Let $R$$\subset$$ S$ be commutative noetherian rings,and $I$ is an ideal of $S$.
We now that $I$ is a $R-$module.
Do we have $grade_{R}(I)$ $\le$ $ grade_{S}(I)$?
Thank you!
$\begingroup$
$\endgroup$
15
-
4$\begingroup$ What do you mean by "an ideal of $R$ and $S$"? $\endgroup$– abxCommented Jan 2, 2017 at 7:39
-
$\begingroup$ Sorry, I mean that $R$ is a subring of $S$. $\endgroup$– Paulo RossiCommented Jan 2, 2017 at 18:23
-
2$\begingroup$ But it is very unlikely that an ideal in $R$ is also an ideal in $S$... $\endgroup$– abxCommented Jan 2, 2017 at 19:32
-
1$\begingroup$ Could you give me just one example where this situation occurs? $\endgroup$– abxCommented Jan 2, 2017 at 20:21
-
1$\begingroup$ For an ideal of $R$ which is also an ideal of $S$. $\endgroup$– abxCommented Jan 2, 2017 at 20:44
|
Show 10 more comments
1 Answer
$\begingroup$
$\endgroup$
0
This is false. Take any $R$-module $M$ of grade $>0$, and put $S=R\oplus M$, the product of any two elements of $M$ being $0$. Then $M$ is an ideal of $S$ and $\mathrm{grade}_S(M)=0\ $ (because $\mathrm{Hom}_S(S/M,S)\neq 0$).
