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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.

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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
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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
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1 Answer

up vote 5 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:

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

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Thanks. That's exactly what I needed. –  fmoura2005 Aug 14 '12 at 15:40
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