Maximal ideal in polynomial ring Is it true that the intersection of a maximal ideal in $A[x]$ with $A$ is a maximal ideal in $A$?
Let's say A is Noetherian. I would be surprised if it isn't true but somehow I can't seem to show it. Any help or tip will be appreciated. Thanks!
 A: See also my sci.math post [1] for some further discussion and references
[1] sci.math, 22 Apr 2009
http://groups.google.com/group/sci.math/msg/b00f5e1f7397842f
http://google.com/groups?selm=y8zfxg12ckl.fsf%40nestle.csail.mit.edu
A: No, it's not true in general.  E.g. the pricipal ideal generated by $px - 1$ is maximal
in $\mathbb Z_p[x]$ (for any prime $p$); the quotient $\mathbb Z_p[x]/(p x - 1)$ is precisely the field $\mathbb Q_p$.  However, the intersection of this ideal with $\mathbb Z_p$ is equal to the zero
ideal, which is not maximal.
If the ring $A$ is Jacobson, then the result you want is true.  (For in this case, if $\mathfrak m$ is a maximal ideal in $A[x]$, then
$A[x]/\mathfrak m$ is a finite type $A$-algebra which is a field, therefore is finite over $A$
by the Jacobson hypothesis, and so the image of $A$ in $A[x]/\mathfrak m$ (which is equal to
$A/A\cap \mathfrak m$) is itself a field, and so $A\cap \mathfrak m$ is maximal in $A$.)
A: To amplify on Emerton's answer:
An integral domain $R$ is a Goldman domain if its fraction field is finitely generated as an $R$-algebra.  A prime ideal $\mathfrak{p}$ in a commutative ring is a Goldman ideal if $R/\mathfrak{p}$ is a Goldman domain.  A commutative ring $R$ is a Hilbert ring if every Goldman ideal is maximal.
It is not hard to show that a one-dimensional Noetherian domain is a Hilbert ring iff it has infinitely many prime ideals.  In particular, $\mathbb{Z}$ is a Hilbert domain and $\mathbb{Z}_p$ is not a Hilbert domain.
Theorem: For a commutative ring $R$, the following are equivalent:
(i) $R$ is a Hilbert ring.
(ii) $R$ is a Jacobson ring -- every prime ideal is the intersection of maximal ideals.
(iii) For every maximal ideal $\mathfrak{m}$ of $R[t]$, $\mathfrak{m} \cap R$ is a maximal ideal of $R$.
I learned this theorem from Kaplansky's book on commutative rings.  For the proof, see Theorem 132 on p. 84 of
http://alpha.math.uga.edu/~pete/integral.pdf
Putting these two results together, Emerton's counterexample can be predicted.
Also see the Exercise on p. 113 of my notes, in which this issue comes up in the context of computing the spectrum of $R[t]$, where $R$ is a PID.
