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!
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$\begingroup$ I think if A is a PID with only finitely many primes, then there is a maximal ideal of A[x] which intersects A trivially. $\endgroup$– Steve DMay 20, 2010 at 16:23
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1$\begingroup$ Perhaps I should clarify that when I write "I think" I mean "I know", and the two statements are equivalent over PIDs. This includes Emerton's example below. $\endgroup$– Steve DMay 20, 2010 at 17:44
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3 Answers
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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 $A$.)
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7$\begingroup$ This is related to a mistake I once saw in a preprint by a well-known mathematician: for sep'td scheme $X$ of finite type over dvr $R$, it was said that a closed point has to be in the special fiber. Above is standard counterexample, and what's really happening is that the affine line is open in the projective line over dvr, and there are $\mathbf{Z}_ p$-points whose special fiber is infinity but generic fiber is in affine line over $\mathbf{Q}_ p$. So closed section meets open affine line over dvr only in generic pt of section, "explaining" closedness of that pt in affine line over dvr. $\endgroup$– BCnrdMay 20, 2010 at 17:23
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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.
answered May 20, 2010 at 18:10
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$\begingroup$ Thank you for making your notes available! I've never seen the material on Goldman domains before. There is a typo in line $-2$ of the proof of Prop 122: inverse missing. $\endgroup$ May 22, 2010 at 14:33
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$\begingroup$ Thank you for the positive feedback and for pointing out the typo (I have corrected it in my "home" copy, so it will get uploaded with the next revision / addition.) Let me recommend once again Kaplansky's book: it has a lot of material that I have not found in other texts on the subject. $\endgroup$ May 23, 2010 at 0:06
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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
answered Jun 10, 2010 at 3:53