Hartshorne 14.5 Give necessary and sufficient conditions on the field F for the given configuration to exist
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I'm working through Hartshorne's Euclid and Beyond and have hit a wall with the following question, 14.5:
"For each of he following problems, assume that you are working in the Cartesian plane PI over a field F of characteristic 0. Give necessary and sufficient conditions on the field F for the given configuration to exist. Assume that all lines that appear to be parallel are parallel, and apparent right angles are right angles."
He gives the solution as (13)^(1/2)
I've set up my axes as usual, and unit points at F=(0,1) vertically and B=(1,0) horizontally. I've been focused on using HC || IG, equating the slopes, etc. But it's unclear to me where he's getting his result.
Any suggestions on how to proceed with this?
Problem statement
Set up of axes and points
geometry algebraic-geometry
asked Nov 20 at 20:32
Raymond Moore
133
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Raymond Moore is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |
up vote
2
down vote
favorite
I'm working through Hartshorne's Euclid and Beyond and have hit a wall with the following question, 14.5:
"For each of he following problems, assume that you are working in the Cartesian plane PI over a field F of characteristic 0. Give necessary and sufficient conditions on the field F for the given configuration to exist. Assume that all lines that appear to be parallel are parallel, and apparent right angles are right angles."
He gives the solution as (13)^(1/2)
I've set up my axes as usual, and unit points at F=(0,1) vertically and B=(1,0) horizontally. I've been focused on using HC || IG, equating the slopes, etc. But it's unclear to me where he's getting his result.
Any suggestions on how to proceed with this?
Problem statement
Set up of axes and points
geometry algebraic-geometry
asked Nov 20 at 20:32
Raymond Moore
133
New contributor
Raymond Moore is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
Please include the full statement of the problem in your question: right now the "given configuration" is nowhere to be found in your post, which makes writing good answers very difficult.
– KReiser
Nov 20 at 20:36
Added as links. Hartshorne's statement as well as the setup I'm working with. Thanks... first post and had to sort out how to add graphics.
– Raymond Moore
Nov 20 at 20:43
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I'm working through Hartshorne's Euclid and Beyond and have hit a wall with the following question, 14.5:
"For each of he following problems, assume that you are working in the Cartesian plane PI over a field F of characteristic 0. Give necessary and sufficient conditions on the field F for the given configuration to exist. Assume that all lines that appear to be parallel are parallel, and apparent right angles are right angles."
He gives the solution as (13)^(1/2)
I've set up my axes as usual, and unit points at F=(0,1) vertically and B=(1,0) horizontally. I've been focused on using HC || IG, equating the slopes, etc. But it's unclear to me where he's getting his result.
Any suggestions on how to proceed with this?
Problem statement
Set up of axes and points
geometry algebraic-geometry
asked Nov 20 at 20:32
Raymond Moore
133
New contributor
Raymond Moore is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I'm working through Hartshorne's Euclid and Beyond and have hit a wall with the following question, 14.5:
"For each of he following problems, assume that you are working in the Cartesian plane PI over a field F of characteristic 0. Give necessary and sufficient conditions on the field F for the given configuration to exist. Assume that all lines that appear to be parallel are parallel, and apparent right angles are right angles."
He gives the solution as (13)^(1/2)
I've set up my axes as usual, and unit points at F=(0,1) vertically and B=(1,0) horizontally. I've been focused on using HC || IG, equating the slopes, etc. But it's unclear to me where he's getting his result.
Any suggestions on how to proceed with this?
Problem statement
Set up of axes and points
geometry algebraic-geometry
geometry algebraic-geometry
asked Nov 20 at 20:32
Raymond Moore
133
New contributor
Raymond Moore is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Nov 20 at 20:32
Raymond Moore
133
New contributor
Raymond Moore is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Nov 20 at 20:41
asked Nov 20 at 20:32
Raymond Moore
133
New contributor
Raymond Moore is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Nov 20 at 20:32
Raymond Moore
133
asked Nov 20 at 20:32
Raymond Moore
133
133
New contributor
Raymond Moore is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Raymond Moore is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Raymond Moore is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
Please include the full statement of the problem in your question: right now the "given configuration" is nowhere to be found in your post, which makes writing good answers very difficult.
– KReiser
Nov 20 at 20:36
Added as links. Hartshorne's statement as well as the setup I'm working with. Thanks... first post and had to sort out how to add graphics.
– Raymond Moore
Nov 20 at 20:43
add a comment |
1
Please include the full statement of the problem in your question: right now the "given configuration" is nowhere to be found in your post, which makes writing good answers very difficult.
– KReiser
Nov 20 at 20:36
Added as links. Hartshorne's statement as well as the setup I'm working with. Thanks... first post and had to sort out how to add graphics.
– Raymond Moore
Nov 20 at 20:43
1
1
Please include the full statement of the problem in your question: right now the "given configuration" is nowhere to be found in your post, which makes writing good answers very difficult.
– KReiser
Nov 20 at 20:36
Please include the full statement of the problem in your question: right now the "given configuration" is nowhere to be found in your post, which makes writing good answers very difficult.
– KReiser
Nov 20 at 20:36
Added as links. Hartshorne's statement as well as the setup I'm working with. Thanks... first post and had to sort out how to add graphics.
– Raymond Moore
Nov 20 at 20:43
Added as links. Hartshorne's statement as well as the setup I'm working with. Thanks... first post and had to sort out how to add graphics.
– Raymond Moore
Nov 20 at 20:43
add a comment |
1 Answer
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0
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I'll use the letter names of the points in your image, but I'll rename one distance: let $I=(0,1+b)$, which makes $0<b<1$.
Look at the three similar triangles, those with hypotenuse highlighted in pink in your uploaded image. These are $EIL,IAG,HKC$. $IAG$ has legs $1+b$ and $a$, $HKC$ has legs $1$
and $1-a$, and $EIL$ has legs $1-b$ and $b$. Together, this gives us the three equations $frac{1+b}{a}=frac{1}{1-a}=frac{1-b}{b}$. Using $frac{1}{1-a}=frac{1-b}{b}$, we may solve for $a=frac{1-2b}{1-b}$. Using $frac{1+b}{a}=frac{1-b}{b}$, we may solve for $a=frac{b(b+1)}{1-b}$.
Putting these together, we have $frac{b(b+1)}{1-b}=frac{1-2b}{1-b}$, or $b^2+b=1-2b$, or $b^2+3b-1=0$. This equation is solved by $b=frac{-3pmsqrt{13}}{2}$ and no other values, so this configuration occurs exactly when $b=frac{-3pmsqrt{13}}{2}$ is in the field we're working over.
answered Nov 21 at 0:44
KReiser
9,04211233
Oh! Thank you very much! I wasn't even looking at the relationship of the triangles. I appreciate your help very much!
– Raymond Moore
Nov 21 at 2:59
Glad to have helped you! If you find this response answers your question, it is customary to click on the green checkmark to the left of the answer to accept it. If you find it useful, it is also customary to upvote the answer using the up arrow next to it.
– KReiser
Nov 21 at 3:22
I clicked the green check, but it's not allowing me to upvote. Thanks for the follow up!
– Raymond Moore
Nov 21 at 3:39
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
accepted
I'll use the letter names of the points in your image, but I'll rename one distance: let $I=(0,1+b)$, which makes $0<b<1$.
Look at the three similar triangles, those with hypotenuse highlighted in pink in your uploaded image. These are $EIL,IAG,HKC$. $IAG$ has legs $1+b$ and $a$, $HKC$ has legs $1$
and $1-a$, and $EIL$ has legs $1-b$ and $b$. Together, this gives us the three equations $frac{1+b}{a}=frac{1}{1-a}=frac{1-b}{b}$. Using $frac{1}{1-a}=frac{1-b}{b}$, we may solve for $a=frac{1-2b}{1-b}$. Using $frac{1+b}{a}=frac{1-b}{b}$, we may solve for $a=frac{b(b+1)}{1-b}$.
Putting these together, we have $frac{b(b+1)}{1-b}=frac{1-2b}{1-b}$, or $b^2+b=1-2b$, or $b^2+3b-1=0$. This equation is solved by $b=frac{-3pmsqrt{13}}{2}$ and no other values, so this configuration occurs exactly when $b=frac{-3pmsqrt{13}}{2}$ is in the field we're working over.
answered Nov 21 at 0:44
KReiser
9,04211233
Oh! Thank you very much! I wasn't even looking at the relationship of the triangles. I appreciate your help very much!
– Raymond Moore
Nov 21 at 2:59
Glad to have helped you! If you find this response answers your question, it is customary to click on the green checkmark to the left of the answer to accept it. If you find it useful, it is also customary to upvote the answer using the up arrow next to it.
– KReiser
Nov 21 at 3:22
I clicked the green check, but it's not allowing me to upvote. Thanks for the follow up!
– Raymond Moore
Nov 21 at 3:39
add a comment |
up vote
0
down vote
accepted
I'll use the letter names of the points in your image, but I'll rename one distance: let $I=(0,1+b)$, which makes $0<b<1$.
Look at the three similar triangles, those with hypotenuse highlighted in pink in your uploaded image. These are $EIL,IAG,HKC$. $IAG$ has legs $1+b$ and $a$, $HKC$ has legs $1$
and $1-a$, and $EIL$ has legs $1-b$ and $b$. Together, this gives us the three equations $frac{1+b}{a}=frac{1}{1-a}=frac{1-b}{b}$. Using $frac{1}{1-a}=frac{1-b}{b}$, we may solve for $a=frac{1-2b}{1-b}$. Using $frac{1+b}{a}=frac{1-b}{b}$, we may solve for $a=frac{b(b+1)}{1-b}$.
Putting these together, we have $frac{b(b+1)}{1-b}=frac{1-2b}{1-b}$, or $b^2+b=1-2b$, or $b^2+3b-1=0$. This equation is solved by $b=frac{-3pmsqrt{13}}{2}$ and no other values, so this configuration occurs exactly when $b=frac{-3pmsqrt{13}}{2}$ is in the field we're working over.
answered Nov 21 at 0:44
KReiser
9,04211233
Oh! Thank you very much! I wasn't even looking at the relationship of the triangles. I appreciate your help very much!
– Raymond Moore
Nov 21 at 2:59
Glad to have helped you! If you find this response answers your question, it is customary to click on the green checkmark to the left of the answer to accept it. If you find it useful, it is also customary to upvote the answer using the up arrow next to it.
– KReiser
Nov 21 at 3:22
I clicked the green check, but it's not allowing me to upvote. Thanks for the follow up!
– Raymond Moore
Nov 21 at 3:39
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
I'll use the letter names of the points in your image, but I'll rename one distance: let $I=(0,1+b)$, which makes $0<b<1$.
Look at the three similar triangles, those with hypotenuse highlighted in pink in your uploaded image. These are $EIL,IAG,HKC$. $IAG$ has legs $1+b$ and $a$, $HKC$ has legs $1$
and $1-a$, and $EIL$ has legs $1-b$ and $b$. Together, this gives us the three equations $frac{1+b}{a}=frac{1}{1-a}=frac{1-b}{b}$. Using $frac{1}{1-a}=frac{1-b}{b}$, we may solve for $a=frac{1-2b}{1-b}$. Using $frac{1+b}{a}=frac{1-b}{b}$, we may solve for $a=frac{b(b+1)}{1-b}$.
Putting these together, we have $frac{b(b+1)}{1-b}=frac{1-2b}{1-b}$, or $b^2+b=1-2b$, or $b^2+3b-1=0$. This equation is solved by $b=frac{-3pmsqrt{13}}{2}$ and no other values, so this configuration occurs exactly when $b=frac{-3pmsqrt{13}}{2}$ is in the field we're working over.
answered Nov 21 at 0:44
KReiser
9,04211233
I'll use the letter names of the points in your image, but I'll rename one distance: let $I=(0,1+b)$, which makes $0<b<1$.
Look at the three similar triangles, those with hypotenuse highlighted in pink in your uploaded image. These are $EIL,IAG,HKC$. $IAG$ has legs $1+b$ and $a$, $HKC$ has legs $1$
and $1-a$, and $EIL$ has legs $1-b$ and $b$. Together, this gives us the three equations $frac{1+b}{a}=frac{1}{1-a}=frac{1-b}{b}$. Using $frac{1}{1-a}=frac{1-b}{b}$, we may solve for $a=frac{1-2b}{1-b}$. Using $frac{1+b}{a}=frac{1-b}{b}$, we may solve for $a=frac{b(b+1)}{1-b}$.
Putting these together, we have $frac{b(b+1)}{1-b}=frac{1-2b}{1-b}$, or $b^2+b=1-2b$, or $b^2+3b-1=0$. This equation is solved by $b=frac{-3pmsqrt{13}}{2}$ and no other values, so this configuration occurs exactly when $b=frac{-3pmsqrt{13}}{2}$ is in the field we're working over.
answered Nov 21 at 0:44
KReiser
9,04211233
answered Nov 21 at 0:44
KReiser
9,04211233
answered Nov 21 at 0:44
KReiser
9,04211233
answered Nov 21 at 0:44
KReiser
9,04211233
9,04211233
Oh! Thank you very much! I wasn't even looking at the relationship of the triangles. I appreciate your help very much!
– Raymond Moore
Nov 21 at 2:59
Glad to have helped you! If you find this response answers your question, it is customary to click on the green checkmark to the left of the answer to accept it. If you find it useful, it is also customary to upvote the answer using the up arrow next to it.
– KReiser
Nov 21 at 3:22
I clicked the green check, but it's not allowing me to upvote. Thanks for the follow up!
– Raymond Moore
Nov 21 at 3:39
add a comment |
Oh! Thank you very much! I wasn't even looking at the relationship of the triangles. I appreciate your help very much!
– Raymond Moore
Nov 21 at 2:59
Glad to have helped you! If you find this response answers your question, it is customary to click on the green checkmark to the left of the answer to accept it. If you find it useful, it is also customary to upvote the answer using the up arrow next to it.
– KReiser
Nov 21 at 3:22
I clicked the green check, but it's not allowing me to upvote. Thanks for the follow up!
– Raymond Moore
Nov 21 at 3:39
Oh! Thank you very much! I wasn't even looking at the relationship of the triangles. I appreciate your help very much!
– Raymond Moore
Nov 21 at 2:59
Oh! Thank you very much! I wasn't even looking at the relationship of the triangles. I appreciate your help very much!
– Raymond Moore
Nov 21 at 2:59
Glad to have helped you! If you find this response answers your question, it is customary to click on the green checkmark to the left of the answer to accept it. If you find it useful, it is also customary to upvote the answer using the up arrow next to it.
– KReiser
Nov 21 at 3:22
Glad to have helped you! If you find this response answers your question, it is customary to click on the green checkmark to the left of the answer to accept it. If you find it useful, it is also customary to upvote the answer using the up arrow next to it.
– KReiser
Nov 21 at 3:22
I clicked the green check, but it's not allowing me to upvote. Thanks for the follow up!
– Raymond Moore
Nov 21 at 3:39
I clicked the green check, but it's not allowing me to upvote. Thanks for the follow up!
– Raymond Moore
Nov 21 at 3:39
add a comment |
Raymond Moore is a new contributor. Be nice, and check out our Code of Conduct.
Raymond Moore is a new contributor. Be nice, and check out our Code of Conduct.
Raymond Moore is a new contributor. Be nice, and check out our Code of Conduct.
Raymond Moore is a new contributor. Be nice, and check out our Code of Conduct.
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1
Please include the full statement of the problem in your question: right now the "given configuration" is nowhere to be found in your post, which makes writing good answers very difficult.
– KReiser
Nov 20 at 20:36
Added as links. Hartshorne's statement as well as the setup I'm working with. Thanks... first post and had to sort out how to add graphics.
– Raymond Moore
Nov 20 at 20:43