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The following is a question I have asked here without receiving any comments, therefore I post it here:

Let $A \subseteq B$ be commutative rings, $m$ a maximal ideal of $A$. When $mB \neq B$?

This is true when $A \subseteq B$ is faithfully flat. (If I am not wrong, this is also true when $A \subseteq B$ is integral). Any other ideas are welcome.

Please notice: A similar (but not identical, I think) question is: https://math.stackexchange.com/questions/194261/when-does-mathfrakab-cap-a-mathfraka/194306#194306, since the property I am talking about is slightly more general then $IB \cap A =I$, for every ideal $I$ of $A$. Indeed, let $m$ be a maximal ideal of $A$. Then, in particular, $mB \cap A=m$ and if $mB=B$ we would get $A= B \cap A = mB \cap A=m$, a contradiction to the maximality of $m$. So, "$IB \cap A =I$ implies $mB \neq B$". (An exercise in Atiyah-MacDonald, which was mentioned in the second answer of https://math.stackexchange.com/questions/194261/when-does-mathfrakab-cap-a-mathfraka/194306#194306 shows that for a flat extension those two properties are equivalent).

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  • $\begingroup$ Question is not so fundamental. Let A be ANY field, so m = 0 is its maximal ideal. And for any extension B, we have mB = 0. $\endgroup$
    – Algirdas Rugys
    Commented Aug 6, 2015 at 13:26
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    $\begingroup$ Tautologically, your condition means that $m$ belongs to the image of the morphism $\mathrm{Spec}(B)\rightarrow \mathrm{Spec}(A)$. This morphism is surjective indeed if $B$ is faithfully flat or integral over $A$; it is not in other situations, e.g. if $B$ is a localization of $A$. I don't think you can say much more. $\endgroup$
    – abx
    Commented Aug 6, 2015 at 14:06
  • $\begingroup$ Thank you very much. Please, do you claim that TFAE: (1) $mB \neq B$ for every maximal ideal $m$ of $A$. (2) For every prime ideal $p$ of $A$ there exists a prime ideal $q$ of $B$ such that $p= q \cap A$. $\endgroup$
    – user237522
    Commented Aug 6, 2015 at 15:43
  • $\begingroup$ No, I didn't say that. $\endgroup$
    – abx
    Commented Aug 6, 2015 at 15:58
  • $\begingroup$ Thanks. Did you mean: If $mB \neq B$, then there exists a prime ideal $q$ of $B$ such that $m=q \cap A$? $\endgroup$
    – user237522
    Commented Aug 6, 2015 at 21:58

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Here are a few comments that may help.

  1. As abx explains, a good way to think about it is in terms whether or not $m$ lies in image of the map on spectra.

  2. As you surmised, either condition of faithful flatness or integrality are sufficient,

  3. but not necessary! For example, take $A=k[x,y]$ and the affine blow up $B=k[x,y,z]/(y-zt)$. I'll let you convince yourself that it is surjective on spectra but neither flat nor integral (think about the fibres).

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  • $\begingroup$ Thanks!! When I wrote "Any other ideas are welcome", I meant that I suspect there exist other cases (=not f.f. not integral) where $mB \neq B$; your nice example is (probably) one of them, so it may serve as an answer. However, it would be even nicer to know if one can characterize all (commutative) ring extensions $A \subseteq B$ such that $mB \neq B$. Namely, if one can say that faithfully flat, integral and ... are precisely the extensions that satisfy $mB \neq B$. That's what I originally tried to ask (and hoped there exists a paper on this question, which I am not aware of) $\endgroup$
    – user237522
    Commented Aug 8, 2015 at 23:21
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This is a bit of a non-answer, but it doesn't fit in a comment, so I put it here. You write "If I am not wrong, this is also true when $A\subseteq B$ is integral".

I think you are mistaken here. Let $B = \mathbb{R}$ (which as a field is clearly an integral domain) and $A = \mathbb{Z} \subseteq \mathbb{R}$. This is a maximal ideal of $A = \mathbb{Z}$ (but of course it is not an ideal of $B=\mathbb{R}$ because $\mathbb{R}$ as a field doesn't have non-trivial ideals).

Let $m = 2\mathbb{Z}\subseteq A$. Then we have $\frac{1}{2}\in \mathbb{R}=B$ and therefore $\frac{1}{2}\cdot 2 = 1\in Bm = mB$ therefore $mB = B$.

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    $\begingroup$ I think you are confused about the meaning of "integral". It means that any element of $B$ is annihilated by a monic polynomial with coefficients in $A$. $\mathbb{R}$ is certainly not integral over $\mathbb{Z}$. $\endgroup$
    – abx
    Commented Aug 6, 2015 at 14:03

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