Mixing
Sufficient conditions for a projective curve $X$ over $k$ to satisfy $H^0(X,mathcal{O}_X) = k$?
$begingroup$
Let $X$ be a one-dimensional proper (resp. projective) scheme over a field $k$. By Tag 0BY5 it is necessary that $X$ is Cohen-Macaulay, connected and equidimensional.
What are the weakest assumptions on $X$ you know such that $H^0(X,mathcal{O}_X) = k$ holds?
algebraic-geometry algebraic-curves projective-schemes
asked Feb 1 at 10:40
windsheafwindsheaf
607312
$endgroup$
add a comment |
$begingroup$
Let $X$ be a one-dimensional proper (resp. projective) scheme over a field $k$. By Tag 0BY5 it is necessary that $X$ is Cohen-Macaulay, connected and equidimensional.
What are the weakest assumptions on $X$ you know such that $H^0(X,mathcal{O}_X) = k$ holds?
algebraic-geometry algebraic-curves projective-schemes
asked Feb 1 at 10:40
windsheafwindsheaf
607312
$endgroup$
2
$begingroup$
I found one sufficient condition: If $X$ is a proper reduced scheme of dimension one over $k$, then $H^0(X,mathcal{O}_X)$ is a field, which is a finite extension of $k$. If $X$ is also geometrically connected and geometrically reduced, then $H^0(X,mathcal{O}_X) = k$. This is Cor. 3.3.21 in Liu's book.
$endgroup$
– windsheaf
Feb 1 at 11:28
add a comment |
$begingroup$
Let $X$ be a one-dimensional proper (resp. projective) scheme over a field $k$. By Tag 0BY5 it is necessary that $X$ is Cohen-Macaulay, connected and equidimensional.
What are the weakest assumptions on $X$ you know such that $H^0(X,mathcal{O}_X) = k$ holds?
algebraic-geometry algebraic-curves projective-schemes
asked Feb 1 at 10:40
windsheafwindsheaf
607312
$endgroup$
Let $X$ be a one-dimensional proper (resp. projective) scheme over a field $k$. By Tag 0BY5 it is necessary that $X$ is Cohen-Macaulay, connected and equidimensional.
What are the weakest assumptions on $X$ you know such that $H^0(X,mathcal{O}_X) = k$ holds?
algebraic-geometry algebraic-curves projective-schemes
algebraic-geometry algebraic-curves projective-schemes
asked Feb 1 at 10:40
windsheafwindsheaf
607312
asked Feb 1 at 10:40
windsheafwindsheaf
607312
asked Feb 1 at 10:40
windsheafwindsheaf
607312
asked Feb 1 at 10:40
windsheafwindsheaf
607312
asked Feb 1 at 10:40
windsheafwindsheaf
607312
607312
2
$begingroup$
I found one sufficient condition: If $X$ is a proper reduced scheme of dimension one over $k$, then $H^0(X,mathcal{O}_X)$ is a field, which is a finite extension of $k$. If $X$ is also geometrically connected and geometrically reduced, then $H^0(X,mathcal{O}_X) = k$. This is Cor. 3.3.21 in Liu's book.
$endgroup$
– windsheaf
Feb 1 at 11:28
add a comment |
2
$begingroup$
I found one sufficient condition: If $X$ is a proper reduced scheme of dimension one over $k$, then $H^0(X,mathcal{O}_X)$ is a field, which is a finite extension of $k$. If $X$ is also geometrically connected and geometrically reduced, then $H^0(X,mathcal{O}_X) = k$. This is Cor. 3.3.21 in Liu's book.
$endgroup$
– windsheaf
Feb 1 at 11:28
2
2
$begingroup$
I found one sufficient condition: If $X$ is a proper reduced scheme of dimension one over $k$, then $H^0(X,mathcal{O}_X)$ is a field, which is a finite extension of $k$. If $X$ is also geometrically connected and geometrically reduced, then $H^0(X,mathcal{O}_X) = k$. This is Cor. 3.3.21 in Liu's book.
$endgroup$
– windsheaf
Feb 1 at 11:28
$begingroup$
I found one sufficient condition: If $X$ is a proper reduced scheme of dimension one over $k$, then $H^0(X,mathcal{O}_X)$ is a field, which is a finite extension of $k$. If $X$ is also geometrically connected and geometrically reduced, then $H^0(X,mathcal{O}_X) = k$. This is Cor. 3.3.21 in Liu's book.
$endgroup$
– windsheaf
Feb 1 at 11:28
add a comment |
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$begingroup$
I found one sufficient condition: If $X$ is a proper reduced scheme of dimension one over $k$, then $H^0(X,mathcal{O}_X)$ is a field, which is a finite extension of $k$. If $X$ is also geometrically connected and geometrically reduced, then $H^0(X,mathcal{O}_X) = k$. This is Cor. 3.3.21 in Liu's book.
$endgroup$
– windsheaf
Feb 1 at 11:28