When can $varphi in mathcal{O}_{X}(U)$ be written globally as quotients of polynomials in $k[x_1, dots x_n]$?











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I am reading Gathmann's online notes.



Here, we define the ring of regular functions on an open neighborhood $U subset X$ by
$$mathcal{O}_{X}(U):= bigcap_{x in U} mathcal{O}_{X,P}$$



where $mathcal{O}_{X,p}:={f/g :f,g in k[X], g(p) neq 0}$.



Note that this does not mean that $f/g$ are globally quotients of polynomials. In particular consider $X=V(x_1x_4-x_2x_3)$ and $U$ is the subset where $x_2 neq 0$ or $x_4 neq 0$. In this case,
$frac{x_1}{x_2} =frac{x_3}{x_4}$ are both regular functions that are given "locally" as quotients of polynomials.



For $U:= mathbb A^2 setminus {0}$, we can write every regular function globally as quotients of polynomials since $X=mathbb A^2$.



I want to know if there is a systematic way of telling when you can write these things as globally regular functions.





My guess is that if we take some $U subset mathbb A^n$ and consider $i:X to mathbb A^n$, then $i^{-1}(U):=V$ is open in $X$. There is an induced map



$$i^*:mathcal{O}_{mathbb A^n}(U) to mathcal{O}_X(V).$$



I guess the induced map is $ psi mapsto psi circ i$. Am I just asking when this map is a surjection?










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  • Here $X$ is assumed to be an irreducible affine variety?
    – Eric Wofsey
    2 days ago










  • @EricWofsey sure, I don't see why not. Is the answer particularly easy in this case?
    – Andres Mejia
    2 days ago










  • Well you seemed to be assuming that already, since some of your notation doesn't make sense otherwise.
    – Eric Wofsey
    2 days ago










  • True, I'm not sure what $f/g$ with $f,g in k[X]$ would mean otherwise haha.
    – Andres Mejia
    2 days ago

















up vote
2
down vote

favorite












I am reading Gathmann's online notes.



Here, we define the ring of regular functions on an open neighborhood $U subset X$ by
$$mathcal{O}_{X}(U):= bigcap_{x in U} mathcal{O}_{X,P}$$



where $mathcal{O}_{X,p}:={f/g :f,g in k[X], g(p) neq 0}$.



Note that this does not mean that $f/g$ are globally quotients of polynomials. In particular consider $X=V(x_1x_4-x_2x_3)$ and $U$ is the subset where $x_2 neq 0$ or $x_4 neq 0$. In this case,
$frac{x_1}{x_2} =frac{x_3}{x_4}$ are both regular functions that are given "locally" as quotients of polynomials.



For $U:= mathbb A^2 setminus {0}$, we can write every regular function globally as quotients of polynomials since $X=mathbb A^2$.



I want to know if there is a systematic way of telling when you can write these things as globally regular functions.





My guess is that if we take some $U subset mathbb A^n$ and consider $i:X to mathbb A^n$, then $i^{-1}(U):=V$ is open in $X$. There is an induced map



$$i^*:mathcal{O}_{mathbb A^n}(U) to mathcal{O}_X(V).$$



I guess the induced map is $ psi mapsto psi circ i$. Am I just asking when this map is a surjection?










share|cite|improve this question
























  • Here $X$ is assumed to be an irreducible affine variety?
    – Eric Wofsey
    2 days ago










  • @EricWofsey sure, I don't see why not. Is the answer particularly easy in this case?
    – Andres Mejia
    2 days ago










  • Well you seemed to be assuming that already, since some of your notation doesn't make sense otherwise.
    – Eric Wofsey
    2 days ago










  • True, I'm not sure what $f/g$ with $f,g in k[X]$ would mean otherwise haha.
    – Andres Mejia
    2 days ago















up vote
2
down vote

favorite









up vote
2
down vote

favorite











I am reading Gathmann's online notes.



Here, we define the ring of regular functions on an open neighborhood $U subset X$ by
$$mathcal{O}_{X}(U):= bigcap_{x in U} mathcal{O}_{X,P}$$



where $mathcal{O}_{X,p}:={f/g :f,g in k[X], g(p) neq 0}$.



Note that this does not mean that $f/g$ are globally quotients of polynomials. In particular consider $X=V(x_1x_4-x_2x_3)$ and $U$ is the subset where $x_2 neq 0$ or $x_4 neq 0$. In this case,
$frac{x_1}{x_2} =frac{x_3}{x_4}$ are both regular functions that are given "locally" as quotients of polynomials.



For $U:= mathbb A^2 setminus {0}$, we can write every regular function globally as quotients of polynomials since $X=mathbb A^2$.



I want to know if there is a systematic way of telling when you can write these things as globally regular functions.





My guess is that if we take some $U subset mathbb A^n$ and consider $i:X to mathbb A^n$, then $i^{-1}(U):=V$ is open in $X$. There is an induced map



$$i^*:mathcal{O}_{mathbb A^n}(U) to mathcal{O}_X(V).$$



I guess the induced map is $ psi mapsto psi circ i$. Am I just asking when this map is a surjection?










share|cite|improve this question















I am reading Gathmann's online notes.



Here, we define the ring of regular functions on an open neighborhood $U subset X$ by
$$mathcal{O}_{X}(U):= bigcap_{x in U} mathcal{O}_{X,P}$$



where $mathcal{O}_{X,p}:={f/g :f,g in k[X], g(p) neq 0}$.



Note that this does not mean that $f/g$ are globally quotients of polynomials. In particular consider $X=V(x_1x_4-x_2x_3)$ and $U$ is the subset where $x_2 neq 0$ or $x_4 neq 0$. In this case,
$frac{x_1}{x_2} =frac{x_3}{x_4}$ are both regular functions that are given "locally" as quotients of polynomials.



For $U:= mathbb A^2 setminus {0}$, we can write every regular function globally as quotients of polynomials since $X=mathbb A^2$.



I want to know if there is a systematic way of telling when you can write these things as globally regular functions.





My guess is that if we take some $U subset mathbb A^n$ and consider $i:X to mathbb A^n$, then $i^{-1}(U):=V$ is open in $X$. There is an induced map



$$i^*:mathcal{O}_{mathbb A^n}(U) to mathcal{O}_X(V).$$



I guess the induced map is $ psi mapsto psi circ i$. Am I just asking when this map is a surjection?







algebraic-geometry






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Andres Mejia

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  • Here $X$ is assumed to be an irreducible affine variety?
    – Eric Wofsey
    2 days ago










  • @EricWofsey sure, I don't see why not. Is the answer particularly easy in this case?
    – Andres Mejia
    2 days ago










  • Well you seemed to be assuming that already, since some of your notation doesn't make sense otherwise.
    – Eric Wofsey
    2 days ago










  • True, I'm not sure what $f/g$ with $f,g in k[X]$ would mean otherwise haha.
    – Andres Mejia
    2 days ago




















  • Here $X$ is assumed to be an irreducible affine variety?
    – Eric Wofsey
    2 days ago










  • @EricWofsey sure, I don't see why not. Is the answer particularly easy in this case?
    – Andres Mejia
    2 days ago










  • Well you seemed to be assuming that already, since some of your notation doesn't make sense otherwise.
    – Eric Wofsey
    2 days ago










  • True, I'm not sure what $f/g$ with $f,g in k[X]$ would mean otherwise haha.
    – Andres Mejia
    2 days ago


















Here $X$ is assumed to be an irreducible affine variety?
– Eric Wofsey
2 days ago




Here $X$ is assumed to be an irreducible affine variety?
– Eric Wofsey
2 days ago












@EricWofsey sure, I don't see why not. Is the answer particularly easy in this case?
– Andres Mejia
2 days ago




@EricWofsey sure, I don't see why not. Is the answer particularly easy in this case?
– Andres Mejia
2 days ago












Well you seemed to be assuming that already, since some of your notation doesn't make sense otherwise.
– Eric Wofsey
2 days ago




Well you seemed to be assuming that already, since some of your notation doesn't make sense otherwise.
– Eric Wofsey
2 days ago












True, I'm not sure what $f/g$ with $f,g in k[X]$ would mean otherwise haha.
– Andres Mejia
2 days ago






True, I'm not sure what $f/g$ with $f,g in k[X]$ would mean otherwise haha.
– Andres Mejia
2 days ago

















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