Why non surjective polynomial map between $A^n$ and $A^m$ demand image closed algebraic set?
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Let $f_iin C[x_1,dots, x_n]$ with $ileq m$. Suppose $f_i$ are algebraically independent functions. Consider the map $phi:A^nto A^m$ by $phi(x_1,dots, x_n)=(f_i(x_1,dots, x_n))_i$. Suppose $phi$ is non surjection. Then $Im(phi)$ is closed algebraic subset of $A^n$.
$textbf{Q:}$ Why is $Im(phi)$ a closed subset if $phi$ is not surjection? I could imagine $A^2to A^2$ by $(x,y)to (x,xy)$. Clearly $x,xy$ are algebraically independent. Furhtermore, the map is not surjection as part of $x=0$ axis is missing. It is worse that the image is not closed algebraic set. Have I misunderstood the statement?
Context: The book tries to explain $f_i$ algebraic independence/essential parameters is equivalent to surjectivity of the map $phi$ above.
Ref. Beltrametti M.C., et al. - Lectures on curves, surfaces and projective varieties Sec 3.2 of Chpt 3, Pg 62
abstract-algebra algebraic-geometry
asked Dec 27 '18 at 0:19
user45765user45765
2,7332722
$endgroup$
add a comment |
$begingroup$
Let $f_iin C[x_1,dots, x_n]$ with $ileq m$. Suppose $f_i$ are algebraically independent functions. Consider the map $phi:A^nto A^m$ by $phi(x_1,dots, x_n)=(f_i(x_1,dots, x_n))_i$. Suppose $phi$ is non surjection. Then $Im(phi)$ is closed algebraic subset of $A^n$.
$textbf{Q:}$ Why is $Im(phi)$ a closed subset if $phi$ is not surjection? I could imagine $A^2to A^2$ by $(x,y)to (x,xy)$. Clearly $x,xy$ are algebraically independent. Furhtermore, the map is not surjection as part of $x=0$ axis is missing. It is worse that the image is not closed algebraic set. Have I misunderstood the statement?
Context: The book tries to explain $f_i$ algebraic independence/essential parameters is equivalent to surjectivity of the map $phi$ above.
Ref. Beltrametti M.C., et al. - Lectures on curves, surfaces and projective varieties Sec 3.2 of Chpt 3, Pg 62
abstract-algebra algebraic-geometry
asked Dec 27 '18 at 0:19
user45765user45765
2,7332722
$endgroup$
add a comment |
$begingroup$
Let $f_iin C[x_1,dots, x_n]$ with $ileq m$. Suppose $f_i$ are algebraically independent functions. Consider the map $phi:A^nto A^m$ by $phi(x_1,dots, x_n)=(f_i(x_1,dots, x_n))_i$. Suppose $phi$ is non surjection. Then $Im(phi)$ is closed algebraic subset of $A^n$.
$textbf{Q:}$ Why is $Im(phi)$ a closed subset if $phi$ is not surjection? I could imagine $A^2to A^2$ by $(x,y)to (x,xy)$. Clearly $x,xy$ are algebraically independent. Furhtermore, the map is not surjection as part of $x=0$ axis is missing. It is worse that the image is not closed algebraic set. Have I misunderstood the statement?
Context: The book tries to explain $f_i$ algebraic independence/essential parameters is equivalent to surjectivity of the map $phi$ above.
Ref. Beltrametti M.C., et al. - Lectures on curves, surfaces and projective varieties Sec 3.2 of Chpt 3, Pg 62
abstract-algebra algebraic-geometry
asked Dec 27 '18 at 0:19
user45765user45765
2,7332722
$endgroup$
Let $f_iin C[x_1,dots, x_n]$ with $ileq m$. Suppose $f_i$ are algebraically independent functions. Consider the map $phi:A^nto A^m$ by $phi(x_1,dots, x_n)=(f_i(x_1,dots, x_n))_i$. Suppose $phi$ is non surjection. Then $Im(phi)$ is closed algebraic subset of $A^n$.
$textbf{Q:}$ Why is $Im(phi)$ a closed subset if $phi$ is not surjection? I could imagine $A^2to A^2$ by $(x,y)to (x,xy)$. Clearly $x,xy$ are algebraically independent. Furhtermore, the map is not surjection as part of $x=0$ axis is missing. It is worse that the image is not closed algebraic set. Have I misunderstood the statement?
Context: The book tries to explain $f_i$ algebraic independence/essential parameters is equivalent to surjectivity of the map $phi$ above.
Ref. Beltrametti M.C., et al. - Lectures on curves, surfaces and projective varieties Sec 3.2 of Chpt 3, Pg 62
abstract-algebra algebraic-geometry
abstract-algebra algebraic-geometry
asked Dec 27 '18 at 0:19
user45765user45765
2,7332722
asked Dec 27 '18 at 0:19
user45765user45765
2,7332722
asked Dec 27 '18 at 0:19
user45765user45765
2,7332722
asked Dec 27 '18 at 0:19
user45765user45765
2,7332722
asked Dec 27 '18 at 0:19
user45765user45765
2,7332722
2,7332722
add a comment |
add a comment |
1 Answer
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This is just an error in the text. You are correct that the image of $varphi$ need not be closed. The result being proved (that $varphi$ is surjective iff the $f_i$ are algebraically independent) is simply wrong, as your example also shows. The correct statement is that $varphi$ is dominant (has dense image) iff the $f_i$ are algebraically independent.
answered Dec 27 '18 at 0:47
Eric WofseyEric Wofsey
186k14215342
$endgroup$
$begingroup$
Ah in that case, it seems very reasonable.
$endgroup$
– user45765
Dec 27 '18 at 0:48
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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active
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active
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votes
$begingroup$
This is just an error in the text. You are correct that the image of $varphi$ need not be closed. The result being proved (that $varphi$ is surjective iff the $f_i$ are algebraically independent) is simply wrong, as your example also shows. The correct statement is that $varphi$ is dominant (has dense image) iff the $f_i$ are algebraically independent.
answered Dec 27 '18 at 0:47
Eric WofseyEric Wofsey
186k14215342
$endgroup$
$begingroup$
Ah in that case, it seems very reasonable.
$endgroup$
– user45765
Dec 27 '18 at 0:48
add a comment |
$begingroup$
This is just an error in the text. You are correct that the image of $varphi$ need not be closed. The result being proved (that $varphi$ is surjective iff the $f_i$ are algebraically independent) is simply wrong, as your example also shows. The correct statement is that $varphi$ is dominant (has dense image) iff the $f_i$ are algebraically independent.
answered Dec 27 '18 at 0:47
Eric WofseyEric Wofsey
186k14215342
$endgroup$
$begingroup$
Ah in that case, it seems very reasonable.
$endgroup$
– user45765
Dec 27 '18 at 0:48
add a comment |
$begingroup$
This is just an error in the text. You are correct that the image of $varphi$ need not be closed. The result being proved (that $varphi$ is surjective iff the $f_i$ are algebraically independent) is simply wrong, as your example also shows. The correct statement is that $varphi$ is dominant (has dense image) iff the $f_i$ are algebraically independent.
answered Dec 27 '18 at 0:47
Eric WofseyEric Wofsey
186k14215342
$endgroup$
This is just an error in the text. You are correct that the image of $varphi$ need not be closed. The result being proved (that $varphi$ is surjective iff the $f_i$ are algebraically independent) is simply wrong, as your example also shows. The correct statement is that $varphi$ is dominant (has dense image) iff the $f_i$ are algebraically independent.
answered Dec 27 '18 at 0:47
Eric WofseyEric Wofsey
186k14215342
answered Dec 27 '18 at 0:47
Eric WofseyEric Wofsey
186k14215342
answered Dec 27 '18 at 0:47
Eric WofseyEric Wofsey
186k14215342
answered Dec 27 '18 at 0:47
Eric WofseyEric Wofsey
186k14215342
186k14215342
$begingroup$
Ah in that case, it seems very reasonable.
$endgroup$
– user45765
Dec 27 '18 at 0:48
add a comment |
$begingroup$
Ah in that case, it seems very reasonable.
$endgroup$
– user45765
Dec 27 '18 at 0:48
$begingroup$
Ah in that case, it seems very reasonable.
$endgroup$
– user45765
Dec 27 '18 at 0:48
$begingroup$
Ah in that case, it seems very reasonable.
$endgroup$
– user45765
Dec 27 '18 at 0:48
add a comment |
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