Proper pushforward
1
I am studying a some intersection theory. Now I am stuck on the proof of the Lemma 02R6 here. More precisely, I don't understand why the coeffiecient of $Z$ in $[f_astmathcal{F}]_k$ is equal to $text{length}_{R_{mathfrak p}}(M_{mathfrak p})$. Can you help me?
algebraic-geometry proof-explanation
asked Nov 21 '18 at 16:46
Vincenzo Zaccaro
1,239719
add a comment |
1
I am studying a some intersection theory. Now I am stuck on the proof of the Lemma 02R6 here. More precisely, I don't understand why the coeffiecient of $Z$ in $[f_astmathcal{F}]_k$ is equal to $text{length}_{R_{mathfrak p}}(M_{mathfrak p})$. Can you help me?
algebraic-geometry proof-explanation
asked Nov 21 '18 at 16:46
Vincenzo Zaccaro
1,239719
1
This just follows from the definition of $[f_*mathcal{F}]$ - note here $mathfrak{p}$ is the generic point of $Z$.
– loch
Nov 21 '18 at 19:19
I don't understand because it is not $deg(Z/f(Z))text{length}_{R_{mathfrak p}}(M_{mathfrak p})$. Can you expand your answer?
– Vincenzo Zaccaro
Nov 21 '18 at 19:22
1
The degree of $f$ doesn't really matter here - it would show up if you were computing $f_*[mathcal{F}]_k$ ( and in particular you see this expression later), but here you're looking at $[f_*mathcal{F}]_k$! In other words what's happening is you have a coherent sheaf ($mathcal{G}:= f_*mathcal{F})$ on $Y$ and you're just trying to compute its corresponding cycle.
– loch
Nov 21 '18 at 19:32
add a comment |
1
1
1
I am studying a some intersection theory. Now I am stuck on the proof of the Lemma 02R6 here. More precisely, I don't understand why the coeffiecient of $Z$ in $[f_astmathcal{F}]_k$ is equal to $text{length}_{R_{mathfrak p}}(M_{mathfrak p})$. Can you help me?
algebraic-geometry proof-explanation
asked Nov 21 '18 at 16:46
Vincenzo Zaccaro
1,239719
I am studying a some intersection theory. Now I am stuck on the proof of the Lemma 02R6 here. More precisely, I don't understand why the coeffiecient of $Z$ in $[f_astmathcal{F}]_k$ is equal to $text{length}_{R_{mathfrak p}}(M_{mathfrak p})$. Can you help me?
algebraic-geometry proof-explanation
algebraic-geometry proof-explanation
asked Nov 21 '18 at 16:46
Vincenzo Zaccaro
1,239719
asked Nov 21 '18 at 16:46
Vincenzo Zaccaro
1,239719
asked Nov 21 '18 at 16:46
Vincenzo Zaccaro
1,239719
asked Nov 21 '18 at 16:46
Vincenzo Zaccaro
1,239719
asked Nov 21 '18 at 16:46
Vincenzo Zaccaro
1,239719
1,239719
1
This just follows from the definition of $[f_*mathcal{F}]$ - note here $mathfrak{p}$ is the generic point of $Z$.
– loch
Nov 21 '18 at 19:19
I don't understand because it is not $deg(Z/f(Z))text{length}_{R_{mathfrak p}}(M_{mathfrak p})$. Can you expand your answer?
– Vincenzo Zaccaro
Nov 21 '18 at 19:22
1
The degree of $f$ doesn't really matter here - it would show up if you were computing $f_*[mathcal{F}]_k$ ( and in particular you see this expression later), but here you're looking at $[f_*mathcal{F}]_k$! In other words what's happening is you have a coherent sheaf ($mathcal{G}:= f_*mathcal{F})$ on $Y$ and you're just trying to compute its corresponding cycle.
– loch
Nov 21 '18 at 19:32
add a comment |
1
This just follows from the definition of $[f_*mathcal{F}]$ - note here $mathfrak{p}$ is the generic point of $Z$.
– loch
Nov 21 '18 at 19:19
I don't understand because it is not $deg(Z/f(Z))text{length}_{R_{mathfrak p}}(M_{mathfrak p})$. Can you expand your answer?
– Vincenzo Zaccaro
Nov 21 '18 at 19:22
1
The degree of $f$ doesn't really matter here - it would show up if you were computing $f_*[mathcal{F}]_k$ ( and in particular you see this expression later), but here you're looking at $[f_*mathcal{F}]_k$! In other words what's happening is you have a coherent sheaf ($mathcal{G}:= f_*mathcal{F})$ on $Y$ and you're just trying to compute its corresponding cycle.
– loch
Nov 21 '18 at 19:32
1
1
This just follows from the definition of $[f_*mathcal{F}]$ - note here $mathfrak{p}$ is the generic point of $Z$.
– loch
Nov 21 '18 at 19:19
This just follows from the definition of $[f_*mathcal{F}]$ - note here $mathfrak{p}$ is the generic point of $Z$.
– loch
Nov 21 '18 at 19:19
I don't understand because it is not $deg(Z/f(Z))text{length}_{R_{mathfrak p}}(M_{mathfrak p})$. Can you expand your answer?
– Vincenzo Zaccaro
Nov 21 '18 at 19:22
I don't understand because it is not $deg(Z/f(Z))text{length}_{R_{mathfrak p}}(M_{mathfrak p})$. Can you expand your answer?
– Vincenzo Zaccaro
Nov 21 '18 at 19:22
1
1
The degree of $f$ doesn't really matter here - it would show up if you were computing $f_*[mathcal{F}]_k$ ( and in particular you see this expression later), but here you're looking at $[f_*mathcal{F}]_k$! In other words what's happening is you have a coherent sheaf ($mathcal{G}:= f_*mathcal{F})$ on $Y$ and you're just trying to compute its corresponding cycle.
– loch
Nov 21 '18 at 19:32
The degree of $f$ doesn't really matter here - it would show up if you were computing $f_*[mathcal{F}]_k$ ( and in particular you see this expression later), but here you're looking at $[f_*mathcal{F}]_k$! In other words what's happening is you have a coherent sheaf ($mathcal{G}:= f_*mathcal{F})$ on $Y$ and you're just trying to compute its corresponding cycle.
– loch
Nov 21 '18 at 19:32
add a comment |
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1
This just follows from the definition of $[f_*mathcal{F}]$ - note here $mathfrak{p}$ is the generic point of $Z$.
– loch
Nov 21 '18 at 19:19
I don't understand because it is not $deg(Z/f(Z))text{length}_{R_{mathfrak p}}(M_{mathfrak p})$. Can you expand your answer?
– Vincenzo Zaccaro
Nov 21 '18 at 19:22
1
The degree of $f$ doesn't really matter here - it would show up if you were computing $f_*[mathcal{F}]_k$ ( and in particular you see this expression later), but here you're looking at $[f_*mathcal{F}]_k$! In other words what's happening is you have a coherent sheaf ($mathcal{G}:= f_*mathcal{F})$ on $Y$ and you're just trying to compute its corresponding cycle.
– loch
Nov 21 '18 at 19:32