Prove that a map is open in the Zariski topology
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Say $k$ is an algebraically closed field and define the equivalence relation on $k^{n+1}$ given by $x sim y iff x=lambda y $ for some $lambda in mathbb{k}^{times}$. Clearly $mathbb{P}^{n} = k^{n+1}/sim$. Let the map $q:k^{n+1} setminus 0 longrightarrow mathbb{P}^{n}$ be the quotient map that sends $x$ to its equivalence class. I have to show that it is an open map (for the Zariski topology). Any hint is appreciated.
general-topology algebraic-geometry projective-geometry
asked Jan 25 at 11:47
DalamarDalamar
465410
$endgroup$
add a comment |
$begingroup$
Say $k$ is an algebraically closed field and define the equivalence relation on $k^{n+1}$ given by $x sim y iff x=lambda y $ for some $lambda in mathbb{k}^{times}$. Clearly $mathbb{P}^{n} = k^{n+1}/sim$. Let the map $q:k^{n+1} setminus 0 longrightarrow mathbb{P}^{n}$ be the quotient map that sends $x$ to its equivalence class. I have to show that it is an open map (for the Zariski topology). Any hint is appreciated.
general-topology algebraic-geometry projective-geometry
asked Jan 25 at 11:47
DalamarDalamar
465410
$endgroup$
$begingroup$
What is your definition for the topology on $mathbb{P^n}$?
$endgroup$
– trii
Jan 25 at 16:07
$begingroup$
Quotient topology
$endgroup$
– Dalamar
Jan 25 at 16:12
add a comment |
$begingroup$
Say $k$ is an algebraically closed field and define the equivalence relation on $k^{n+1}$ given by $x sim y iff x=lambda y $ for some $lambda in mathbb{k}^{times}$. Clearly $mathbb{P}^{n} = k^{n+1}/sim$. Let the map $q:k^{n+1} setminus 0 longrightarrow mathbb{P}^{n}$ be the quotient map that sends $x$ to its equivalence class. I have to show that it is an open map (for the Zariski topology). Any hint is appreciated.
general-topology algebraic-geometry projective-geometry
asked Jan 25 at 11:47
DalamarDalamar
465410
$endgroup$
Say $k$ is an algebraically closed field and define the equivalence relation on $k^{n+1}$ given by $x sim y iff x=lambda y $ for some $lambda in mathbb{k}^{times}$. Clearly $mathbb{P}^{n} = k^{n+1}/sim$. Let the map $q:k^{n+1} setminus 0 longrightarrow mathbb{P}^{n}$ be the quotient map that sends $x$ to its equivalence class. I have to show that it is an open map (for the Zariski topology). Any hint is appreciated.
general-topology algebraic-geometry projective-geometry
general-topology algebraic-geometry projective-geometry
asked Jan 25 at 11:47
DalamarDalamar
465410
asked Jan 25 at 11:47
DalamarDalamar
465410
asked Jan 25 at 11:47
DalamarDalamar
465410
asked Jan 25 at 11:47
DalamarDalamar
465410
asked Jan 25 at 11:47
DalamarDalamar
465410
465410
$begingroup$
What is your definition for the topology on $mathbb{P^n}$?
$endgroup$
– trii
Jan 25 at 16:07
$begingroup$
Quotient topology
$endgroup$
– Dalamar
Jan 25 at 16:12
add a comment |
$begingroup$
What is your definition for the topology on $mathbb{P^n}$?
$endgroup$
– trii
Jan 25 at 16:07
$begingroup$
Quotient topology
$endgroup$
– Dalamar
Jan 25 at 16:12
$begingroup$
What is your definition for the topology on $mathbb{P^n}$?
$endgroup$
– trii
Jan 25 at 16:07
$begingroup$
What is your definition for the topology on $mathbb{P^n}$?
$endgroup$
– trii
Jan 25 at 16:07
$begingroup$
Quotient topology
$endgroup$
– Dalamar
Jan 25 at 16:12
$begingroup$
Quotient topology
$endgroup$
– Dalamar
Jan 25 at 16:12
add a comment |
1 Answer
1
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$begingroup$
Hint:
$q^{-1}(q(U))=bigcup_{lambdain k^{times}}lambda U$
Multiplication by $lambdaneq 0$ is a homeomorphism as for basic opens $lambda D(f)=D(f(frac{1}{lambda}(cdot))) $
answered Jan 25 at 15:28
triitrii
2985
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add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint:
$q^{-1}(q(U))=bigcup_{lambdain k^{times}}lambda U$
Multiplication by $lambdaneq 0$ is a homeomorphism as for basic opens $lambda D(f)=D(f(frac{1}{lambda}(cdot))) $
answered Jan 25 at 15:28
triitrii
2985
$endgroup$
add a comment |
$begingroup$
Hint:
$q^{-1}(q(U))=bigcup_{lambdain k^{times}}lambda U$
Multiplication by $lambdaneq 0$ is a homeomorphism as for basic opens $lambda D(f)=D(f(frac{1}{lambda}(cdot))) $
answered Jan 25 at 15:28
triitrii
2985
$endgroup$
add a comment |
$begingroup$
Hint:
$q^{-1}(q(U))=bigcup_{lambdain k^{times}}lambda U$
Multiplication by $lambdaneq 0$ is a homeomorphism as for basic opens $lambda D(f)=D(f(frac{1}{lambda}(cdot))) $
answered Jan 25 at 15:28
triitrii
2985
$endgroup$
Hint:
$q^{-1}(q(U))=bigcup_{lambdain k^{times}}lambda U$
Multiplication by $lambdaneq 0$ is a homeomorphism as for basic opens $lambda D(f)=D(f(frac{1}{lambda}(cdot))) $
answered Jan 25 at 15:28
triitrii
2985
edited Jan 25 at 16:14
answered Jan 25 at 15:28
triitrii
2985
answered Jan 25 at 15:28
triitrii
2985
answered Jan 25 at 15:28
triitrii
2985
2985
add a comment |
add a comment |
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$begingroup$
What is your definition for the topology on $mathbb{P^n}$?
$endgroup$
– trii
Jan 25 at 16:07
$begingroup$
Quotient topology
$endgroup$
– Dalamar
Jan 25 at 16:12