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edited Apr 13, 2017 at 12:19

The following is a question I have asked here 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: http://math.stackexchange.com/questions/194261/when-does-mathfrakab-cap-a-mathfraka/194306#194306 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 http://math.stackexchange.com/questions/194261/when-does-mathfrakab-cap-a-mathfraka/194306#194306 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).

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: http://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 http://math.stackexchange.com/questions/194261/when-does-mathfrakab-cap-a-mathfraka/194306#194306 shows that for a flat extension those two properties are equivalent).

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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asked Aug 6, 2015 at 10:27

user237522

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When $mB \neq B$? $m$ is a maximal ideal of $A$, $A \subseteq B$

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: http://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 http://math.stackexchange.com/questions/194261/when-does-mathfrakab-cap-a-mathfraka/194306#194306 shows that for a flat extension those two properties are equivalent).