Justification for $f_ig_j equiv 0$ in $U cup V = mathbb C^n$ for varieties $U$ and $V$.
$begingroup$
Artin Algebra Chapter 11
Here are solutions by Takumi Murayama here and Brian Bi. Here is the definition of a variety in $mathbb C^n$
Here is Takumi Murayama's solution for $U cup V = mathbb C^n$
- I have questions about the finite zeroes part.
Why are there finite zeroes? Is this a generalization of the following theorem for 2 polynomials and $mathbb C^2$ to finitely many polynomials and $mathbb C^n$
?
- I have a question about a different argument.
Can we instead argue that the only such polynomial in the variety $mathbb C^n$ is the zero polynomial?
My argument is that $mathbb C^n$ as a variety is defined by $$mathbb C^n := {text{the zero polynomial}=text{the number zero}}$$
So without references to finiteness, we must have $f_ig_j equiv text{the zero polynomial} forall i,j$. Then we conclude $U=mathbb C^n$ or $V = mathbb C^n$.
abstract-algebra geometry algebraic-geometry polynomials ring-theory
$endgroup$
add a comment |
$begingroup$
Artin Algebra Chapter 11
Here are solutions by Takumi Murayama here and Brian Bi. Here is the definition of a variety in $mathbb C^n$
Here is Takumi Murayama's solution for $U cup V = mathbb C^n$
- I have questions about the finite zeroes part.
Why are there finite zeroes? Is this a generalization of the following theorem for 2 polynomials and $mathbb C^2$ to finitely many polynomials and $mathbb C^n$
?
- I have a question about a different argument.
Can we instead argue that the only such polynomial in the variety $mathbb C^n$ is the zero polynomial?
My argument is that $mathbb C^n$ as a variety is defined by $$mathbb C^n := {text{the zero polynomial}=text{the number zero}}$$
So without references to finiteness, we must have $f_ig_j equiv text{the zero polynomial} forall i,j$. Then we conclude $U=mathbb C^n$ or $V = mathbb C^n$.
abstract-algebra geometry algebraic-geometry polynomials ring-theory
$endgroup$
add a comment |
$begingroup$
Artin Algebra Chapter 11
Here are solutions by Takumi Murayama here and Brian Bi. Here is the definition of a variety in $mathbb C^n$
Here is Takumi Murayama's solution for $U cup V = mathbb C^n$
- I have questions about the finite zeroes part.
Why are there finite zeroes? Is this a generalization of the following theorem for 2 polynomials and $mathbb C^2$ to finitely many polynomials and $mathbb C^n$
?
- I have a question about a different argument.
Can we instead argue that the only such polynomial in the variety $mathbb C^n$ is the zero polynomial?
My argument is that $mathbb C^n$ as a variety is defined by $$mathbb C^n := {text{the zero polynomial}=text{the number zero}}$$
So without references to finiteness, we must have $f_ig_j equiv text{the zero polynomial} forall i,j$. Then we conclude $U=mathbb C^n$ or $V = mathbb C^n$.
abstract-algebra geometry algebraic-geometry polynomials ring-theory
$endgroup$
Artin Algebra Chapter 11
Here are solutions by Takumi Murayama here and Brian Bi. Here is the definition of a variety in $mathbb C^n$
Here is Takumi Murayama's solution for $U cup V = mathbb C^n$
- I have questions about the finite zeroes part.
Why are there finite zeroes? Is this a generalization of the following theorem for 2 polynomials and $mathbb C^2$ to finitely many polynomials and $mathbb C^n$
?
- I have a question about a different argument.
Can we instead argue that the only such polynomial in the variety $mathbb C^n$ is the zero polynomial?
My argument is that $mathbb C^n$ as a variety is defined by $$mathbb C^n := {text{the zero polynomial}=text{the number zero}}$$
So without references to finiteness, we must have $f_ig_j equiv text{the zero polynomial} forall i,j$. Then we conclude $U=mathbb C^n$ or $V = mathbb C^n$.
abstract-algebra geometry algebraic-geometry polynomials ring-theory
abstract-algebra geometry algebraic-geometry polynomials ring-theory
edited Dec 29 '18 at 9:07
asked Dec 29 '18 at 9:00
user198044
add a comment |
add a comment |
1 Answer
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$begingroup$
Murayama's solution is not correct as stated: in $Bbb C^2$, the polynomial $x$ has the $y$-axis as it's zeroes, which is an infinite set. A statement which is close to Murayama's which is correct would be that the zeroes of a nonconstant polynomial consist of a finite number of irreducible components. These irreducible components are small, and it cannot be the case that a union of finitely many of them is the whole of $Bbb C^n$ - there are many different proofs (they're meager and one may apply the Baire category theorem, there are proofs from the perspective of basic ring theory, dimension-theoretic proofs, etc).
For your second question, there are some issues with the language, but the idea is essentially correct. It would be best to say that "the only set* of polynomials which defines $Bbb C^n$ is the set consisting of the zero polynomial". This statement is correct, and the justification is that every nonconstant polynomial over $Bbb C$ must be nonvanishing at some point (for example, via the fundamental theorem of algebra), and therefore cannot be in the set defining $Bbb C^n$.
(*) soon you will probably be more used to speaking about ideals defining varieties than sets of polynomials, and the correct adjustment to the language is that the ideal of $Bbb C^n$ is exactly the zero ideal.
answered Dec 29 '18 at 9:18
KReiserKReiser
10.2k21435
$endgroup$
$begingroup$
Thank you KReiser!
$endgroup$
– user198044
Dec 29 '18 at 9:35
add a comment |
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1 Answer
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$begingroup$
Murayama's solution is not correct as stated: in $Bbb C^2$, the polynomial $x$ has the $y$-axis as it's zeroes, which is an infinite set. A statement which is close to Murayama's which is correct would be that the zeroes of a nonconstant polynomial consist of a finite number of irreducible components. These irreducible components are small, and it cannot be the case that a union of finitely many of them is the whole of $Bbb C^n$ - there are many different proofs (they're meager and one may apply the Baire category theorem, there are proofs from the perspective of basic ring theory, dimension-theoretic proofs, etc).
For your second question, there are some issues with the language, but the idea is essentially correct. It would be best to say that "the only set* of polynomials which defines $Bbb C^n$ is the set consisting of the zero polynomial". This statement is correct, and the justification is that every nonconstant polynomial over $Bbb C$ must be nonvanishing at some point (for example, via the fundamental theorem of algebra), and therefore cannot be in the set defining $Bbb C^n$.
(*) soon you will probably be more used to speaking about ideals defining varieties than sets of polynomials, and the correct adjustment to the language is that the ideal of $Bbb C^n$ is exactly the zero ideal.
answered Dec 29 '18 at 9:18
KReiserKReiser
10.2k21435
$endgroup$
$begingroup$
Thank you KReiser!
$endgroup$
– user198044
Dec 29 '18 at 9:35
add a comment |
$begingroup$
Murayama's solution is not correct as stated: in $Bbb C^2$, the polynomial $x$ has the $y$-axis as it's zeroes, which is an infinite set. A statement which is close to Murayama's which is correct would be that the zeroes of a nonconstant polynomial consist of a finite number of irreducible components. These irreducible components are small, and it cannot be the case that a union of finitely many of them is the whole of $Bbb C^n$ - there are many different proofs (they're meager and one may apply the Baire category theorem, there are proofs from the perspective of basic ring theory, dimension-theoretic proofs, etc).
For your second question, there are some issues with the language, but the idea is essentially correct. It would be best to say that "the only set* of polynomials which defines $Bbb C^n$ is the set consisting of the zero polynomial". This statement is correct, and the justification is that every nonconstant polynomial over $Bbb C$ must be nonvanishing at some point (for example, via the fundamental theorem of algebra), and therefore cannot be in the set defining $Bbb C^n$.
(*) soon you will probably be more used to speaking about ideals defining varieties than sets of polynomials, and the correct adjustment to the language is that the ideal of $Bbb C^n$ is exactly the zero ideal.
answered Dec 29 '18 at 9:18
KReiserKReiser
10.2k21435
$endgroup$
$begingroup$
Thank you KReiser!
$endgroup$
– user198044
Dec 29 '18 at 9:35
add a comment |
$begingroup$
Murayama's solution is not correct as stated: in $Bbb C^2$, the polynomial $x$ has the $y$-axis as it's zeroes, which is an infinite set. A statement which is close to Murayama's which is correct would be that the zeroes of a nonconstant polynomial consist of a finite number of irreducible components. These irreducible components are small, and it cannot be the case that a union of finitely many of them is the whole of $Bbb C^n$ - there are many different proofs (they're meager and one may apply the Baire category theorem, there are proofs from the perspective of basic ring theory, dimension-theoretic proofs, etc).
For your second question, there are some issues with the language, but the idea is essentially correct. It would be best to say that "the only set* of polynomials which defines $Bbb C^n$ is the set consisting of the zero polynomial". This statement is correct, and the justification is that every nonconstant polynomial over $Bbb C$ must be nonvanishing at some point (for example, via the fundamental theorem of algebra), and therefore cannot be in the set defining $Bbb C^n$.
(*) soon you will probably be more used to speaking about ideals defining varieties than sets of polynomials, and the correct adjustment to the language is that the ideal of $Bbb C^n$ is exactly the zero ideal.
answered Dec 29 '18 at 9:18
KReiserKReiser
10.2k21435
$endgroup$
Murayama's solution is not correct as stated: in $Bbb C^2$, the polynomial $x$ has the $y$-axis as it's zeroes, which is an infinite set. A statement which is close to Murayama's which is correct would be that the zeroes of a nonconstant polynomial consist of a finite number of irreducible components. These irreducible components are small, and it cannot be the case that a union of finitely many of them is the whole of $Bbb C^n$ - there are many different proofs (they're meager and one may apply the Baire category theorem, there are proofs from the perspective of basic ring theory, dimension-theoretic proofs, etc).
For your second question, there are some issues with the language, but the idea is essentially correct. It would be best to say that "the only set* of polynomials which defines $Bbb C^n$ is the set consisting of the zero polynomial". This statement is correct, and the justification is that every nonconstant polynomial over $Bbb C$ must be nonvanishing at some point (for example, via the fundamental theorem of algebra), and therefore cannot be in the set defining $Bbb C^n$.
(*) soon you will probably be more used to speaking about ideals defining varieties than sets of polynomials, and the correct adjustment to the language is that the ideal of $Bbb C^n$ is exactly the zero ideal.
answered Dec 29 '18 at 9:18
KReiserKReiser
10.2k21435
answered Dec 29 '18 at 9:18
KReiserKReiser
10.2k21435
answered Dec 29 '18 at 9:18
KReiserKReiser
10.2k21435
answered Dec 29 '18 at 9:18
KReiserKReiser
10.2k21435
10.2k21435
$begingroup$
Thank you KReiser!
$endgroup$
– user198044
Dec 29 '18 at 9:35
add a comment |
$begingroup$
Thank you KReiser!
$endgroup$
– user198044
Dec 29 '18 at 9:35
$begingroup$
Thank you KReiser!
$endgroup$
– user198044
Dec 29 '18 at 9:35
$begingroup$
Thank you KReiser!
$endgroup$
– user198044
Dec 29 '18 at 9:35
add a comment |
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