Proof that every codimension 1 subvariety of $mathbbP^n$ is $V((f))$

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I've been told that
Theorem. Every codimension 1 subvariety of $mathbbP^n$ is $V((f))$, where $f$ is some prime homogeneous polynomial.
I'm under the impression that this is true over any algebraically closed field.
However, I'm unable to locate a proof. Does anyone know where a proof of this theorem can be located? A version of this for affine space is here, but a bit light on the details.
Also, in this context, does $V((f))$ mean $V$ of the principal ideal generated by $f$, or is the double bracketing indicative of some kind of a power series construction?
algebraic-geometry projective-space
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up vote
2
down vote
favorite
I've been told that
Theorem. Every codimension 1 subvariety of $mathbbP^n$ is $V((f))$, where $f$ is some prime homogeneous polynomial.
I'm under the impression that this is true over any algebraically closed field.
However, I'm unable to locate a proof. Does anyone know where a proof of this theorem can be located? A version of this for affine space is here, but a bit light on the details.
Also, in this context, does $V((f))$ mean $V$ of the principal ideal generated by $f$, or is the double bracketing indicative of some kind of a power series construction?
algebraic-geometry projective-space
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I've been told that
Theorem. Every codimension 1 subvariety of $mathbbP^n$ is $V((f))$, where $f$ is some prime homogeneous polynomial.
I'm under the impression that this is true over any algebraically closed field.
However, I'm unable to locate a proof. Does anyone know where a proof of this theorem can be located? A version of this for affine space is here, but a bit light on the details.
Also, in this context, does $V((f))$ mean $V$ of the principal ideal generated by $f$, or is the double bracketing indicative of some kind of a power series construction?
algebraic-geometry projective-space
I've been told that
Theorem. Every codimension 1 subvariety of $mathbbP^n$ is $V((f))$, where $f$ is some prime homogeneous polynomial.
I'm under the impression that this is true over any algebraically closed field.
However, I'm unable to locate a proof. Does anyone know where a proof of this theorem can be located? A version of this for affine space is here, but a bit light on the details.
Also, in this context, does $V((f))$ mean $V$ of the principal ideal generated by $f$, or is the double bracketing indicative of some kind of a power series construction?
algebraic-geometry projective-space
algebraic-geometry projective-space
asked Sep 10 at 7:15
goblin
35.9k1155182
35.9k1155182
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1 Answer
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This theorem does hold over any algebraically closed field; notably we need the fact that the dimension of an affine variety, as given by the length of a maximal chain of irreducible subvarieties, is the same as the krull dimension of its coordinate ring. Here is a nice pdf by Andreas Gathmann on dimension theory: http://www.mathematik.uni-kl.de/~gathmann/class/commalg-2013/commalg-2013-c11.pdf .
For an irreducible affine variety, the answer to your question is given by remark 11.18. In the projective case, it suffices to notice that the dimension of a space is the same as that of an open dense subset. So intersecting your variety $V$ by a standard open affine (so that the intersection is non empty) will give you a polynomial $f in k[x_1,dots,x_n]$ such that $V$ is now given by the vanishing set of the homogenization of $f$.
In regards to $V((f))$, your first interpretation is correct.
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
This theorem does hold over any algebraically closed field; notably we need the fact that the dimension of an affine variety, as given by the length of a maximal chain of irreducible subvarieties, is the same as the krull dimension of its coordinate ring. Here is a nice pdf by Andreas Gathmann on dimension theory: http://www.mathematik.uni-kl.de/~gathmann/class/commalg-2013/commalg-2013-c11.pdf .
For an irreducible affine variety, the answer to your question is given by remark 11.18. In the projective case, it suffices to notice that the dimension of a space is the same as that of an open dense subset. So intersecting your variety $V$ by a standard open affine (so that the intersection is non empty) will give you a polynomial $f in k[x_1,dots,x_n]$ such that $V$ is now given by the vanishing set of the homogenization of $f$.
In regards to $V((f))$, your first interpretation is correct.
add a comment |Â
up vote
2
down vote
accepted
This theorem does hold over any algebraically closed field; notably we need the fact that the dimension of an affine variety, as given by the length of a maximal chain of irreducible subvarieties, is the same as the krull dimension of its coordinate ring. Here is a nice pdf by Andreas Gathmann on dimension theory: http://www.mathematik.uni-kl.de/~gathmann/class/commalg-2013/commalg-2013-c11.pdf .
For an irreducible affine variety, the answer to your question is given by remark 11.18. In the projective case, it suffices to notice that the dimension of a space is the same as that of an open dense subset. So intersecting your variety $V$ by a standard open affine (so that the intersection is non empty) will give you a polynomial $f in k[x_1,dots,x_n]$ such that $V$ is now given by the vanishing set of the homogenization of $f$.
In regards to $V((f))$, your first interpretation is correct.
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
This theorem does hold over any algebraically closed field; notably we need the fact that the dimension of an affine variety, as given by the length of a maximal chain of irreducible subvarieties, is the same as the krull dimension of its coordinate ring. Here is a nice pdf by Andreas Gathmann on dimension theory: http://www.mathematik.uni-kl.de/~gathmann/class/commalg-2013/commalg-2013-c11.pdf .
For an irreducible affine variety, the answer to your question is given by remark 11.18. In the projective case, it suffices to notice that the dimension of a space is the same as that of an open dense subset. So intersecting your variety $V$ by a standard open affine (so that the intersection is non empty) will give you a polynomial $f in k[x_1,dots,x_n]$ such that $V$ is now given by the vanishing set of the homogenization of $f$.
In regards to $V((f))$, your first interpretation is correct.
This theorem does hold over any algebraically closed field; notably we need the fact that the dimension of an affine variety, as given by the length of a maximal chain of irreducible subvarieties, is the same as the krull dimension of its coordinate ring. Here is a nice pdf by Andreas Gathmann on dimension theory: http://www.mathematik.uni-kl.de/~gathmann/class/commalg-2013/commalg-2013-c11.pdf .
For an irreducible affine variety, the answer to your question is given by remark 11.18. In the projective case, it suffices to notice that the dimension of a space is the same as that of an open dense subset. So intersecting your variety $V$ by a standard open affine (so that the intersection is non empty) will give you a polynomial $f in k[x_1,dots,x_n]$ such that $V$ is now given by the vanishing set of the homogenization of $f$.
In regards to $V((f))$, your first interpretation is correct.
answered Sep 10 at 10:25
Gabriel Micolet
1141
1141
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