Mixing
Zariski closure and affine transformations commute
$begingroup$
Let $mathbb K$ be a field, $Xsubseteq mathbb K^n$ a subset, and $ f:mathbb K^nrightarrow mathbb K^m$ an affine map. Using the standard notation $V$, $I$ for algebraic geometry, it should hold that
$$V(I(f(X)))=f(V(I(X))),$$
i.e. that the Zariski closure $VI$ commutes with the affine transformation $f$. Why is that?
(To specify the notation more: for a set $Y$, we mean by $I(Y)$ the set of all polynomials vanishing on $Y$; and for a set of polynomials $J$, we mean by $V(J)$ the set of all zeros common to all polynomials in $J$.)
Note that I do NOT assume $f$ to be bijective, and that $X$ does not need to be an algebraic set.
The equality obviously doesn't hold for every polynomial mapping $ f$, as that would mean that an image of Zariski closed set under polynomial mapping is always Zariski closed, which is not true (e.g. $f(x)=x^2:mathbb Rrightarrow mathbb R$ and $X=mathbb R$).
EDIT:
As pointed out by reuns, using the general topology fact that $overline{f(X)}supseteq f(overline{X})$ for continuous maps, it is enough to show that $f(V(I(X)))$ is Zariski closed (i.e. an algebraic set).
algebraic-geometry polynomials
asked Jan 21 at 12:38
OnDragiOnDragi
656
$endgroup$
|
show 3 more comments
$begingroup$
Let $mathbb K$ be a field, $Xsubseteq mathbb K^n$ a subset, and $ f:mathbb K^nrightarrow mathbb K^m$ an affine map. Using the standard notation $V$, $I$ for algebraic geometry, it should hold that
$$V(I(f(X)))=f(V(I(X))),$$
i.e. that the Zariski closure $VI$ commutes with the affine transformation $f$. Why is that?
(To specify the notation more: for a set $Y$, we mean by $I(Y)$ the set of all polynomials vanishing on $Y$; and for a set of polynomials $J$, we mean by $V(J)$ the set of all zeros common to all polynomials in $J$.)
Note that I do NOT assume $f$ to be bijective, and that $X$ does not need to be an algebraic set.
The equality obviously doesn't hold for every polynomial mapping $ f$, as that would mean that an image of Zariski closed set under polynomial mapping is always Zariski closed, which is not true (e.g. $f(x)=x^2:mathbb Rrightarrow mathbb R$ and $X=mathbb R$).
EDIT:
As pointed out by reuns, using the general topology fact that $overline{f(X)}supseteq f(overline{X})$ for continuous maps, it is enough to show that $f(V(I(X)))$ is Zariski closed (i.e. an algebraic set).
algebraic-geometry polynomials
asked Jan 21 at 12:38
OnDragiOnDragi
656
$endgroup$
$begingroup$
The reformulation of the question would be "if $f$ is $textbf{affine}$ then $overline{f(X)}=f(overline{X})$". The equality does not hold for $f$ continuous in general, as my example shows. If $f(x)=x^2$, we have $overline{f(R)}=overline{[0,inf)}=R neq [0,inf)=f(R)=f(overline{R})$. But you are right that $supseteq$ holds in general, so that shows one direction.
$endgroup$
– OnDragi
Jan 21 at 13:26
$begingroup$
In what sense is $f : mathbb{R} to mathbb{R}, x to x^2$ continuous, is it continuous for the topology induced by $V(I(X))$ ?
$endgroup$
– reuns
Jan 21 at 13:40
$begingroup$
Yes, it is a polynomial map, and polynomial maps are continuous w.r.t. Zariski topology.
$endgroup$
– OnDragi
Jan 21 at 13:42
$begingroup$
$overline{f(X)} = f(overline{X})$ is true whenever $f$ maps closed sets to closed sets (that's part of your definition of affine right ?). Here you are saying $f^{-1}(U)$ is open whenever $U$ is open (this would make $f$ continuous) but $f(X)$ isn't closed whenever $X$ ? Isn't the situation different for polynomial maps $f : mathbb{C} to mathbb{C}$.
$endgroup$
– reuns
Jan 21 at 13:53
$begingroup$
Indeed, that is the case. I am not sure whether the situation is different for algebraically closed fields.
$endgroup$
– OnDragi
Jan 21 at 14:00
|
show 3 more comments
$begingroup$
Let $mathbb K$ be a field, $Xsubseteq mathbb K^n$ a subset, and $ f:mathbb K^nrightarrow mathbb K^m$ an affine map. Using the standard notation $V$, $I$ for algebraic geometry, it should hold that
$$V(I(f(X)))=f(V(I(X))),$$
i.e. that the Zariski closure $VI$ commutes with the affine transformation $f$. Why is that?
(To specify the notation more: for a set $Y$, we mean by $I(Y)$ the set of all polynomials vanishing on $Y$; and for a set of polynomials $J$, we mean by $V(J)$ the set of all zeros common to all polynomials in $J$.)
Note that I do NOT assume $f$ to be bijective, and that $X$ does not need to be an algebraic set.
The equality obviously doesn't hold for every polynomial mapping $ f$, as that would mean that an image of Zariski closed set under polynomial mapping is always Zariski closed, which is not true (e.g. $f(x)=x^2:mathbb Rrightarrow mathbb R$ and $X=mathbb R$).
EDIT:
As pointed out by reuns, using the general topology fact that $overline{f(X)}supseteq f(overline{X})$ for continuous maps, it is enough to show that $f(V(I(X)))$ is Zariski closed (i.e. an algebraic set).
algebraic-geometry polynomials
asked Jan 21 at 12:38
OnDragiOnDragi
656
$endgroup$
Let $mathbb K$ be a field, $Xsubseteq mathbb K^n$ a subset, and $ f:mathbb K^nrightarrow mathbb K^m$ an affine map. Using the standard notation $V$, $I$ for algebraic geometry, it should hold that
$$V(I(f(X)))=f(V(I(X))),$$
i.e. that the Zariski closure $VI$ commutes with the affine transformation $f$. Why is that?
(To specify the notation more: for a set $Y$, we mean by $I(Y)$ the set of all polynomials vanishing on $Y$; and for a set of polynomials $J$, we mean by $V(J)$ the set of all zeros common to all polynomials in $J$.)
Note that I do NOT assume $f$ to be bijective, and that $X$ does not need to be an algebraic set.
The equality obviously doesn't hold for every polynomial mapping $ f$, as that would mean that an image of Zariski closed set under polynomial mapping is always Zariski closed, which is not true (e.g. $f(x)=x^2:mathbb Rrightarrow mathbb R$ and $X=mathbb R$).
EDIT:
As pointed out by reuns, using the general topology fact that $overline{f(X)}supseteq f(overline{X})$ for continuous maps, it is enough to show that $f(V(I(X)))$ is Zariski closed (i.e. an algebraic set).
algebraic-geometry polynomials
algebraic-geometry polynomials
asked Jan 21 at 12:38
OnDragiOnDragi
656
asked Jan 21 at 12:38
OnDragiOnDragi
656
edited Jan 21 at 14:05
OnDragi
asked Jan 21 at 12:38
OnDragiOnDragi
656
asked Jan 21 at 12:38
OnDragiOnDragi
656
asked Jan 21 at 12:38
OnDragiOnDragi
656
656
$begingroup$
The reformulation of the question would be "if $f$ is $textbf{affine}$ then $overline{f(X)}=f(overline{X})$". The equality does not hold for $f$ continuous in general, as my example shows. If $f(x)=x^2$, we have $overline{f(R)}=overline{[0,inf)}=R neq [0,inf)=f(R)=f(overline{R})$. But you are right that $supseteq$ holds in general, so that shows one direction.
$endgroup$
– OnDragi
Jan 21 at 13:26
$begingroup$
In what sense is $f : mathbb{R} to mathbb{R}, x to x^2$ continuous, is it continuous for the topology induced by $V(I(X))$ ?
$endgroup$
– reuns
Jan 21 at 13:40
$begingroup$
Yes, it is a polynomial map, and polynomial maps are continuous w.r.t. Zariski topology.
$endgroup$
– OnDragi
Jan 21 at 13:42
$begingroup$
$overline{f(X)} = f(overline{X})$ is true whenever $f$ maps closed sets to closed sets (that's part of your definition of affine right ?). Here you are saying $f^{-1}(U)$ is open whenever $U$ is open (this would make $f$ continuous) but $f(X)$ isn't closed whenever $X$ ? Isn't the situation different for polynomial maps $f : mathbb{C} to mathbb{C}$.
$endgroup$
– reuns
Jan 21 at 13:53
$begingroup$
Indeed, that is the case. I am not sure whether the situation is different for algebraically closed fields.
$endgroup$
– OnDragi
Jan 21 at 14:00
|
show 3 more comments
$begingroup$
The reformulation of the question would be "if $f$ is $textbf{affine}$ then $overline{f(X)}=f(overline{X})$". The equality does not hold for $f$ continuous in general, as my example shows. If $f(x)=x^2$, we have $overline{f(R)}=overline{[0,inf)}=R neq [0,inf)=f(R)=f(overline{R})$. But you are right that $supseteq$ holds in general, so that shows one direction.
$endgroup$
– OnDragi
Jan 21 at 13:26
$begingroup$
In what sense is $f : mathbb{R} to mathbb{R}, x to x^2$ continuous, is it continuous for the topology induced by $V(I(X))$ ?
$endgroup$
– reuns
Jan 21 at 13:40
$begingroup$
Yes, it is a polynomial map, and polynomial maps are continuous w.r.t. Zariski topology.
$endgroup$
– OnDragi
Jan 21 at 13:42
$begingroup$
$overline{f(X)} = f(overline{X})$ is true whenever $f$ maps closed sets to closed sets (that's part of your definition of affine right ?). Here you are saying $f^{-1}(U)$ is open whenever $U$ is open (this would make $f$ continuous) but $f(X)$ isn't closed whenever $X$ ? Isn't the situation different for polynomial maps $f : mathbb{C} to mathbb{C}$.
$endgroup$
– reuns
Jan 21 at 13:53
$begingroup$
Indeed, that is the case. I am not sure whether the situation is different for algebraically closed fields.
$endgroup$
– OnDragi
Jan 21 at 14:00
$begingroup$
The reformulation of the question would be "if $f$ is $textbf{affine}$ then $overline{f(X)}=f(overline{X})$". The equality does not hold for $f$ continuous in general, as my example shows. If $f(x)=x^2$, we have $overline{f(R)}=overline{[0,inf)}=R neq [0,inf)=f(R)=f(overline{R})$. But you are right that $supseteq$ holds in general, so that shows one direction.
$endgroup$
– OnDragi
Jan 21 at 13:26
$begingroup$
The reformulation of the question would be "if $f$ is $textbf{affine}$ then $overline{f(X)}=f(overline{X})$". The equality does not hold for $f$ continuous in general, as my example shows. If $f(x)=x^2$, we have $overline{f(R)}=overline{[0,inf)}=R neq [0,inf)=f(R)=f(overline{R})$. But you are right that $supseteq$ holds in general, so that shows one direction.
$endgroup$
– OnDragi
Jan 21 at 13:26
$begingroup$
In what sense is $f : mathbb{R} to mathbb{R}, x to x^2$ continuous, is it continuous for the topology induced by $V(I(X))$ ?
$endgroup$
– reuns
Jan 21 at 13:40
$begingroup$
In what sense is $f : mathbb{R} to mathbb{R}, x to x^2$ continuous, is it continuous for the topology induced by $V(I(X))$ ?
$endgroup$
– reuns
Jan 21 at 13:40
$begingroup$
Yes, it is a polynomial map, and polynomial maps are continuous w.r.t. Zariski topology.
$endgroup$
– OnDragi
Jan 21 at 13:42
$begingroup$
Yes, it is a polynomial map, and polynomial maps are continuous w.r.t. Zariski topology.
$endgroup$
– OnDragi
Jan 21 at 13:42
$begingroup$
$overline{f(X)} = f(overline{X})$ is true whenever $f$ maps closed sets to closed sets (that's part of your definition of affine right ?). Here you are saying $f^{-1}(U)$ is open whenever $U$ is open (this would make $f$ continuous) but $f(X)$ isn't closed whenever $X$ ? Isn't the situation different for polynomial maps $f : mathbb{C} to mathbb{C}$.
$endgroup$
– reuns
Jan 21 at 13:53
$begingroup$
$overline{f(X)} = f(overline{X})$ is true whenever $f$ maps closed sets to closed sets (that's part of your definition of affine right ?). Here you are saying $f^{-1}(U)$ is open whenever $U$ is open (this would make $f$ continuous) but $f(X)$ isn't closed whenever $X$ ? Isn't the situation different for polynomial maps $f : mathbb{C} to mathbb{C}$.
$endgroup$
– reuns
Jan 21 at 13:53
$begingroup$
Indeed, that is the case. I am not sure whether the situation is different for algebraically closed fields.
$endgroup$
– OnDragi
Jan 21 at 14:00
$begingroup$
Indeed, that is the case. I am not sure whether the situation is different for algebraically closed fields.
$endgroup$
– OnDragi
Jan 21 at 14:00
|
show 3 more comments
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$begingroup$
The reformulation of the question would be "if $f$ is $textbf{affine}$ then $overline{f(X)}=f(overline{X})$". The equality does not hold for $f$ continuous in general, as my example shows. If $f(x)=x^2$, we have $overline{f(R)}=overline{[0,inf)}=R neq [0,inf)=f(R)=f(overline{R})$. But you are right that $supseteq$ holds in general, so that shows one direction.
$endgroup$
– OnDragi
Jan 21 at 13:26
$begingroup$
In what sense is $f : mathbb{R} to mathbb{R}, x to x^2$ continuous, is it continuous for the topology induced by $V(I(X))$ ?
$endgroup$
– reuns
Jan 21 at 13:40
$begingroup$
Yes, it is a polynomial map, and polynomial maps are continuous w.r.t. Zariski topology.
$endgroup$
– OnDragi
Jan 21 at 13:42
$begingroup$
$overline{f(X)} = f(overline{X})$ is true whenever $f$ maps closed sets to closed sets (that's part of your definition of affine right ?). Here you are saying $f^{-1}(U)$ is open whenever $U$ is open (this would make $f$ continuous) but $f(X)$ isn't closed whenever $X$ ? Isn't the situation different for polynomial maps $f : mathbb{C} to mathbb{C}$.
$endgroup$
– reuns
Jan 21 at 13:53
$begingroup$
Indeed, that is the case. I am not sure whether the situation is different for algebraically closed fields.
$endgroup$
– OnDragi
Jan 21 at 14:00