 at 15:40

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.