Desingularisation of curves (Lorenzini-Invitiation to Arithmetic Geometry, chap 6,ex 7)
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Given a nonsingular complete curve over algebraically closed $bar{k}$, which is interpreted as a field $bar{k}(X)$ of transcendence degree 1 and its set of valuations trivial on $bar{k}$, we may choose elements $x,y$ such that $x$ is trancendental, and $bar{k}(X)=bar{k}(x)(y)$. Letting $F$ be the homogenization of the minimal polynomial of $y$ over $bar{k}(x)$, using these coordinates we get a function from the set of valuations to the points of the projective curve $X_F(bar{k})$, given by mapping a valuation $v$ to the unique point $P$ of $X_F(bar{k})$ such that $O_P subset O_v$ and inclusion of maximal ideals.
The first question is to show that if $x,y$ are in $O_v$, then the valuation $v$ is mapped to is $(x(v):y(v):1)$, where $x(v),y(v)$ are the cosets of $x,y$ in $O_v / M_vcong bar{k}$.
For this, my thought was that if $psi =G/H$ with $H(x(v):y(v):1)neq 0$, then we get the coset of $H(x,y,1)$ in $O_v / M_v$ is nonzero, so we can conclude $G/H$ is defined at $v$, in the sense of having nonnegative valuation.
I feel okay with this, and the homogenisation/dehomogenisation is just from our isomorphism of fields $bar{k}(X)cong bar{k}(X_F)$, but the next part I am feeling stuck, which is to show that if $ynotin O_v, x/yin O_v$, then the point is $((x/y)(v):1:0)$.
Any assistance with how to think about this would also be very much appreciated.
algebraic-geometry algebraic-curves projective-geometry valuation-theory
asked Jan 5 at 23:49
user277182user277182
561212
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add a comment |
$begingroup$
Given a nonsingular complete curve over algebraically closed $bar{k}$, which is interpreted as a field $bar{k}(X)$ of transcendence degree 1 and its set of valuations trivial on $bar{k}$, we may choose elements $x,y$ such that $x$ is trancendental, and $bar{k}(X)=bar{k}(x)(y)$. Letting $F$ be the homogenization of the minimal polynomial of $y$ over $bar{k}(x)$, using these coordinates we get a function from the set of valuations to the points of the projective curve $X_F(bar{k})$, given by mapping a valuation $v$ to the unique point $P$ of $X_F(bar{k})$ such that $O_P subset O_v$ and inclusion of maximal ideals.
The first question is to show that if $x,y$ are in $O_v$, then the valuation $v$ is mapped to is $(x(v):y(v):1)$, where $x(v),y(v)$ are the cosets of $x,y$ in $O_v / M_vcong bar{k}$.
For this, my thought was that if $psi =G/H$ with $H(x(v):y(v):1)neq 0$, then we get the coset of $H(x,y,1)$ in $O_v / M_v$ is nonzero, so we can conclude $G/H$ is defined at $v$, in the sense of having nonnegative valuation.
I feel okay with this, and the homogenisation/dehomogenisation is just from our isomorphism of fields $bar{k}(X)cong bar{k}(X_F)$, but the next part I am feeling stuck, which is to show that if $ynotin O_v, x/yin O_v$, then the point is $((x/y)(v):1:0)$.
Any assistance with how to think about this would also be very much appreciated.
algebraic-geometry algebraic-curves projective-geometry valuation-theory
asked Jan 5 at 23:49
user277182user277182
561212
$endgroup$
add a comment |
$begingroup$
Given a nonsingular complete curve over algebraically closed $bar{k}$, which is interpreted as a field $bar{k}(X)$ of transcendence degree 1 and its set of valuations trivial on $bar{k}$, we may choose elements $x,y$ such that $x$ is trancendental, and $bar{k}(X)=bar{k}(x)(y)$. Letting $F$ be the homogenization of the minimal polynomial of $y$ over $bar{k}(x)$, using these coordinates we get a function from the set of valuations to the points of the projective curve $X_F(bar{k})$, given by mapping a valuation $v$ to the unique point $P$ of $X_F(bar{k})$ such that $O_P subset O_v$ and inclusion of maximal ideals.
The first question is to show that if $x,y$ are in $O_v$, then the valuation $v$ is mapped to is $(x(v):y(v):1)$, where $x(v),y(v)$ are the cosets of $x,y$ in $O_v / M_vcong bar{k}$.
For this, my thought was that if $psi =G/H$ with $H(x(v):y(v):1)neq 0$, then we get the coset of $H(x,y,1)$ in $O_v / M_v$ is nonzero, so we can conclude $G/H$ is defined at $v$, in the sense of having nonnegative valuation.
I feel okay with this, and the homogenisation/dehomogenisation is just from our isomorphism of fields $bar{k}(X)cong bar{k}(X_F)$, but the next part I am feeling stuck, which is to show that if $ynotin O_v, x/yin O_v$, then the point is $((x/y)(v):1:0)$.
Any assistance with how to think about this would also be very much appreciated.
algebraic-geometry algebraic-curves projective-geometry valuation-theory
asked Jan 5 at 23:49
user277182user277182
561212
$endgroup$
Given a nonsingular complete curve over algebraically closed $bar{k}$, which is interpreted as a field $bar{k}(X)$ of transcendence degree 1 and its set of valuations trivial on $bar{k}$, we may choose elements $x,y$ such that $x$ is trancendental, and $bar{k}(X)=bar{k}(x)(y)$. Letting $F$ be the homogenization of the minimal polynomial of $y$ over $bar{k}(x)$, using these coordinates we get a function from the set of valuations to the points of the projective curve $X_F(bar{k})$, given by mapping a valuation $v$ to the unique point $P$ of $X_F(bar{k})$ such that $O_P subset O_v$ and inclusion of maximal ideals.
The first question is to show that if $x,y$ are in $O_v$, then the valuation $v$ is mapped to is $(x(v):y(v):1)$, where $x(v),y(v)$ are the cosets of $x,y$ in $O_v / M_vcong bar{k}$.
For this, my thought was that if $psi =G/H$ with $H(x(v):y(v):1)neq 0$, then we get the coset of $H(x,y,1)$ in $O_v / M_v$ is nonzero, so we can conclude $G/H$ is defined at $v$, in the sense of having nonnegative valuation.
I feel okay with this, and the homogenisation/dehomogenisation is just from our isomorphism of fields $bar{k}(X)cong bar{k}(X_F)$, but the next part I am feeling stuck, which is to show that if $ynotin O_v, x/yin O_v$, then the point is $((x/y)(v):1:0)$.
Any assistance with how to think about this would also be very much appreciated.
algebraic-geometry algebraic-curves projective-geometry valuation-theory
algebraic-geometry algebraic-curves projective-geometry valuation-theory
asked Jan 5 at 23:49
user277182user277182
561212
asked Jan 5 at 23:49
user277182user277182
561212
asked Jan 5 at 23:49
user277182user277182
561212
asked Jan 5 at 23:49
user277182user277182
561212
asked Jan 5 at 23:49
user277182user277182
561212
561212
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