ideal sheaf modulo square of ideal sheaf equals restriction?
$begingroup$
Assume we have schemes $Yhookrightarrow X$. Then we can define the ideal sheaf of $Y$ in $X$, denoted $I=I_{Y/X}$. I read at several stages that $I/I^2=Imid_Y$. 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!
abstract-algebra algebraic-geometry
$endgroup$
add a comment |
$begingroup$
Assume we have schemes $Yhookrightarrow X$. Then we can define the ideal sheaf of $Y$ in $X$, denoted $I=I_{Y/X}$. I read at several stages that $I/I^2=Imid_Y$. 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!
abstract-algebra algebraic-geometry
$endgroup$
add a comment |
$begingroup$
Assume we have schemes $Yhookrightarrow X$. Then we can define the ideal sheaf of $Y$ in $X$, denoted $I=I_{Y/X}$. I read at several stages that $I/I^2=Imid_Y$. 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!
abstract-algebra algebraic-geometry
$endgroup$
Assume we have schemes $Yhookrightarrow X$. Then we can define the ideal sheaf of $Y$ in $X$, denoted $I=I_{Y/X}$. I read at several stages that $I/I^2=Imid_Y$. 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!
abstract-algebra algebraic-geometry
abstract-algebra algebraic-geometry
asked Dec 3 '18 at 9:40
user620217
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1 Answer
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$begingroup$
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
$$
I/I^2 cong I otimes O_Y cong Ivert_Y.
$$
answered Dec 3 '18 at 11:45
SashaSasha
4,518139
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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
$$
I/I^2 cong I otimes O_Y cong Ivert_Y.
$$
answered Dec 3 '18 at 11:45
SashaSasha
4,518139
$endgroup$
add a comment |
$begingroup$
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
$$
I/I^2 cong I otimes O_Y cong Ivert_Y.
$$
answered Dec 3 '18 at 11:45
SashaSasha
4,518139
$endgroup$
add a comment |
$begingroup$
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
$$
I/I^2 cong I otimes O_Y cong Ivert_Y.
$$
answered Dec 3 '18 at 11:45
SashaSasha
4,518139
$endgroup$
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
$$
I/I^2 cong I otimes O_Y cong Ivert_Y.
$$
answered Dec 3 '18 at 11:45
SashaSasha
4,518139
answered Dec 3 '18 at 11:45
SashaSasha
4,518139
answered Dec 3 '18 at 11:45
SashaSasha
4,518139
answered Dec 3 '18 at 11:45
SashaSasha
4,518139
4,518139
add a comment |
add a comment |
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