Shafarevich, locally regular $Rightarrow$ globally regular
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So I am confused with this argument in 3rd Ed, pg 47, Basic Alg. Geo. 1.
Definition 1: if $X subseteq Bbb P^n$ is a quasiprojective variety, $x in X$, and $f=P/Q$ is a homogenous function of degree $0$ with $Q(x)not=0$. A function on $X$ that is regular at all points $x in X$ is a regular function on $X$.
Definition 2: If $A$ is a variety of $Bbb A^n$ then $f:A rightarrow k$ is regular if exists poylnomial function $F$ such that $F|A=f$.
So Shafarevich sets out to prove definition 1 for a closed subset $X$ of an affine space coincides with 2.
Proof: By assumption each point $x in X$ has a nhood $U_x$ with $q_x not=0 $ on $U_x$, in which $f=p_x/q_x$. So
$$q_x f= p_x $$
on $U_x$. We can assume that this holds over whole of $X$ by multiplying a regular function equal to $0$ on $Xsetminus U_x$ and nonzero at $x$.
What regular function is referred here? I suppose it is definition 2. By how does such a function exist?
algebraic-geometry affine-varieties projective-varieties
asked Nov 25 '18 at 22:34
CL.CL.
2,1902824
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add a comment |
$begingroup$
So I am confused with this argument in 3rd Ed, pg 47, Basic Alg. Geo. 1.
Definition 1: if $X subseteq Bbb P^n$ is a quasiprojective variety, $x in X$, and $f=P/Q$ is a homogenous function of degree $0$ with $Q(x)not=0$. A function on $X$ that is regular at all points $x in X$ is a regular function on $X$.
Definition 2: If $A$ is a variety of $Bbb A^n$ then $f:A rightarrow k$ is regular if exists poylnomial function $F$ such that $F|A=f$.
So Shafarevich sets out to prove definition 1 for a closed subset $X$ of an affine space coincides with 2.
Proof: By assumption each point $x in X$ has a nhood $U_x$ with $q_x not=0 $ on $U_x$, in which $f=p_x/q_x$. So
$$q_x f= p_x $$
on $U_x$. We can assume that this holds over whole of $X$ by multiplying a regular function equal to $0$ on $Xsetminus U_x$ and nonzero at $x$.
What regular function is referred here? I suppose it is definition 2. By how does such a function exist?
algebraic-geometry affine-varieties projective-varieties
asked Nov 25 '18 at 22:34
CL.CL.
2,1902824
$endgroup$
add a comment |
$begingroup$
So I am confused with this argument in 3rd Ed, pg 47, Basic Alg. Geo. 1.
Definition 1: if $X subseteq Bbb P^n$ is a quasiprojective variety, $x in X$, and $f=P/Q$ is a homogenous function of degree $0$ with $Q(x)not=0$. A function on $X$ that is regular at all points $x in X$ is a regular function on $X$.
Definition 2: If $A$ is a variety of $Bbb A^n$ then $f:A rightarrow k$ is regular if exists poylnomial function $F$ such that $F|A=f$.
So Shafarevich sets out to prove definition 1 for a closed subset $X$ of an affine space coincides with 2.
Proof: By assumption each point $x in X$ has a nhood $U_x$ with $q_x not=0 $ on $U_x$, in which $f=p_x/q_x$. So
$$q_x f= p_x $$
on $U_x$. We can assume that this holds over whole of $X$ by multiplying a regular function equal to $0$ on $Xsetminus U_x$ and nonzero at $x$.
What regular function is referred here? I suppose it is definition 2. By how does such a function exist?
algebraic-geometry affine-varieties projective-varieties
asked Nov 25 '18 at 22:34
CL.CL.
2,1902824
$endgroup$
So I am confused with this argument in 3rd Ed, pg 47, Basic Alg. Geo. 1.
Definition 1: if $X subseteq Bbb P^n$ is a quasiprojective variety, $x in X$, and $f=P/Q$ is a homogenous function of degree $0$ with $Q(x)not=0$. A function on $X$ that is regular at all points $x in X$ is a regular function on $X$.
Definition 2: If $A$ is a variety of $Bbb A^n$ then $f:A rightarrow k$ is regular if exists poylnomial function $F$ such that $F|A=f$.
So Shafarevich sets out to prove definition 1 for a closed subset $X$ of an affine space coincides with 2.
Proof: By assumption each point $x in X$ has a nhood $U_x$ with $q_x not=0 $ on $U_x$, in which $f=p_x/q_x$. So
$$q_x f= p_x $$
on $U_x$. We can assume that this holds over whole of $X$ by multiplying a regular function equal to $0$ on $Xsetminus U_x$ and nonzero at $x$.
What regular function is referred here? I suppose it is definition 2. By how does such a function exist?
algebraic-geometry affine-varieties projective-varieties
algebraic-geometry affine-varieties projective-varieties
asked Nov 25 '18 at 22:34
CL.CL.
2,1902824
asked Nov 25 '18 at 22:34
CL.CL.
2,1902824
edited Nov 26 '18 at 7:26
CL.
asked Nov 25 '18 at 22:34
CL.CL.
2,1902824
asked Nov 25 '18 at 22:34
CL.CL.
2,1902824
asked Nov 25 '18 at 22:34
CL.CL.
2,1902824
2,1902824
add a comment |
add a comment |
1 Answer
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Note that $Z:=Xsmallsetminus U_x$ is a closed subset of $X$ (and hence of $Bbb A^n$). If we take elements $f_1,dots,f_r$ which generate the ideal $I(Z)subset k[x_1,dots,x_n]$ (I'm not sure if this is the notation Shafarevich uses, but $I(Z)$ is the ideal given by polynomials vanishing on all of $Z$, and it is finitely generated because $k[x_1,dots,x_n]$ is noetherian).
If each $f_i$ vanished at $x$, then $x$ would have to be in $Z$ because $Z$ is the vanishing locus of $(f_1,dots,f_r)$, but this is not the case. So some $f_i$ does not vanish at $x$ and, furthermore, it must vanish on all of $Xsmallsetminus U_x$ because it is an element of $I(Z)$. So $f_i$ is the element you want.
answered Nov 26 '18 at 8:02
Alex MathersAlex Mathers
10.8k21344
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$begingroup$
I also have another definition question post, I hope you don't mind giving some guidance.
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– CL.
Nov 26 '18 at 8:19
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@CL. I've gone ahead and responded, let me know if you have further questions.
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– Alex Mathers
Nov 26 '18 at 18:10
add a comment |
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$begingroup$
Note that $Z:=Xsmallsetminus U_x$ is a closed subset of $X$ (and hence of $Bbb A^n$). If we take elements $f_1,dots,f_r$ which generate the ideal $I(Z)subset k[x_1,dots,x_n]$ (I'm not sure if this is the notation Shafarevich uses, but $I(Z)$ is the ideal given by polynomials vanishing on all of $Z$, and it is finitely generated because $k[x_1,dots,x_n]$ is noetherian).
If each $f_i$ vanished at $x$, then $x$ would have to be in $Z$ because $Z$ is the vanishing locus of $(f_1,dots,f_r)$, but this is not the case. So some $f_i$ does not vanish at $x$ and, furthermore, it must vanish on all of $Xsmallsetminus U_x$ because it is an element of $I(Z)$. So $f_i$ is the element you want.
answered Nov 26 '18 at 8:02
Alex MathersAlex Mathers
10.8k21344
$endgroup$
$begingroup$
I also have another definition question post, I hope you don't mind giving some guidance.
$endgroup$
– CL.
Nov 26 '18 at 8:19
$begingroup$
@CL. I've gone ahead and responded, let me know if you have further questions.
$endgroup$
– Alex Mathers
Nov 26 '18 at 18:10
add a comment |
$begingroup$
Note that $Z:=Xsmallsetminus U_x$ is a closed subset of $X$ (and hence of $Bbb A^n$). If we take elements $f_1,dots,f_r$ which generate the ideal $I(Z)subset k[x_1,dots,x_n]$ (I'm not sure if this is the notation Shafarevich uses, but $I(Z)$ is the ideal given by polynomials vanishing on all of $Z$, and it is finitely generated because $k[x_1,dots,x_n]$ is noetherian).
If each $f_i$ vanished at $x$, then $x$ would have to be in $Z$ because $Z$ is the vanishing locus of $(f_1,dots,f_r)$, but this is not the case. So some $f_i$ does not vanish at $x$ and, furthermore, it must vanish on all of $Xsmallsetminus U_x$ because it is an element of $I(Z)$. So $f_i$ is the element you want.
answered Nov 26 '18 at 8:02
Alex MathersAlex Mathers
10.8k21344
$endgroup$
$begingroup$
I also have another definition question post, I hope you don't mind giving some guidance.
$endgroup$
– CL.
Nov 26 '18 at 8:19
$begingroup$
@CL. I've gone ahead and responded, let me know if you have further questions.
$endgroup$
– Alex Mathers
Nov 26 '18 at 18:10
add a comment |
$begingroup$
Note that $Z:=Xsmallsetminus U_x$ is a closed subset of $X$ (and hence of $Bbb A^n$). If we take elements $f_1,dots,f_r$ which generate the ideal $I(Z)subset k[x_1,dots,x_n]$ (I'm not sure if this is the notation Shafarevich uses, but $I(Z)$ is the ideal given by polynomials vanishing on all of $Z$, and it is finitely generated because $k[x_1,dots,x_n]$ is noetherian).
If each $f_i$ vanished at $x$, then $x$ would have to be in $Z$ because $Z$ is the vanishing locus of $(f_1,dots,f_r)$, but this is not the case. So some $f_i$ does not vanish at $x$ and, furthermore, it must vanish on all of $Xsmallsetminus U_x$ because it is an element of $I(Z)$. So $f_i$ is the element you want.
answered Nov 26 '18 at 8:02
Alex MathersAlex Mathers
10.8k21344
$endgroup$
Note that $Z:=Xsmallsetminus U_x$ is a closed subset of $X$ (and hence of $Bbb A^n$). If we take elements $f_1,dots,f_r$ which generate the ideal $I(Z)subset k[x_1,dots,x_n]$ (I'm not sure if this is the notation Shafarevich uses, but $I(Z)$ is the ideal given by polynomials vanishing on all of $Z$, and it is finitely generated because $k[x_1,dots,x_n]$ is noetherian).
If each $f_i$ vanished at $x$, then $x$ would have to be in $Z$ because $Z$ is the vanishing locus of $(f_1,dots,f_r)$, but this is not the case. So some $f_i$ does not vanish at $x$ and, furthermore, it must vanish on all of $Xsmallsetminus U_x$ because it is an element of $I(Z)$. So $f_i$ is the element you want.
answered Nov 26 '18 at 8:02
Alex MathersAlex Mathers
10.8k21344
answered Nov 26 '18 at 8:02
Alex MathersAlex Mathers
10.8k21344
answered Nov 26 '18 at 8:02
Alex MathersAlex Mathers
10.8k21344
answered Nov 26 '18 at 8:02
Alex MathersAlex Mathers
10.8k21344
10.8k21344
$begingroup$
I also have another definition question post, I hope you don't mind giving some guidance.
$endgroup$
– CL.
Nov 26 '18 at 8:19
$begingroup$
@CL. I've gone ahead and responded, let me know if you have further questions.
$endgroup$
– Alex Mathers
Nov 26 '18 at 18:10
add a comment |
$begingroup$
I also have another definition question post, I hope you don't mind giving some guidance.
$endgroup$
– CL.
Nov 26 '18 at 8:19
$begingroup$
@CL. I've gone ahead and responded, let me know if you have further questions.
$endgroup$
– Alex Mathers
Nov 26 '18 at 18:10
$begingroup$
I also have another definition question post, I hope you don't mind giving some guidance.
$endgroup$
– CL.
Nov 26 '18 at 8:19
$begingroup$
I also have another definition question post, I hope you don't mind giving some guidance.
$endgroup$
– CL.
Nov 26 '18 at 8:19
$begingroup$
@CL. I've gone ahead and responded, let me know if you have further questions.
$endgroup$
– Alex Mathers
Nov 26 '18 at 18:10
$begingroup$
@CL. I've gone ahead and responded, let me know if you have further questions.
$endgroup$
– Alex Mathers
Nov 26 '18 at 18:10
add a comment |
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