ideal sheaf modulo square of ideal sheaf equals restriction?

Question

Assume we have schemes $Y\hookrightarrow X$. Then we can define the ideal sheaf of $Y$ in $X$, denoted $I=I_{Y/X}$. I read at several stages that $I/I^2=I\mid_Y$. There never is given any kind of reference so it is probably a very stupid question... Anyway, I just don't see why this holds at the moment. Please help!

Answer 1

If you tensor the standard exact sequence $$ 0 \to I \to O_X \to O_Y \to 0 $$ with $I$, you get $$ I \otimes I \to I \to I \otimes O_Y \to 0. $$ Note that the first map takes $f \otimes g$ to $fg$, hence its image is $I^2$. This proves that $$ I/I^2 \cong I \otimes O_Y \cong I\vert_Y. $$