$K neq sum K^2 implies K$ 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
1288
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
1288
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
1288
$K$ 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
1288
asked Nov 14 at 20:53
Hikicianka
1288
edited Nov 14 at 20:59
asked Nov 14 at 20:53
Hikicianka
1288
asked Nov 14 at 20:53
Hikicianka
1288
asked Nov 14 at 20:53
Hikicianka
1288
1288
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
0
down vote
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
6,7902518
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
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
6,7902518
add a comment |
up vote
0
down vote
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
6,7902518
add a comment |
up vote
0
down vote
up vote
0
down vote
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
6,7902518
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
6,7902518
answered Nov 17 at 9:10
William Elliot
6,7902518
answered Nov 17 at 9:10
William Elliot
6,7902518
answered Nov 17 at 9:10
William Elliot
6,7902518
6,7902518
add a comment |
add a comment |
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2998803%2fk-neq-sum-k2-implies-k-admits-an-ordering%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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