Degree of a Map between $k$ Schemes
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Let $S$ a surface (therefore a $2$ dimensional, proper $k$-scheme) and $C$ a curve (" $1$ dim " ). Suppose that both schemes are irreducible so we can talk about canonically defined function fields as localizations at generic points.
Consider a morphism $f:S to C$.
How to see that in this case $deg(f)=0$?
The degree of $f$ is defined as field extension degrees $$deg(f):= [k(S):f^*k(C)]$$
for function fields $k(S) $ of $S$ (resp $k(C)$ of $C$).
Since $S$ is a surface and $B$ a curve intuitively I guess that then $k(S)$ is an infinite extension of $f^*k(C)$ of transcendence degree $1$, therefore we must have $deg(f)= infty$.
Where is the error in my reasonings?
algebraic-geometry schemes
asked Aug 19 at 11:49
KarlPeter
544313
add a comment |Â
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0
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Let $S$ a surface (therefore a $2$ dimensional, proper $k$-scheme) and $C$ a curve (" $1$ dim " ). Suppose that both schemes are irreducible so we can talk about canonically defined function fields as localizations at generic points.
Consider a morphism $f:S to C$.
How to see that in this case $deg(f)=0$?
The degree of $f$ is defined as field extension degrees $$deg(f):= [k(S):f^*k(C)]$$
for function fields $k(S) $ of $S$ (resp $k(C)$ of $C$).
Since $S$ is a surface and $B$ a curve intuitively I guess that then $k(S)$ is an infinite extension of $f^*k(C)$ of transcendence degree $1$, therefore we must have $deg(f)= infty$.
Where is the error in my reasonings?
algebraic-geometry schemes
asked Aug 19 at 11:49
KarlPeter
544313
2
Where is your definition of degree from? Usually degree is a concept associated to a finite morphism.
– Tabes Bridges
Aug 19 at 19:14
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $S$ a surface (therefore a $2$ dimensional, proper $k$-scheme) and $C$ a curve (" $1$ dim " ). Suppose that both schemes are irreducible so we can talk about canonically defined function fields as localizations at generic points.
Consider a morphism $f:S to C$.
How to see that in this case $deg(f)=0$?
The degree of $f$ is defined as field extension degrees $$deg(f):= [k(S):f^*k(C)]$$
for function fields $k(S) $ of $S$ (resp $k(C)$ of $C$).
Since $S$ is a surface and $B$ a curve intuitively I guess that then $k(S)$ is an infinite extension of $f^*k(C)$ of transcendence degree $1$, therefore we must have $deg(f)= infty$.
Where is the error in my reasonings?
algebraic-geometry schemes
asked Aug 19 at 11:49
KarlPeter
544313
Let $S$ a surface (therefore a $2$ dimensional, proper $k$-scheme) and $C$ a curve (" $1$ dim " ). Suppose that both schemes are irreducible so we can talk about canonically defined function fields as localizations at generic points.
Consider a morphism $f:S to C$.
How to see that in this case $deg(f)=0$?
The degree of $f$ is defined as field extension degrees $$deg(f):= [k(S):f^*k(C)]$$
for function fields $k(S) $ of $S$ (resp $k(C)$ of $C$).
Since $S$ is a surface and $B$ a curve intuitively I guess that then $k(S)$ is an infinite extension of $f^*k(C)$ of transcendence degree $1$, therefore we must have $deg(f)= infty$.
Where is the error in my reasonings?
algebraic-geometry schemes
asked Aug 19 at 11:49
KarlPeter
544313
asked Aug 19 at 11:49
KarlPeter
544313
asked Aug 19 at 11:49
KarlPeter
544313
asked Aug 19 at 11:49
KarlPeter
544313
544313
2
Where is your definition of degree from? Usually degree is a concept associated to a finite morphism.
– Tabes Bridges
Aug 19 at 19:14
add a comment |Â
2
Where is your definition of degree from? Usually degree is a concept associated to a finite morphism.
– Tabes Bridges
Aug 19 at 19:14
2
2
Where is your definition of degree from? Usually degree is a concept associated to a finite morphism.
– Tabes Bridges
Aug 19 at 19:14
Where is your definition of degree from? Usually degree is a concept associated to a finite morphism.
– Tabes Bridges
Aug 19 at 19:14
add a comment |Â
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2
Where is your definition of degree from? Usually degree is a concept associated to a finite morphism.
– Tabes Bridges
Aug 19 at 19:14