$K neq sum K^2 implies K$ admits an ordering
up vote
1
down vote
favorite
$K$ is a field. $K^2 = {a^2 | a in K }$, $sum K^2$ is set of all sum of squares.
I have to prove an implication:
$K neq sum K^2 implies K$ admits an ordering
$K neq sum K^2$, so $-1 notin sum K^2$, so $K$ is formally real. It implies that $sum K^2$ is pre-positive cone. I am not sure what to do now. I know that relation $a<b iff b-ain P$, where $P$ is a positive cone, satisfies conditions for being an order relation. I know also taht every positive cone of field is a pre-positive cone.
real-analysis order-theory ordered-fields
asked Nov 14 at 20:53
Hikicianka
1358
add a comment |
up vote
1
down vote
favorite
$K$ is a field. $K^2 = {a^2 | a in K }$, $sum K^2$ is set of all sum of squares.
I have to prove an implication:
$K neq sum K^2 implies K$ admits an ordering
$K neq sum K^2$, so $-1 notin sum K^2$, so $K$ is formally real. It implies that $sum K^2$ is pre-positive cone. I am not sure what to do now. I know that relation $a<b iff b-ain P$, where $P$ is a positive cone, satisfies conditions for being an order relation. I know also taht every positive cone of field is a pre-positive cone.
real-analysis order-theory ordered-fields
asked Nov 14 at 20:53
Hikicianka
1358
What does "all sum of squares" mean?
– William Elliot
Nov 15 at 9:42
If an element is in $sum K^2$ it means that it can be written as a sum of squares.
– Hikicianka
Nov 15 at 11:09
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
$K$ is a field. $K^2 = {a^2 | a in K }$, $sum K^2$ is set of all sum of squares.
I have to prove an implication:
$K neq sum K^2 implies K$ admits an ordering
$K neq sum K^2$, so $-1 notin sum K^2$, so $K$ is formally real. It implies that $sum K^2$ is pre-positive cone. I am not sure what to do now. I know that relation $a<b iff b-ain P$, where $P$ is a positive cone, satisfies conditions for being an order relation. I know also taht every positive cone of field is a pre-positive cone.
real-analysis order-theory ordered-fields
asked Nov 14 at 20:53
Hikicianka
1358
$K$ is a field. $K^2 = {a^2 | a in K }$, $sum K^2$ is set of all sum of squares.
I have to prove an implication:
$K neq sum K^2 implies K$ admits an ordering
$K neq sum K^2$, so $-1 notin sum K^2$, so $K$ is formally real. It implies that $sum K^2$ is pre-positive cone. I am not sure what to do now. I know that relation $a<b iff b-ain P$, where $P$ is a positive cone, satisfies conditions for being an order relation. I know also taht every positive cone of field is a pre-positive cone.
real-analysis order-theory ordered-fields
real-analysis order-theory ordered-fields
asked Nov 14 at 20:53
Hikicianka
1358
asked Nov 14 at 20:53
Hikicianka
1358
edited Nov 14 at 20:59
asked Nov 14 at 20:53
Hikicianka
1358
asked Nov 14 at 20:53
Hikicianka
1358
asked Nov 14 at 20:53
Hikicianka
1358
1358
What does "all sum of squares" mean?
– William Elliot
Nov 15 at 9:42
If an element is in $sum K^2$ it means that it can be written as a sum of squares.
– Hikicianka
Nov 15 at 11:09
add a comment |
What does "all sum of squares" mean?
– William Elliot
Nov 15 at 9:42
If an element is in $sum K^2$ it means that it can be written as a sum of squares.
– Hikicianka
Nov 15 at 11:09
What does "all sum of squares" mean?
– William Elliot
Nov 15 at 9:42
What does "all sum of squares" mean?
– William Elliot
Nov 15 at 9:42
If an element is in $sum K^2$ it means that it can be written as a sum of squares.
– Hikicianka
Nov 15 at 11:09
If an element is in $sum K^2$ it means that it can be written as a sum of squares.
– Hikicianka
Nov 15 at 11:09
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
There is no sum of all squares.
Let C be the set of all finite sums of squares.
Show C is an order cone.
Let K be a maximal order cone with C subset K.
Show that K is a linear cone.
If you want to work with positive cones,then remove
0 from the numbers that are being squared.
answered Nov 17 at 9:10
William Elliot
6,8702518
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
There is no sum of all squares.
Let C be the set of all finite sums of squares.
Show C is an order cone.
Let K be a maximal order cone with C subset K.
Show that K is a linear cone.
If you want to work with positive cones,then remove
0 from the numbers that are being squared.
answered Nov 17 at 9:10
William Elliot
6,8702518
add a comment |
up vote
1
down vote
accepted
There is no sum of all squares.
Let C be the set of all finite sums of squares.
Show C is an order cone.
Let K be a maximal order cone with C subset K.
Show that K is a linear cone.
If you want to work with positive cones,then remove
0 from the numbers that are being squared.
answered Nov 17 at 9:10
William Elliot
6,8702518
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
There is no sum of all squares.
Let C be the set of all finite sums of squares.
Show C is an order cone.
Let K be a maximal order cone with C subset K.
Show that K is a linear cone.
If you want to work with positive cones,then remove
0 from the numbers that are being squared.
answered Nov 17 at 9:10
William Elliot
6,8702518
There is no sum of all squares.
Let C be the set of all finite sums of squares.
Show C is an order cone.
Let K be a maximal order cone with C subset K.
Show that K is a linear cone.
If you want to work with positive cones,then remove
0 from the numbers that are being squared.
answered Nov 17 at 9:10
William Elliot
6,8702518
answered Nov 17 at 9:10
William Elliot
6,8702518
answered Nov 17 at 9:10
William Elliot
6,8702518
answered Nov 17 at 9:10
William Elliot
6,8702518
6,8702518
add a comment |
add a comment |
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What does "all sum of squares" mean?
– William Elliot
Nov 15 at 9:42
If an element is in $sum K^2$ it means that it can be written as a sum of squares.
– Hikicianka
Nov 15 at 11:09