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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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    @uncookedfalcon. Yeah,$I$was worried about this: primes go to primes?$I$think you're right; under pullback yes. Thanks.2012-08-14

1 Answers 1

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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.2012-08-14