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!



Depth or Grade of an ideal
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 $\mathrm{Hom}\_S(S/M,S)\neq 0$).


