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For an isogeny of abelian varieties $f : X to Y$, is $Y = X/ operatornamekerf$?
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Let $k$ be a field, $f : X to Y$ be an isogeny of $k$-abelian varieties.
Then there exists the canonical separable isogeny $pi : X to X/operatornamekerf$, such that $X/operatornamekerf$ is a $k$-abelian variety, and on $overlinek$-valued points, $pi$ is the natural quotient map of groups with the kernel $operatornamekerf$.
By III.4.1 of Silverman's The Arithmetic of Elliptic Curves, there exists an isogeny $g : X/operatornamekerf to Y$ such that $f = g circ pi$.
Now, is $g$ an isomorphism (of varieties)?
The Corollary 1 of section 12 of Mumford's Abelian Varieties says this is true.
However, if so, I think that every isogeny becomes separable: $pi$ is separable, and $g$ is an isomorphism, in particular separable, thus $f = g circ pi$ is also separable.
algebraic-geometry abelian-varieties
asked 2 days ago
k.j.
1788
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up vote
2
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Let $k$ be a field, $f : X to Y$ be an isogeny of $k$-abelian varieties.
Then there exists the canonical separable isogeny $pi : X to X/operatornamekerf$, such that $X/operatornamekerf$ is a $k$-abelian variety, and on $overlinek$-valued points, $pi$ is the natural quotient map of groups with the kernel $operatornamekerf$.
By III.4.1 of Silverman's The Arithmetic of Elliptic Curves, there exists an isogeny $g : X/operatornamekerf to Y$ such that $f = g circ pi$.
Now, is $g$ an isomorphism (of varieties)?
The Corollary 1 of section 12 of Mumford's Abelian Varieties says this is true.
However, if so, I think that every isogeny becomes separable: $pi$ is separable, and $g$ is an isomorphism, in particular separable, thus $f = g circ pi$ is also separable.
algebraic-geometry abelian-varieties
asked 2 days ago
k.j.
1788
1
It depens if you are calling ker$(f)$ the group scheme or the points over $bar k$= the étale part (this last case is what it seems from your sentences). I don't have Mumford or Silverman at hand, but may be the problem is that they use the same name for different things.
– xarles
2 days ago
@xarles Thank you very much! I found what you say in Mumford. I'll try it.
– k.j.
yesterday
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $k$ be a field, $f : X to Y$ be an isogeny of $k$-abelian varieties.
Then there exists the canonical separable isogeny $pi : X to X/operatornamekerf$, such that $X/operatornamekerf$ is a $k$-abelian variety, and on $overlinek$-valued points, $pi$ is the natural quotient map of groups with the kernel $operatornamekerf$.
By III.4.1 of Silverman's The Arithmetic of Elliptic Curves, there exists an isogeny $g : X/operatornamekerf to Y$ such that $f = g circ pi$.
Now, is $g$ an isomorphism (of varieties)?
The Corollary 1 of section 12 of Mumford's Abelian Varieties says this is true.
However, if so, I think that every isogeny becomes separable: $pi$ is separable, and $g$ is an isomorphism, in particular separable, thus $f = g circ pi$ is also separable.
algebraic-geometry abelian-varieties
asked 2 days ago
k.j.
1788
Let $k$ be a field, $f : X to Y$ be an isogeny of $k$-abelian varieties.
Then there exists the canonical separable isogeny $pi : X to X/operatornamekerf$, such that $X/operatornamekerf$ is a $k$-abelian variety, and on $overlinek$-valued points, $pi$ is the natural quotient map of groups with the kernel $operatornamekerf$.
By III.4.1 of Silverman's The Arithmetic of Elliptic Curves, there exists an isogeny $g : X/operatornamekerf to Y$ such that $f = g circ pi$.
Now, is $g$ an isomorphism (of varieties)?
The Corollary 1 of section 12 of Mumford's Abelian Varieties says this is true.
However, if so, I think that every isogeny becomes separable: $pi$ is separable, and $g$ is an isomorphism, in particular separable, thus $f = g circ pi$ is also separable.
algebraic-geometry abelian-varieties
asked 2 days ago
k.j.
1788
edited 2 days ago
asked 2 days ago
k.j.
1788
asked 2 days ago
k.j.
1788
asked 2 days ago
k.j.
1788
1788
1
It depens if you are calling ker$(f)$ the group scheme or the points over $bar k$= the étale part (this last case is what it seems from your sentences). I don't have Mumford or Silverman at hand, but may be the problem is that they use the same name for different things.
– xarles
2 days ago
@xarles Thank you very much! I found what you say in Mumford. I'll try it.
– k.j.
yesterday
add a comment |Â
1
It depens if you are calling ker$(f)$ the group scheme or the points over $bar k$= the étale part (this last case is what it seems from your sentences). I don't have Mumford or Silverman at hand, but may be the problem is that they use the same name for different things.
– xarles
2 days ago
@xarles Thank you very much! I found what you say in Mumford. I'll try it.
– k.j.
yesterday
1
1
It depens if you are calling ker$(f)$ the group scheme or the points over $bar k$= the étale part (this last case is what it seems from your sentences). I don't have Mumford or Silverman at hand, but may be the problem is that they use the same name for different things.
– xarles
2 days ago
It depens if you are calling ker$(f)$ the group scheme or the points over $bar k$= the étale part (this last case is what it seems from your sentences). I don't have Mumford or Silverman at hand, but may be the problem is that they use the same name for different things.
– xarles
2 days ago
@xarles Thank you very much! I found what you say in Mumford. I'll try it.
– k.j.
yesterday
@xarles Thank you very much! I found what you say in Mumford. I'll try it.
– k.j.
yesterday
add a comment |Â
1 Answer
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I confirm the part corresponding to Silverman's book: in there $operatornameKer(f)$ for $f: Xto Y$ an isogeny of elliptic curves means the set of $bar K$-points of $X$ that go to $0in Y$. This is a finite (étale) subgroup scheme of $X$, and there exists a unique separable isogeny $g: Xto X'$ such that $operatornameKer(f)=operatornameKer(g)$ (by proposition III.4.12 in Silverman's). Moreover, by corollary 4.11, there exists a unique isogeny $h:X'to X$ such that $f=hcirc g$. This isogeny $h$ is purely inseparable, and $operatornameKer(h)=0$, but $h$ is not aan isomorphism.
On the other hand, in books with scheme theoretic flavour (such as Mumford's book), $operatornameKer(f)$ means the kernel group-scheme, which is a finite subgroup scheme of $X$. One has that $operatornameKer(f)(bar K)$ is what Silverman called $operatornameKer(f)$ above.
For example, if $K$ has characteristic $p$, and $f=[p]$ is multiplication by $p$, then $operatornameKer(f)(bar K)$ has either $p$ points (if $E$ is ordinary) or $0$ points (if it is supersingular). In the first case $g:Xto X'$ has degree $p$, and also $h$. In the second case $g=id$ is the identity, and $h=f$.
By the way, the fact that $f=hcirc g$, with $g$ separable and $h$ purely inseparable is a version of the result that says that any finite field extension $L'/L$, there exists an intermediate extension $L'/L_s/L$ with $L_s/L$ separable and $L'/L_s$ purely inseparable: $L_s$ is the set of separable elements in $L'$.
answered 14 hours ago
xarles
83548
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
I confirm the part corresponding to Silverman's book: in there $operatornameKer(f)$ for $f: Xto Y$ an isogeny of elliptic curves means the set of $bar K$-points of $X$ that go to $0in Y$. This is a finite (étale) subgroup scheme of $X$, and there exists a unique separable isogeny $g: Xto X'$ such that $operatornameKer(f)=operatornameKer(g)$ (by proposition III.4.12 in Silverman's). Moreover, by corollary 4.11, there exists a unique isogeny $h:X'to X$ such that $f=hcirc g$. This isogeny $h$ is purely inseparable, and $operatornameKer(h)=0$, but $h$ is not aan isomorphism.
On the other hand, in books with scheme theoretic flavour (such as Mumford's book), $operatornameKer(f)$ means the kernel group-scheme, which is a finite subgroup scheme of $X$. One has that $operatornameKer(f)(bar K)$ is what Silverman called $operatornameKer(f)$ above.
For example, if $K$ has characteristic $p$, and $f=[p]$ is multiplication by $p$, then $operatornameKer(f)(bar K)$ has either $p$ points (if $E$ is ordinary) or $0$ points (if it is supersingular). In the first case $g:Xto X'$ has degree $p$, and also $h$. In the second case $g=id$ is the identity, and $h=f$.
By the way, the fact that $f=hcirc g$, with $g$ separable and $h$ purely inseparable is a version of the result that says that any finite field extension $L'/L$, there exists an intermediate extension $L'/L_s/L$ with $L_s/L$ separable and $L'/L_s$ purely inseparable: $L_s$ is the set of separable elements in $L'$.
answered 14 hours ago
xarles
83548
add a comment |Â
up vote
3
down vote
accepted
I confirm the part corresponding to Silverman's book: in there $operatornameKer(f)$ for $f: Xto Y$ an isogeny of elliptic curves means the set of $bar K$-points of $X$ that go to $0in Y$. This is a finite (étale) subgroup scheme of $X$, and there exists a unique separable isogeny $g: Xto X'$ such that $operatornameKer(f)=operatornameKer(g)$ (by proposition III.4.12 in Silverman's). Moreover, by corollary 4.11, there exists a unique isogeny $h:X'to X$ such that $f=hcirc g$. This isogeny $h$ is purely inseparable, and $operatornameKer(h)=0$, but $h$ is not aan isomorphism.
On the other hand, in books with scheme theoretic flavour (such as Mumford's book), $operatornameKer(f)$ means the kernel group-scheme, which is a finite subgroup scheme of $X$. One has that $operatornameKer(f)(bar K)$ is what Silverman called $operatornameKer(f)$ above.
For example, if $K$ has characteristic $p$, and $f=[p]$ is multiplication by $p$, then $operatornameKer(f)(bar K)$ has either $p$ points (if $E$ is ordinary) or $0$ points (if it is supersingular). In the first case $g:Xto X'$ has degree $p$, and also $h$. In the second case $g=id$ is the identity, and $h=f$.
By the way, the fact that $f=hcirc g$, with $g$ separable and $h$ purely inseparable is a version of the result that says that any finite field extension $L'/L$, there exists an intermediate extension $L'/L_s/L$ with $L_s/L$ separable and $L'/L_s$ purely inseparable: $L_s$ is the set of separable elements in $L'$.
answered 14 hours ago
xarles
83548
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
I confirm the part corresponding to Silverman's book: in there $operatornameKer(f)$ for $f: Xto Y$ an isogeny of elliptic curves means the set of $bar K$-points of $X$ that go to $0in Y$. This is a finite (étale) subgroup scheme of $X$, and there exists a unique separable isogeny $g: Xto X'$ such that $operatornameKer(f)=operatornameKer(g)$ (by proposition III.4.12 in Silverman's). Moreover, by corollary 4.11, there exists a unique isogeny $h:X'to X$ such that $f=hcirc g$. This isogeny $h$ is purely inseparable, and $operatornameKer(h)=0$, but $h$ is not aan isomorphism.
On the other hand, in books with scheme theoretic flavour (such as Mumford's book), $operatornameKer(f)$ means the kernel group-scheme, which is a finite subgroup scheme of $X$. One has that $operatornameKer(f)(bar K)$ is what Silverman called $operatornameKer(f)$ above.
For example, if $K$ has characteristic $p$, and $f=[p]$ is multiplication by $p$, then $operatornameKer(f)(bar K)$ has either $p$ points (if $E$ is ordinary) or $0$ points (if it is supersingular). In the first case $g:Xto X'$ has degree $p$, and also $h$. In the second case $g=id$ is the identity, and $h=f$.
By the way, the fact that $f=hcirc g$, with $g$ separable and $h$ purely inseparable is a version of the result that says that any finite field extension $L'/L$, there exists an intermediate extension $L'/L_s/L$ with $L_s/L$ separable and $L'/L_s$ purely inseparable: $L_s$ is the set of separable elements in $L'$.
answered 14 hours ago
xarles
83548
I confirm the part corresponding to Silverman's book: in there $operatornameKer(f)$ for $f: Xto Y$ an isogeny of elliptic curves means the set of $bar K$-points of $X$ that go to $0in Y$. This is a finite (étale) subgroup scheme of $X$, and there exists a unique separable isogeny $g: Xto X'$ such that $operatornameKer(f)=operatornameKer(g)$ (by proposition III.4.12 in Silverman's). Moreover, by corollary 4.11, there exists a unique isogeny $h:X'to X$ such that $f=hcirc g$. This isogeny $h$ is purely inseparable, and $operatornameKer(h)=0$, but $h$ is not aan isomorphism.
On the other hand, in books with scheme theoretic flavour (such as Mumford's book), $operatornameKer(f)$ means the kernel group-scheme, which is a finite subgroup scheme of $X$. One has that $operatornameKer(f)(bar K)$ is what Silverman called $operatornameKer(f)$ above.
For example, if $K$ has characteristic $p$, and $f=[p]$ is multiplication by $p$, then $operatornameKer(f)(bar K)$ has either $p$ points (if $E$ is ordinary) or $0$ points (if it is supersingular). In the first case $g:Xto X'$ has degree $p$, and also $h$. In the second case $g=id$ is the identity, and $h=f$.
By the way, the fact that $f=hcirc g$, with $g$ separable and $h$ purely inseparable is a version of the result that says that any finite field extension $L'/L$, there exists an intermediate extension $L'/L_s/L$ with $L_s/L$ separable and $L'/L_s$ purely inseparable: $L_s$ is the set of separable elements in $L'$.
answered 14 hours ago
xarles
83548
answered 14 hours ago
xarles
83548
answered 14 hours ago
xarles
83548
answered 14 hours ago
xarles
83548
83548
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1
It depens if you are calling ker$(f)$ the group scheme or the points over $bar k$= the étale part (this last case is what it seems from your sentences). I don't have Mumford or Silverman at hand, but may be the problem is that they use the same name for different things.
– xarles
2 days ago
@xarles Thank you very much! I found what you say in Mumford. I'll try it.
– k.j.
yesterday