Degree and codimension of nondegenerate varieties
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A projective variety $X subset mathbb{P}^r$ (i.e. reduced, irreducible closed subscheme) is called nonegenerate, if it is not contained in any hypersurface $H subset mathbb{P}^r$. Equivalently, the smallest linear subspace containing $X$ is $mathbb{P}^n$ itself.
I'm having trouble regarding the proof of the following standard lemma.
Lemma If $Xsubset mathbb{P}^r$ is a nondegenerate variety, then $text{deg } X geq text{codim }X + 1$.
This can be found in the paper On Varieties of Minimal Degree by Eisenbud and Harris. The proof is as follows:
- If $text{codim }X = 1$, then $text{deg }X = 1$ if and only if $X$ is a hyperplane. $X$ is nondegenerate, so this is not the case. Hence $text{deg }X geq 2 = text{codim }X + 1$.
- If $text{codim }X geq 2$, choose a "general point of $X$", and project $X$ from that point to $mathbb{P}^{r-1}$.
- Observe that the degree of the projection of $X$ is $< text{deg }X$, and
similarly the codimension of the projection is $< text{codim }X$. - Proceed by induction on $text{codim } X$.
Question How can I see that codimension and degree of $X$ both drop by at least one?
In a class on surfaces I'm taking my professor proved the Lemma even in the case that $X$ is not irreducible, by intersecting $X$ with a general hyperplane, reducing the codimension of $X$ by one, while leaving the degree as is is. Here the induction goes over the dimension of $X$, and the basis is $n+1$ (or more) points, which need to have degree $geq n+1$.
Question How can I see that there exists a hyperplane, such that the codimension drops, and the degree stays the same?
I thought maybe this has something to do with the generalized Bézout's theorem, but I only know this for irreducible varieties. Is there a version of this for only reduced ones?
algebraic-geometry projective-geometry projective-varieties
asked Dec 13 '18 at 19:20
red_trumpetred_trumpet
959319
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add a comment |
$begingroup$
A projective variety $X subset mathbb{P}^r$ (i.e. reduced, irreducible closed subscheme) is called nonegenerate, if it is not contained in any hypersurface $H subset mathbb{P}^r$. Equivalently, the smallest linear subspace containing $X$ is $mathbb{P}^n$ itself.
I'm having trouble regarding the proof of the following standard lemma.
Lemma If $Xsubset mathbb{P}^r$ is a nondegenerate variety, then $text{deg } X geq text{codim }X + 1$.
This can be found in the paper On Varieties of Minimal Degree by Eisenbud and Harris. The proof is as follows:
- If $text{codim }X = 1$, then $text{deg }X = 1$ if and only if $X$ is a hyperplane. $X$ is nondegenerate, so this is not the case. Hence $text{deg }X geq 2 = text{codim }X + 1$.
- If $text{codim }X geq 2$, choose a "general point of $X$", and project $X$ from that point to $mathbb{P}^{r-1}$.
- Observe that the degree of the projection of $X$ is $< text{deg }X$, and
similarly the codimension of the projection is $< text{codim }X$. - Proceed by induction on $text{codim } X$.
Question How can I see that codimension and degree of $X$ both drop by at least one?
In a class on surfaces I'm taking my professor proved the Lemma even in the case that $X$ is not irreducible, by intersecting $X$ with a general hyperplane, reducing the codimension of $X$ by one, while leaving the degree as is is. Here the induction goes over the dimension of $X$, and the basis is $n+1$ (or more) points, which need to have degree $geq n+1$.
Question How can I see that there exists a hyperplane, such that the codimension drops, and the degree stays the same?
I thought maybe this has something to do with the generalized Bézout's theorem, but I only know this for irreducible varieties. Is there a version of this for only reduced ones?
algebraic-geometry projective-geometry projective-varieties
asked Dec 13 '18 at 19:20
red_trumpetred_trumpet
959319
$endgroup$
$begingroup$
How about following the exercise 7.7 from hartshorne Chapter 1 section 7? It seems that this exercise does actually help you complete the induction step.
$endgroup$
– random123
Dec 14 '18 at 10:54
$begingroup$
Ah, good catch, thanks! I will try to work through the exercise.
$endgroup$
– red_trumpet
Dec 14 '18 at 17:55
add a comment |
$begingroup$
A projective variety $X subset mathbb{P}^r$ (i.e. reduced, irreducible closed subscheme) is called nonegenerate, if it is not contained in any hypersurface $H subset mathbb{P}^r$. Equivalently, the smallest linear subspace containing $X$ is $mathbb{P}^n$ itself.
I'm having trouble regarding the proof of the following standard lemma.
Lemma If $Xsubset mathbb{P}^r$ is a nondegenerate variety, then $text{deg } X geq text{codim }X + 1$.
This can be found in the paper On Varieties of Minimal Degree by Eisenbud and Harris. The proof is as follows:
- If $text{codim }X = 1$, then $text{deg }X = 1$ if and only if $X$ is a hyperplane. $X$ is nondegenerate, so this is not the case. Hence $text{deg }X geq 2 = text{codim }X + 1$.
- If $text{codim }X geq 2$, choose a "general point of $X$", and project $X$ from that point to $mathbb{P}^{r-1}$.
- Observe that the degree of the projection of $X$ is $< text{deg }X$, and
similarly the codimension of the projection is $< text{codim }X$. - Proceed by induction on $text{codim } X$.
Question How can I see that codimension and degree of $X$ both drop by at least one?
In a class on surfaces I'm taking my professor proved the Lemma even in the case that $X$ is not irreducible, by intersecting $X$ with a general hyperplane, reducing the codimension of $X$ by one, while leaving the degree as is is. Here the induction goes over the dimension of $X$, and the basis is $n+1$ (or more) points, which need to have degree $geq n+1$.
Question How can I see that there exists a hyperplane, such that the codimension drops, and the degree stays the same?
I thought maybe this has something to do with the generalized Bézout's theorem, but I only know this for irreducible varieties. Is there a version of this for only reduced ones?
algebraic-geometry projective-geometry projective-varieties
asked Dec 13 '18 at 19:20
red_trumpetred_trumpet
959319
$endgroup$
A projective variety $X subset mathbb{P}^r$ (i.e. reduced, irreducible closed subscheme) is called nonegenerate, if it is not contained in any hypersurface $H subset mathbb{P}^r$. Equivalently, the smallest linear subspace containing $X$ is $mathbb{P}^n$ itself.
I'm having trouble regarding the proof of the following standard lemma.
Lemma If $Xsubset mathbb{P}^r$ is a nondegenerate variety, then $text{deg } X geq text{codim }X + 1$.
This can be found in the paper On Varieties of Minimal Degree by Eisenbud and Harris. The proof is as follows:
- If $text{codim }X = 1$, then $text{deg }X = 1$ if and only if $X$ is a hyperplane. $X$ is nondegenerate, so this is not the case. Hence $text{deg }X geq 2 = text{codim }X + 1$.
- If $text{codim }X geq 2$, choose a "general point of $X$", and project $X$ from that point to $mathbb{P}^{r-1}$.
- Observe that the degree of the projection of $X$ is $< text{deg }X$, and
similarly the codimension of the projection is $< text{codim }X$. - Proceed by induction on $text{codim } X$.
Question How can I see that codimension and degree of $X$ both drop by at least one?
In a class on surfaces I'm taking my professor proved the Lemma even in the case that $X$ is not irreducible, by intersecting $X$ with a general hyperplane, reducing the codimension of $X$ by one, while leaving the degree as is is. Here the induction goes over the dimension of $X$, and the basis is $n+1$ (or more) points, which need to have degree $geq n+1$.
Question How can I see that there exists a hyperplane, such that the codimension drops, and the degree stays the same?
I thought maybe this has something to do with the generalized Bézout's theorem, but I only know this for irreducible varieties. Is there a version of this for only reduced ones?
algebraic-geometry projective-geometry projective-varieties
algebraic-geometry projective-geometry projective-varieties
asked Dec 13 '18 at 19:20
red_trumpetred_trumpet
959319
asked Dec 13 '18 at 19:20
red_trumpetred_trumpet
959319
asked Dec 13 '18 at 19:20
red_trumpetred_trumpet
959319
asked Dec 13 '18 at 19:20
red_trumpetred_trumpet
959319
asked Dec 13 '18 at 19:20
red_trumpetred_trumpet
959319
959319
$begingroup$
How about following the exercise 7.7 from hartshorne Chapter 1 section 7? It seems that this exercise does actually help you complete the induction step.
$endgroup$
– random123
Dec 14 '18 at 10:54
$begingroup$
Ah, good catch, thanks! I will try to work through the exercise.
$endgroup$
– red_trumpet
Dec 14 '18 at 17:55
add a comment |
$begingroup$
How about following the exercise 7.7 from hartshorne Chapter 1 section 7? It seems that this exercise does actually help you complete the induction step.
$endgroup$
– random123
Dec 14 '18 at 10:54
$begingroup$
Ah, good catch, thanks! I will try to work through the exercise.
$endgroup$
– red_trumpet
Dec 14 '18 at 17:55
$begingroup$
How about following the exercise 7.7 from hartshorne Chapter 1 section 7? It seems that this exercise does actually help you complete the induction step.
$endgroup$
– random123
Dec 14 '18 at 10:54
$begingroup$
How about following the exercise 7.7 from hartshorne Chapter 1 section 7? It seems that this exercise does actually help you complete the induction step.
$endgroup$
– random123
Dec 14 '18 at 10:54
$begingroup$
Ah, good catch, thanks! I will try to work through the exercise.
$endgroup$
– red_trumpet
Dec 14 '18 at 17:55
$begingroup$
Ah, good catch, thanks! I will try to work through the exercise.
$endgroup$
– red_trumpet
Dec 14 '18 at 17:55
add a comment |
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$begingroup$
How about following the exercise 7.7 from hartshorne Chapter 1 section 7? It seems that this exercise does actually help you complete the induction step.
$endgroup$
– random123
Dec 14 '18 at 10:54
$begingroup$
Ah, good catch, thanks! I will try to work through the exercise.
$endgroup$
– red_trumpet
Dec 14 '18 at 17:55