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Dimension of irreducible components of single homogeneous polynomial
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I'm currently working my way through Algebraic Geometry by Andreas Gathmann and could use a hint to exercise 7.9:
Let $ f : mathbb{P}^n to mathbb{P}^m$ be a morphism. Prove:
If $ X subset mathbb{P}^m $ is the zero locus of a single homogeneous polynomial in $K[x_0,...,x_m]$ then every irreducible component of $f^{-1}(X)$ has dimension at least $n-1$.
I know that the irreducible components of $X$ in $mathbb{P}^m$ have dimension $m-1$, but how do i get the needed dimension in $mathbb{P}^n$.
Help appreciated
algebraic-geometry commutative-algebra
asked Jan 29 at 19:47
ASPASP
224
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add a comment |
$begingroup$
I'm currently working my way through Algebraic Geometry by Andreas Gathmann and could use a hint to exercise 7.9:
Let $ f : mathbb{P}^n to mathbb{P}^m$ be a morphism. Prove:
If $ X subset mathbb{P}^m $ is the zero locus of a single homogeneous polynomial in $K[x_0,...,x_m]$ then every irreducible component of $f^{-1}(X)$ has dimension at least $n-1$.
I know that the irreducible components of $X$ in $mathbb{P}^m$ have dimension $m-1$, but how do i get the needed dimension in $mathbb{P}^n$.
Help appreciated
algebraic-geometry commutative-algebra
asked Jan 29 at 19:47
ASPASP
224
$endgroup$
add a comment |
$begingroup$
I'm currently working my way through Algebraic Geometry by Andreas Gathmann and could use a hint to exercise 7.9:
Let $ f : mathbb{P}^n to mathbb{P}^m$ be a morphism. Prove:
If $ X subset mathbb{P}^m $ is the zero locus of a single homogeneous polynomial in $K[x_0,...,x_m]$ then every irreducible component of $f^{-1}(X)$ has dimension at least $n-1$.
I know that the irreducible components of $X$ in $mathbb{P}^m$ have dimension $m-1$, but how do i get the needed dimension in $mathbb{P}^n$.
Help appreciated
algebraic-geometry commutative-algebra
asked Jan 29 at 19:47
ASPASP
224
$endgroup$
I'm currently working my way through Algebraic Geometry by Andreas Gathmann and could use a hint to exercise 7.9:
Let $ f : mathbb{P}^n to mathbb{P}^m$ be a morphism. Prove:
If $ X subset mathbb{P}^m $ is the zero locus of a single homogeneous polynomial in $K[x_0,...,x_m]$ then every irreducible component of $f^{-1}(X)$ has dimension at least $n-1$.
I know that the irreducible components of $X$ in $mathbb{P}^m$ have dimension $m-1$, but how do i get the needed dimension in $mathbb{P}^n$.
Help appreciated
algebraic-geometry commutative-algebra
algebraic-geometry commutative-algebra
asked Jan 29 at 19:47
ASPASP
224
asked Jan 29 at 19:47
ASPASP
224
asked Jan 29 at 19:47
ASPASP
224
asked Jan 29 at 19:47
ASPASP
224
asked Jan 29 at 19:47
ASPASP
224
224
add a comment |
add a comment |
1 Answer
1
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oldest
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Hint: Use the following theorem (together with Krull's principal ideal theorem) to conclude
Theorem: Let $Usubseteq mathbb{P}^n$ be an open set and $f:Urightarrow mathbb{P^m}$ a morphism. Then there are unique (except for a constant) homogeneous polynomials $f_0,dots,f_min k[x_0,dots,x_n]$ of the same degree such that $f$ is globally defined by them, i.e,
$$f(x_0:dots:x_n)=(f_1(x_0:dots:x_n):dots:f_m(x_0:dots:x_n)) ; ; forall ,(x_0:dots:x_n)in U$$
Hint for the theorem: First prove the uniqueness, then as there is a cover $U=bigcup_{i=1}^n U_i$ such that $f$ is defined by homogeneous polynomials over each $U_i$ (this should be either your definition of morphism or something already proved) by the uniqueness over $U_icap U_j$ you can glue all this polynomials.
answered Jan 29 at 21:19
yamete kudasaiyamete kudasai
1,160818
$endgroup$
$begingroup$
Hello, thank you for providing a hint. The Theorem you mentioned is already proven in the notes ( lemma 7.5 ). But i still fail to see how i can get the needed proof. I would be thankful if you could extent your hint.
$endgroup$
– ASP
Jan 31 at 19:37
$begingroup$
The idea is that if $X=V(h)$ then $xin f^{-1}(X) iff$ $f(x)in V(h) iff$ $h(f(x))=0 iff xin V(hcirc f)$ So $f^{-1}(X)$ is defined by a single equation and hence is of pure codimension 1. The composition $hcirc f$ is a polynomial because it is the composition of polynomials thanks to the theorem.
$endgroup$
– yamete kudasai
Jan 31 at 20:31
$begingroup$
Also, I have just read Lemma 7.5 and is not quite the same. What this lemma tells you is that for a projective variety $Xsubset mathbb{P}^n$ you can construct morphisms with domain open subsets of $X$ using homogeneous elements. What the theorem above tells you is that when $X=mathbb{P}^n$ this are actually all the morphisms you can possible construct with domain an open set. This can fail when $X$ is not the entire projective space.
$endgroup$
– yamete kudasai
Jan 31 at 20:58
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Hint: Use the following theorem (together with Krull's principal ideal theorem) to conclude
Theorem: Let $Usubseteq mathbb{P}^n$ be an open set and $f:Urightarrow mathbb{P^m}$ a morphism. Then there are unique (except for a constant) homogeneous polynomials $f_0,dots,f_min k[x_0,dots,x_n]$ of the same degree such that $f$ is globally defined by them, i.e,
$$f(x_0:dots:x_n)=(f_1(x_0:dots:x_n):dots:f_m(x_0:dots:x_n)) ; ; forall ,(x_0:dots:x_n)in U$$
Hint for the theorem: First prove the uniqueness, then as there is a cover $U=bigcup_{i=1}^n U_i$ such that $f$ is defined by homogeneous polynomials over each $U_i$ (this should be either your definition of morphism or something already proved) by the uniqueness over $U_icap U_j$ you can glue all this polynomials.
answered Jan 29 at 21:19
yamete kudasaiyamete kudasai
1,160818
$endgroup$
$begingroup$
Hello, thank you for providing a hint. The Theorem you mentioned is already proven in the notes ( lemma 7.5 ). But i still fail to see how i can get the needed proof. I would be thankful if you could extent your hint.
$endgroup$
– ASP
Jan 31 at 19:37
$begingroup$
The idea is that if $X=V(h)$ then $xin f^{-1}(X) iff$ $f(x)in V(h) iff$ $h(f(x))=0 iff xin V(hcirc f)$ So $f^{-1}(X)$ is defined by a single equation and hence is of pure codimension 1. The composition $hcirc f$ is a polynomial because it is the composition of polynomials thanks to the theorem.
$endgroup$
– yamete kudasai
Jan 31 at 20:31
$begingroup$
Also, I have just read Lemma 7.5 and is not quite the same. What this lemma tells you is that for a projective variety $Xsubset mathbb{P}^n$ you can construct morphisms with domain open subsets of $X$ using homogeneous elements. What the theorem above tells you is that when $X=mathbb{P}^n$ this are actually all the morphisms you can possible construct with domain an open set. This can fail when $X$ is not the entire projective space.
$endgroup$
– yamete kudasai
Jan 31 at 20:58
add a comment |
$begingroup$
Hint: Use the following theorem (together with Krull's principal ideal theorem) to conclude
Theorem: Let $Usubseteq mathbb{P}^n$ be an open set and $f:Urightarrow mathbb{P^m}$ a morphism. Then there are unique (except for a constant) homogeneous polynomials $f_0,dots,f_min k[x_0,dots,x_n]$ of the same degree such that $f$ is globally defined by them, i.e,
$$f(x_0:dots:x_n)=(f_1(x_0:dots:x_n):dots:f_m(x_0:dots:x_n)) ; ; forall ,(x_0:dots:x_n)in U$$
Hint for the theorem: First prove the uniqueness, then as there is a cover $U=bigcup_{i=1}^n U_i$ such that $f$ is defined by homogeneous polynomials over each $U_i$ (this should be either your definition of morphism or something already proved) by the uniqueness over $U_icap U_j$ you can glue all this polynomials.
answered Jan 29 at 21:19
yamete kudasaiyamete kudasai
1,160818
$endgroup$
$begingroup$
Hello, thank you for providing a hint. The Theorem you mentioned is already proven in the notes ( lemma 7.5 ). But i still fail to see how i can get the needed proof. I would be thankful if you could extent your hint.
$endgroup$
– ASP
Jan 31 at 19:37
$begingroup$
The idea is that if $X=V(h)$ then $xin f^{-1}(X) iff$ $f(x)in V(h) iff$ $h(f(x))=0 iff xin V(hcirc f)$ So $f^{-1}(X)$ is defined by a single equation and hence is of pure codimension 1. The composition $hcirc f$ is a polynomial because it is the composition of polynomials thanks to the theorem.
$endgroup$
– yamete kudasai
Jan 31 at 20:31
$begingroup$
Also, I have just read Lemma 7.5 and is not quite the same. What this lemma tells you is that for a projective variety $Xsubset mathbb{P}^n$ you can construct morphisms with domain open subsets of $X$ using homogeneous elements. What the theorem above tells you is that when $X=mathbb{P}^n$ this are actually all the morphisms you can possible construct with domain an open set. This can fail when $X$ is not the entire projective space.
$endgroup$
– yamete kudasai
Jan 31 at 20:58
add a comment |
$begingroup$
Hint: Use the following theorem (together with Krull's principal ideal theorem) to conclude
Theorem: Let $Usubseteq mathbb{P}^n$ be an open set and $f:Urightarrow mathbb{P^m}$ a morphism. Then there are unique (except for a constant) homogeneous polynomials $f_0,dots,f_min k[x_0,dots,x_n]$ of the same degree such that $f$ is globally defined by them, i.e,
$$f(x_0:dots:x_n)=(f_1(x_0:dots:x_n):dots:f_m(x_0:dots:x_n)) ; ; forall ,(x_0:dots:x_n)in U$$
Hint for the theorem: First prove the uniqueness, then as there is a cover $U=bigcup_{i=1}^n U_i$ such that $f$ is defined by homogeneous polynomials over each $U_i$ (this should be either your definition of morphism or something already proved) by the uniqueness over $U_icap U_j$ you can glue all this polynomials.
answered Jan 29 at 21:19
yamete kudasaiyamete kudasai
1,160818
$endgroup$
Hint: Use the following theorem (together with Krull's principal ideal theorem) to conclude
Theorem: Let $Usubseteq mathbb{P}^n$ be an open set and $f:Urightarrow mathbb{P^m}$ a morphism. Then there are unique (except for a constant) homogeneous polynomials $f_0,dots,f_min k[x_0,dots,x_n]$ of the same degree such that $f$ is globally defined by them, i.e,
$$f(x_0:dots:x_n)=(f_1(x_0:dots:x_n):dots:f_m(x_0:dots:x_n)) ; ; forall ,(x_0:dots:x_n)in U$$
Hint for the theorem: First prove the uniqueness, then as there is a cover $U=bigcup_{i=1}^n U_i$ such that $f$ is defined by homogeneous polynomials over each $U_i$ (this should be either your definition of morphism or something already proved) by the uniqueness over $U_icap U_j$ you can glue all this polynomials.
answered Jan 29 at 21:19
yamete kudasaiyamete kudasai
1,160818
edited Jan 29 at 21:32
answered Jan 29 at 21:19
yamete kudasaiyamete kudasai
1,160818
answered Jan 29 at 21:19
yamete kudasaiyamete kudasai
1,160818
answered Jan 29 at 21:19
yamete kudasaiyamete kudasai
1,160818
1,160818
$begingroup$
Hello, thank you for providing a hint. The Theorem you mentioned is already proven in the notes ( lemma 7.5 ). But i still fail to see how i can get the needed proof. I would be thankful if you could extent your hint.
$endgroup$
– ASP
Jan 31 at 19:37
$begingroup$
The idea is that if $X=V(h)$ then $xin f^{-1}(X) iff$ $f(x)in V(h) iff$ $h(f(x))=0 iff xin V(hcirc f)$ So $f^{-1}(X)$ is defined by a single equation and hence is of pure codimension 1. The composition $hcirc f$ is a polynomial because it is the composition of polynomials thanks to the theorem.
$endgroup$
– yamete kudasai
Jan 31 at 20:31
$begingroup$
Also, I have just read Lemma 7.5 and is not quite the same. What this lemma tells you is that for a projective variety $Xsubset mathbb{P}^n$ you can construct morphisms with domain open subsets of $X$ using homogeneous elements. What the theorem above tells you is that when $X=mathbb{P}^n$ this are actually all the morphisms you can possible construct with domain an open set. This can fail when $X$ is not the entire projective space.
$endgroup$
– yamete kudasai
Jan 31 at 20:58
add a comment |
$begingroup$
Hello, thank you for providing a hint. The Theorem you mentioned is already proven in the notes ( lemma 7.5 ). But i still fail to see how i can get the needed proof. I would be thankful if you could extent your hint.
$endgroup$
– ASP
Jan 31 at 19:37
$begingroup$
The idea is that if $X=V(h)$ then $xin f^{-1}(X) iff$ $f(x)in V(h) iff$ $h(f(x))=0 iff xin V(hcirc f)$ So $f^{-1}(X)$ is defined by a single equation and hence is of pure codimension 1. The composition $hcirc f$ is a polynomial because it is the composition of polynomials thanks to the theorem.
$endgroup$
– yamete kudasai
Jan 31 at 20:31
$begingroup$
Also, I have just read Lemma 7.5 and is not quite the same. What this lemma tells you is that for a projective variety $Xsubset mathbb{P}^n$ you can construct morphisms with domain open subsets of $X$ using homogeneous elements. What the theorem above tells you is that when $X=mathbb{P}^n$ this are actually all the morphisms you can possible construct with domain an open set. This can fail when $X$ is not the entire projective space.
$endgroup$
– yamete kudasai
Jan 31 at 20:58
$begingroup$
Hello, thank you for providing a hint. The Theorem you mentioned is already proven in the notes ( lemma 7.5 ). But i still fail to see how i can get the needed proof. I would be thankful if you could extent your hint.
$endgroup$
– ASP
Jan 31 at 19:37
$begingroup$
Hello, thank you for providing a hint. The Theorem you mentioned is already proven in the notes ( lemma 7.5 ). But i still fail to see how i can get the needed proof. I would be thankful if you could extent your hint.
$endgroup$
– ASP
Jan 31 at 19:37
$begingroup$
The idea is that if $X=V(h)$ then $xin f^{-1}(X) iff$ $f(x)in V(h) iff$ $h(f(x))=0 iff xin V(hcirc f)$ So $f^{-1}(X)$ is defined by a single equation and hence is of pure codimension 1. The composition $hcirc f$ is a polynomial because it is the composition of polynomials thanks to the theorem.
$endgroup$
– yamete kudasai
Jan 31 at 20:31
$begingroup$
The idea is that if $X=V(h)$ then $xin f^{-1}(X) iff$ $f(x)in V(h) iff$ $h(f(x))=0 iff xin V(hcirc f)$ So $f^{-1}(X)$ is defined by a single equation and hence is of pure codimension 1. The composition $hcirc f$ is a polynomial because it is the composition of polynomials thanks to the theorem.
$endgroup$
– yamete kudasai
Jan 31 at 20:31
$begingroup$
Also, I have just read Lemma 7.5 and is not quite the same. What this lemma tells you is that for a projective variety $Xsubset mathbb{P}^n$ you can construct morphisms with domain open subsets of $X$ using homogeneous elements. What the theorem above tells you is that when $X=mathbb{P}^n$ this are actually all the morphisms you can possible construct with domain an open set. This can fail when $X$ is not the entire projective space.
$endgroup$
– yamete kudasai
Jan 31 at 20:58
$begingroup$
Also, I have just read Lemma 7.5 and is not quite the same. What this lemma tells you is that for a projective variety $Xsubset mathbb{P}^n$ you can construct morphisms with domain open subsets of $X$ using homogeneous elements. What the theorem above tells you is that when $X=mathbb{P}^n$ this are actually all the morphisms you can possible construct with domain an open set. This can fail when $X$ is not the entire projective space.
$endgroup$
– yamete kudasai
Jan 31 at 20:58
add a comment |
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