Mixing
Exercse 4.1.1 of Hartshorne: find a rational function on a curve with a pole at one point.
I am struggling with exercise 4.1.1 of Hartshorne. The precise question is as follows:
Let $X$ be a curve (i.e. an integral scheme of dimension 1 that is proper over some algebraically closed field $k$ such that all its local rings are regular), and let $P in X$ be a (closed) point. Then there exists a nonconstant rational function $fin K(X)$ (the function field of $X$), which is regular everywhere except at the point $P$.
I wanted to prove this first for an affine scheme $X=Spec(A)$. Here, $A$ has to be an integral domain of Krull dimension 1 such that all its localizations are regular local rings and $P$ some non-zero prime ideal of $A$. I think I should find some $fin P$ such that $fnotin Q$ for all other prime ideals $Q$. In that case the rational function $frac{1}{f}in Quot(A)$ is a function satisfying the conditions above. (Mimicking the case $A=k[X]$, $P=(x-a)$, $f=x-a$.)
I believe that it would suffice to show that $P$ is principal to show that such $f$ exists. However, I am not sure that this is true.
Does anyone know how to proceed?
algebraic-geometry commutative-algebra
asked Nov 21 '18 at 10:35
PP123
1077
add a comment |
I am struggling with exercise 4.1.1 of Hartshorne. The precise question is as follows:
Let $X$ be a curve (i.e. an integral scheme of dimension 1 that is proper over some algebraically closed field $k$ such that all its local rings are regular), and let $P in X$ be a (closed) point. Then there exists a nonconstant rational function $fin K(X)$ (the function field of $X$), which is regular everywhere except at the point $P$.
I wanted to prove this first for an affine scheme $X=Spec(A)$. Here, $A$ has to be an integral domain of Krull dimension 1 such that all its localizations are regular local rings and $P$ some non-zero prime ideal of $A$. I think I should find some $fin P$ such that $fnotin Q$ for all other prime ideals $Q$. In that case the rational function $frac{1}{f}in Quot(A)$ is a function satisfying the conditions above. (Mimicking the case $A=k[X]$, $P=(x-a)$, $f=x-a$.)
I believe that it would suffice to show that $P$ is principal to show that such $f$ exists. However, I am not sure that this is true.
Does anyone know how to proceed?
algebraic-geometry commutative-algebra
asked Nov 21 '18 at 10:35
PP123
1077
2
Couple thoughts. Firstly, it's been a bit, but I'm suspicious about whether affine schemes can be proper over an algebraically closed field. The universally closed assumption in particular strikes me as not likely to be true for an affine scheme. Secondly, even if you could prove this for an affine scheme that gets you almost no closer to solving it in general, since the question is not a local one.
– jgon
Nov 21 '18 at 16:54
3
Also we have no idea what you're assuming here. This follows more or less trivially from Riemann-Roch as far as I can see.
– jgon
Nov 21 '18 at 17:01
1
As jgon says, Riemann-Roch (and really just Riemann's inequality) implies this immediately: $ell(nP) geq 1 - g + n$, so $mathscr{L}(nP)$ contains a nonconstant function for any $n geq g+1$.
– André 3000
Nov 22 '18 at 4:20
add a comment |
I am struggling with exercise 4.1.1 of Hartshorne. The precise question is as follows:
Let $X$ be a curve (i.e. an integral scheme of dimension 1 that is proper over some algebraically closed field $k$ such that all its local rings are regular), and let $P in X$ be a (closed) point. Then there exists a nonconstant rational function $fin K(X)$ (the function field of $X$), which is regular everywhere except at the point $P$.
I wanted to prove this first for an affine scheme $X=Spec(A)$. Here, $A$ has to be an integral domain of Krull dimension 1 such that all its localizations are regular local rings and $P$ some non-zero prime ideal of $A$. I think I should find some $fin P$ such that $fnotin Q$ for all other prime ideals $Q$. In that case the rational function $frac{1}{f}in Quot(A)$ is a function satisfying the conditions above. (Mimicking the case $A=k[X]$, $P=(x-a)$, $f=x-a$.)
I believe that it would suffice to show that $P$ is principal to show that such $f$ exists. However, I am not sure that this is true.
Does anyone know how to proceed?
algebraic-geometry commutative-algebra
asked Nov 21 '18 at 10:35
PP123
1077
I am struggling with exercise 4.1.1 of Hartshorne. The precise question is as follows:
Let $X$ be a curve (i.e. an integral scheme of dimension 1 that is proper over some algebraically closed field $k$ such that all its local rings are regular), and let $P in X$ be a (closed) point. Then there exists a nonconstant rational function $fin K(X)$ (the function field of $X$), which is regular everywhere except at the point $P$.
I wanted to prove this first for an affine scheme $X=Spec(A)$. Here, $A$ has to be an integral domain of Krull dimension 1 such that all its localizations are regular local rings and $P$ some non-zero prime ideal of $A$. I think I should find some $fin P$ such that $fnotin Q$ for all other prime ideals $Q$. In that case the rational function $frac{1}{f}in Quot(A)$ is a function satisfying the conditions above. (Mimicking the case $A=k[X]$, $P=(x-a)$, $f=x-a$.)
I believe that it would suffice to show that $P$ is principal to show that such $f$ exists. However, I am not sure that this is true.
Does anyone know how to proceed?
algebraic-geometry commutative-algebra
algebraic-geometry commutative-algebra
asked Nov 21 '18 at 10:35
PP123
1077
asked Nov 21 '18 at 10:35
PP123
1077
asked Nov 21 '18 at 10:35
PP123
1077
asked Nov 21 '18 at 10:35
PP123
1077
asked Nov 21 '18 at 10:35
PP123
1077
1077
2
Couple thoughts. Firstly, it's been a bit, but I'm suspicious about whether affine schemes can be proper over an algebraically closed field. The universally closed assumption in particular strikes me as not likely to be true for an affine scheme. Secondly, even if you could prove this for an affine scheme that gets you almost no closer to solving it in general, since the question is not a local one.
– jgon
Nov 21 '18 at 16:54
3
Also we have no idea what you're assuming here. This follows more or less trivially from Riemann-Roch as far as I can see.
– jgon
Nov 21 '18 at 17:01
1
As jgon says, Riemann-Roch (and really just Riemann's inequality) implies this immediately: $ell(nP) geq 1 - g + n$, so $mathscr{L}(nP)$ contains a nonconstant function for any $n geq g+1$.
– André 3000
Nov 22 '18 at 4:20
add a comment |
2
Couple thoughts. Firstly, it's been a bit, but I'm suspicious about whether affine schemes can be proper over an algebraically closed field. The universally closed assumption in particular strikes me as not likely to be true for an affine scheme. Secondly, even if you could prove this for an affine scheme that gets you almost no closer to solving it in general, since the question is not a local one.
– jgon
Nov 21 '18 at 16:54
3
Also we have no idea what you're assuming here. This follows more or less trivially from Riemann-Roch as far as I can see.
– jgon
Nov 21 '18 at 17:01
1
As jgon says, Riemann-Roch (and really just Riemann's inequality) implies this immediately: $ell(nP) geq 1 - g + n$, so $mathscr{L}(nP)$ contains a nonconstant function for any $n geq g+1$.
– André 3000
Nov 22 '18 at 4:20
2
2
Couple thoughts. Firstly, it's been a bit, but I'm suspicious about whether affine schemes can be proper over an algebraically closed field. The universally closed assumption in particular strikes me as not likely to be true for an affine scheme. Secondly, even if you could prove this for an affine scheme that gets you almost no closer to solving it in general, since the question is not a local one.
– jgon
Nov 21 '18 at 16:54
Couple thoughts. Firstly, it's been a bit, but I'm suspicious about whether affine schemes can be proper over an algebraically closed field. The universally closed assumption in particular strikes me as not likely to be true for an affine scheme. Secondly, even if you could prove this for an affine scheme that gets you almost no closer to solving it in general, since the question is not a local one.
– jgon
Nov 21 '18 at 16:54
3
3
Also we have no idea what you're assuming here. This follows more or less trivially from Riemann-Roch as far as I can see.
– jgon
Nov 21 '18 at 17:01
Also we have no idea what you're assuming here. This follows more or less trivially from Riemann-Roch as far as I can see.
– jgon
Nov 21 '18 at 17:01
1
1
As jgon says, Riemann-Roch (and really just Riemann's inequality) implies this immediately: $ell(nP) geq 1 - g + n$, so $mathscr{L}(nP)$ contains a nonconstant function for any $n geq g+1$.
– André 3000
Nov 22 '18 at 4:20
As jgon says, Riemann-Roch (and really just Riemann's inequality) implies this immediately: $ell(nP) geq 1 - g + n$, so $mathscr{L}(nP)$ contains a nonconstant function for any $n geq g+1$.
– André 3000
Nov 22 '18 at 4:20
add a comment |
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2
Couple thoughts. Firstly, it's been a bit, but I'm suspicious about whether affine schemes can be proper over an algebraically closed field. The universally closed assumption in particular strikes me as not likely to be true for an affine scheme. Secondly, even if you could prove this for an affine scheme that gets you almost no closer to solving it in general, since the question is not a local one.
– jgon
Nov 21 '18 at 16:54
3
Also we have no idea what you're assuming here. This follows more or less trivially from Riemann-Roch as far as I can see.
– jgon
Nov 21 '18 at 17:01
1
As jgon says, Riemann-Roch (and really just Riemann's inequality) implies this immediately: $ell(nP) geq 1 - g + n$, so $mathscr{L}(nP)$ contains a nonconstant function for any $n geq g+1$.
– André 3000
Nov 22 '18 at 4:20