8
$\begingroup$

Let $A$ be a commutative ring and $B = S^{-1}A$ be its localization with respect to a certain multiplicative subset of $A$.

Consider the contraction (in $A$) of colon ideals and ideal sums and ideal products (in $B$) as long as they make sense.

Do contracted ideals still possess the original characteristics?

That is, will the contraction of colon ideals (resp. of sums, resp. of products) in $B$ be colon ideals (resp. sums, resp. products) of the corresponding contracted ideals in $A$?

I suspect there are counterexamples if $A$ is not noetherian, but I have no idea how to tackle this.

(Thanks for pointing out obscurity. I hope this time it is more legible.)

  • 0
    Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/36405/discussion-between-richard-and-subhadeep-dey).2016-03-01

1 Answers 1

5

For sums:

No. Let $R=K[X,Y]$, $f=X+Y$, $S=\{1,f,f^2,\dots\}$, $I=(X)$, and $J=(Y)$. Then $S^{-1}I\cap R=I$ and $S^{-1}J\cap R=J$ (since $I,J$ are prime ideals and $I\cap S=J\cap S=\emptyset$), while $(S^{-1}I+S^{-1}J)\cap R=S^{-1}(I+J)\cap R=S^{-1}R\cap R=R$ (since $(I+J)\cap S\ne\emptyset$).

For products:

No. Let $R=K[X,Y,Z]/(XY-Z^2)$, $S=\{1,y,y^2,\dots\}$, and $I=J=(x,z)$. Then $S^{-1}I\cap R=I$ (since $I$ is a prime ideal and $I\cap S=\emptyset$), while $(S^{-1}IS^{-1}J)\cap R=S^{-1}I^2\cap R\supsetneq I^2$ since $x=\frac{xy}{y}=\frac{z^2}{y}\in S^{-1}I^2\cap R$ and $x\notin I^2$.

For colons:

Yes.

$(S^{-1}I:S^{-1}J)\cap R=(S^{-1}I\cap R):(S^{-1}J\cap R).$

  • 0
    Thank you very much for the answer. Especially, the counter examples are wonderful.2016-03-13