Mixing
Zero locus of $r$ homogeneous polynomials in $mathbb{P}^{N}$ where $rleq N$
$begingroup$
On my notes I have the following statement:
Let $F_{1},dots,F_{r}$ be homogeneous polynomials in $k[x_{0},dots,x_{N}]$ ($k$ algebraically closed field) with $rleq N$. Then $X=V(F_{1},dots,F_{r})subseteqmathbb{P}^{N}$ is non empty and every irreducible component of $X$ has dimensione at least $N-r$.
The proof for the non-emptyness is done by induction on $r$. But don't we need some extra hypothesis to guarantee that the locus is non-empty? I.E: let's start with $r=1$. Then $X=V(F_{1})$ is a hypersurface if the polynomial is not constant (and homogeneous). How do we know that from the hypothesis?
Thanks for helping !
algebraic-geometry
asked Feb 1 at 18:40
ElenaElena
64
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add a comment |
$begingroup$
On my notes I have the following statement:
Let $F_{1},dots,F_{r}$ be homogeneous polynomials in $k[x_{0},dots,x_{N}]$ ($k$ algebraically closed field) with $rleq N$. Then $X=V(F_{1},dots,F_{r})subseteqmathbb{P}^{N}$ is non empty and every irreducible component of $X$ has dimensione at least $N-r$.
The proof for the non-emptyness is done by induction on $r$. But don't we need some extra hypothesis to guarantee that the locus is non-empty? I.E: let's start with $r=1$. Then $X=V(F_{1})$ is a hypersurface if the polynomial is not constant (and homogeneous). How do we know that from the hypothesis?
Thanks for helping !
algebraic-geometry
asked Feb 1 at 18:40
ElenaElena
64
$endgroup$
$begingroup$
I'm confused about your question - are you asking about why we know that $V(F_1)$ has a point? If so, for nonconstant $F_1$, this is just the projective nullstellensatz. If you're asking about more/different things, please clarify.
$endgroup$
– KReiser
Feb 1 at 20:15
$begingroup$
Thank you for answering. My question is: since in the hypothesis of the statement, it is not assumed that the polynomials $F_{i}$ are non-constant, is the proposition still true?
$endgroup$
– Elena
Feb 1 at 20:41
$begingroup$
It's probably a minor oversight on the part of the lecturer. If $F_1=1$, for instance, then $X$ is empty. Probably the cleanest to state the proposition is to assume all $F_i$ are nonconstant and deal with the presence of constant polynomials as a side case. Any nonzero nonconstant polynomial makes $X$ empty, and any $F_i$ that's zero can be ignored.
$endgroup$
– KReiser
Feb 1 at 21:01
$begingroup$
Thank you very much, now it's clear :)
$endgroup$
– Elena
Feb 1 at 21:35
add a comment |
$begingroup$
On my notes I have the following statement:
Let $F_{1},dots,F_{r}$ be homogeneous polynomials in $k[x_{0},dots,x_{N}]$ ($k$ algebraically closed field) with $rleq N$. Then $X=V(F_{1},dots,F_{r})subseteqmathbb{P}^{N}$ is non empty and every irreducible component of $X$ has dimensione at least $N-r$.
The proof for the non-emptyness is done by induction on $r$. But don't we need some extra hypothesis to guarantee that the locus is non-empty? I.E: let's start with $r=1$. Then $X=V(F_{1})$ is a hypersurface if the polynomial is not constant (and homogeneous). How do we know that from the hypothesis?
Thanks for helping !
algebraic-geometry
asked Feb 1 at 18:40
ElenaElena
64
$endgroup$
On my notes I have the following statement:
Let $F_{1},dots,F_{r}$ be homogeneous polynomials in $k[x_{0},dots,x_{N}]$ ($k$ algebraically closed field) with $rleq N$. Then $X=V(F_{1},dots,F_{r})subseteqmathbb{P}^{N}$ is non empty and every irreducible component of $X$ has dimensione at least $N-r$.
The proof for the non-emptyness is done by induction on $r$. But don't we need some extra hypothesis to guarantee that the locus is non-empty? I.E: let's start with $r=1$. Then $X=V(F_{1})$ is a hypersurface if the polynomial is not constant (and homogeneous). How do we know that from the hypothesis?
Thanks for helping !
algebraic-geometry
algebraic-geometry
asked Feb 1 at 18:40
ElenaElena
64
asked Feb 1 at 18:40
ElenaElena
64
asked Feb 1 at 18:40
ElenaElena
64
asked Feb 1 at 18:40
ElenaElena
64
asked Feb 1 at 18:40
ElenaElena
64
64
$begingroup$
I'm confused about your question - are you asking about why we know that $V(F_1)$ has a point? If so, for nonconstant $F_1$, this is just the projective nullstellensatz. If you're asking about more/different things, please clarify.
$endgroup$
– KReiser
Feb 1 at 20:15
$begingroup$
Thank you for answering. My question is: since in the hypothesis of the statement, it is not assumed that the polynomials $F_{i}$ are non-constant, is the proposition still true?
$endgroup$
– Elena
Feb 1 at 20:41
$begingroup$
It's probably a minor oversight on the part of the lecturer. If $F_1=1$, for instance, then $X$ is empty. Probably the cleanest to state the proposition is to assume all $F_i$ are nonconstant and deal with the presence of constant polynomials as a side case. Any nonzero nonconstant polynomial makes $X$ empty, and any $F_i$ that's zero can be ignored.
$endgroup$
– KReiser
Feb 1 at 21:01
$begingroup$
Thank you very much, now it's clear :)
$endgroup$
– Elena
Feb 1 at 21:35
add a comment |
$begingroup$
I'm confused about your question - are you asking about why we know that $V(F_1)$ has a point? If so, for nonconstant $F_1$, this is just the projective nullstellensatz. If you're asking about more/different things, please clarify.
$endgroup$
– KReiser
Feb 1 at 20:15
$begingroup$
Thank you for answering. My question is: since in the hypothesis of the statement, it is not assumed that the polynomials $F_{i}$ are non-constant, is the proposition still true?
$endgroup$
– Elena
Feb 1 at 20:41
$begingroup$
It's probably a minor oversight on the part of the lecturer. If $F_1=1$, for instance, then $X$ is empty. Probably the cleanest to state the proposition is to assume all $F_i$ are nonconstant and deal with the presence of constant polynomials as a side case. Any nonzero nonconstant polynomial makes $X$ empty, and any $F_i$ that's zero can be ignored.
$endgroup$
– KReiser
Feb 1 at 21:01
$begingroup$
Thank you very much, now it's clear :)
$endgroup$
– Elena
Feb 1 at 21:35
$begingroup$
I'm confused about your question - are you asking about why we know that $V(F_1)$ has a point? If so, for nonconstant $F_1$, this is just the projective nullstellensatz. If you're asking about more/different things, please clarify.
$endgroup$
– KReiser
Feb 1 at 20:15
$begingroup$
I'm confused about your question - are you asking about why we know that $V(F_1)$ has a point? If so, for nonconstant $F_1$, this is just the projective nullstellensatz. If you're asking about more/different things, please clarify.
$endgroup$
– KReiser
Feb 1 at 20:15
$begingroup$
Thank you for answering. My question is: since in the hypothesis of the statement, it is not assumed that the polynomials $F_{i}$ are non-constant, is the proposition still true?
$endgroup$
– Elena
Feb 1 at 20:41
$begingroup$
Thank you for answering. My question is: since in the hypothesis of the statement, it is not assumed that the polynomials $F_{i}$ are non-constant, is the proposition still true?
$endgroup$
– Elena
Feb 1 at 20:41
$begingroup$
It's probably a minor oversight on the part of the lecturer. If $F_1=1$, for instance, then $X$ is empty. Probably the cleanest to state the proposition is to assume all $F_i$ are nonconstant and deal with the presence of constant polynomials as a side case. Any nonzero nonconstant polynomial makes $X$ empty, and any $F_i$ that's zero can be ignored.
$endgroup$
– KReiser
Feb 1 at 21:01
$begingroup$
It's probably a minor oversight on the part of the lecturer. If $F_1=1$, for instance, then $X$ is empty. Probably the cleanest to state the proposition is to assume all $F_i$ are nonconstant and deal with the presence of constant polynomials as a side case. Any nonzero nonconstant polynomial makes $X$ empty, and any $F_i$ that's zero can be ignored.
$endgroup$
– KReiser
Feb 1 at 21:01
$begingroup$
Thank you very much, now it's clear :)
$endgroup$
– Elena
Feb 1 at 21:35
$begingroup$
Thank you very much, now it's clear :)
$endgroup$
– Elena
Feb 1 at 21:35
add a comment |
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$begingroup$
I'm confused about your question - are you asking about why we know that $V(F_1)$ has a point? If so, for nonconstant $F_1$, this is just the projective nullstellensatz. If you're asking about more/different things, please clarify.
$endgroup$
– KReiser
Feb 1 at 20:15
$begingroup$
Thank you for answering. My question is: since in the hypothesis of the statement, it is not assumed that the polynomials $F_{i}$ are non-constant, is the proposition still true?
$endgroup$
– Elena
Feb 1 at 20:41
$begingroup$
It's probably a minor oversight on the part of the lecturer. If $F_1=1$, for instance, then $X$ is empty. Probably the cleanest to state the proposition is to assume all $F_i$ are nonconstant and deal with the presence of constant polynomials as a side case. Any nonzero nonconstant polynomial makes $X$ empty, and any $F_i$ that's zero can be ignored.
$endgroup$
– KReiser
Feb 1 at 21:01
$begingroup$
Thank you very much, now it's clear :)
$endgroup$
– Elena
Feb 1 at 21:35