Transforming Quadrics in Characteristic 2Simultaneous diagonalization of quadratic formsShould isometries be linear?classification of quadricsDoes irreducible quadric mean that the corresponding bilinear form is nondegenerate?Irreducibility of a quadricIntersection of quadricsQuadrics in characteristic 2Space of quadricsAutomorphism fixing quadrics.Quadratic Form as Sum of Squares
How to answer questions about my characters?
Pinhole Camera with Instant Film
How to deal with a cynical class?
Is it true that real estate prices mainly go up?
Possible Leak In Concrete
How do anti-virus programs start at Windows boot?
Life insurance that covers only simultaneous/dual deaths
Could the Saturn V actually have launched astronauts around Venus?
Professor being mistaken for a grad student
Why do Australian milk farmers need to protest supermarkets' milk price?
Can anyone tell me why this program fails?
Is it normal that my co-workers at a fitness company criticize my food choices?
Know when to turn notes upside-down(eighth notes, sixteen notes, etc.)
Dot in front of file
Why does Deadpool say "You're welcome, Canada," after shooting Ryan Reynolds in the end credits?
Replacing Windows 7 security updates with anti-virus?
Is it possible that AIC = BIC?
Employee lack of ownership
Science-fiction short story where space navy wanted hospital ships and settlers had guns mounted everywhere
Making a sword in the stone, in a medieval world without magic
An Accountant Seeks the Help of a Mathematician
PTIJ: Who should pay for Uber rides: the child or the parent?
How could a scammer know the apps on my phone / iTunes account?
Rejected in 4th interview round citing insufficient years of experience
Transforming Quadrics in Characteristic 2
Simultaneous diagonalization of quadratic formsShould isometries be linear?classification of quadricsDoes irreducible quadric mean that the corresponding bilinear form is nondegenerate?Irreducibility of a quadricIntersection of quadricsQuadrics in characteristic 2Space of quadricsAutomorphism fixing quadrics.Quadratic Form as Sum of Squares
$begingroup$
I’m trying to solve the following problem given in a textbook:
Let $k$ be an algebraically closed field and $Q=V(F)$ a quadric in $mathbbP^3(k)$, where $F$ is an irreducible polynomial in $X,Y,Z,T$, and hence gives rise to a quadratic form on $k^4$ which we assume is non-degenerate.
Show that after some change of coordinates we can write $Q=V(XT-YZ)$.
I’ve solved the case where $textchar(k)neq2$ by diagonalising the quadratic form and then making a suitable change of coordinates. However this process involves using the correspondence between quadratic forms and symmetric bilinear forms which is not valid in characteristic $2$ (and if we could diagonalise $F$ then it would become reducible).
My first issue is that I can’t find a reference defining what it means for a quadratic form to be non-degenerate in characteristic $2$, so if I tried to come up with a counterexample I wouldn’t know if it was valid or not.
Beyond this, I’m not even sure that the result is true, I can’t seem to find anywhere claiming that it is. Has the textbook simply forgotten to specify that $textchar(k)neq2$, or am I missing something?
Any help would be much appreciated.
algebraic-geometry projective-geometry quadratic-forms positive-characteristic quadrics
asked Mar 11 at 9:29
DaveDave
597
$endgroup$
add a comment |
$begingroup$
I’m trying to solve the following problem given in a textbook:
Let $k$ be an algebraically closed field and $Q=V(F)$ a quadric in $mathbbP^3(k)$, where $F$ is an irreducible polynomial in $X,Y,Z,T$, and hence gives rise to a quadratic form on $k^4$ which we assume is non-degenerate.
Show that after some change of coordinates we can write $Q=V(XT-YZ)$.
I’ve solved the case where $textchar(k)neq2$ by diagonalising the quadratic form and then making a suitable change of coordinates. However this process involves using the correspondence between quadratic forms and symmetric bilinear forms which is not valid in characteristic $2$ (and if we could diagonalise $F$ then it would become reducible).
My first issue is that I can’t find a reference defining what it means for a quadratic form to be non-degenerate in characteristic $2$, so if I tried to come up with a counterexample I wouldn’t know if it was valid or not.
Beyond this, I’m not even sure that the result is true, I can’t seem to find anywhere claiming that it is. Has the textbook simply forgotten to specify that $textchar(k)neq2$, or am I missing something?
Any help would be much appreciated.
algebraic-geometry projective-geometry quadratic-forms positive-characteristic quadrics
asked Mar 11 at 9:29
DaveDave
597
$endgroup$
add a comment |
$begingroup$
I’m trying to solve the following problem given in a textbook:
Let $k$ be an algebraically closed field and $Q=V(F)$ a quadric in $mathbbP^3(k)$, where $F$ is an irreducible polynomial in $X,Y,Z,T$, and hence gives rise to a quadratic form on $k^4$ which we assume is non-degenerate.
Show that after some change of coordinates we can write $Q=V(XT-YZ)$.
I’ve solved the case where $textchar(k)neq2$ by diagonalising the quadratic form and then making a suitable change of coordinates. However this process involves using the correspondence between quadratic forms and symmetric bilinear forms which is not valid in characteristic $2$ (and if we could diagonalise $F$ then it would become reducible).
My first issue is that I can’t find a reference defining what it means for a quadratic form to be non-degenerate in characteristic $2$, so if I tried to come up with a counterexample I wouldn’t know if it was valid or not.
Beyond this, I’m not even sure that the result is true, I can’t seem to find anywhere claiming that it is. Has the textbook simply forgotten to specify that $textchar(k)neq2$, or am I missing something?
Any help would be much appreciated.
algebraic-geometry projective-geometry quadratic-forms positive-characteristic quadrics
asked Mar 11 at 9:29
DaveDave
597
$endgroup$
I’m trying to solve the following problem given in a textbook:
Let $k$ be an algebraically closed field and $Q=V(F)$ a quadric in $mathbbP^3(k)$, where $F$ is an irreducible polynomial in $X,Y,Z,T$, and hence gives rise to a quadratic form on $k^4$ which we assume is non-degenerate.
Show that after some change of coordinates we can write $Q=V(XT-YZ)$.
I’ve solved the case where $textchar(k)neq2$ by diagonalising the quadratic form and then making a suitable change of coordinates. However this process involves using the correspondence between quadratic forms and symmetric bilinear forms which is not valid in characteristic $2$ (and if we could diagonalise $F$ then it would become reducible).
My first issue is that I can’t find a reference defining what it means for a quadratic form to be non-degenerate in characteristic $2$, so if I tried to come up with a counterexample I wouldn’t know if it was valid or not.
Beyond this, I’m not even sure that the result is true, I can’t seem to find anywhere claiming that it is. Has the textbook simply forgotten to specify that $textchar(k)neq2$, or am I missing something?
Any help would be much appreciated.
algebraic-geometry projective-geometry quadratic-forms positive-characteristic quadrics
algebraic-geometry projective-geometry quadratic-forms positive-characteristic quadrics
asked Mar 11 at 9:29
DaveDave
597
asked Mar 11 at 9:29
DaveDave
597
edited Mar 11 at 15:20
Dave
asked Mar 11 at 9:29
DaveDave
597
asked Mar 11 at 9:29
DaveDave
597
asked Mar 11 at 9:29
DaveDave
597
597
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Having spent more time on this, I think I have solved the problem, and the result is in fact true.
Arf defines non-singularity for quadratic forms of characteristic $2$ here, explained in English here. From these papers, we see that if $Q$ is non-singular over a field of characteristic $2$, we can write $$Q=(aX^2+XT+bT^2)+(cY^2+YZ+dZ^2)$$ for some $a,b,c,din k$ and an appropriate choice of coordinates. Let’s consider $aX^2+XT+bT^2$.
If $a=b=0$ then we have $XT$ already, if say $b=0$ then we have $X(aX+T)$ so taking the inverse of the transformation sending $Xmapsto X$ and $Tmapsto aX+T$ we have $XT$.
Then we assume $a,bneq0$. Sending $Xmapstofrac1sqrtaX$ and $Tmapstofrac1sqrtbT$ we have $X^2+alpha XT+T^2$ for $alpha=frac1sqrtab$. All square roots exist since $k$ is algebraically closed, and for the same reason we can also find a root $beta$ of the polynomial $x^2+alpha x+1$. Then sending $Xmapsto X+frac1alpha^2T$ and $Tmapstobeta X+fracalpha+betaalpha^2T$ we have $XT$. This transformation is invertible since $$beginvmatrix1 & beta \frac1alpha^2 & fracalpha+betaalpha^2endvmatrix=frac1alphaneq0$$
We can repeat the same process for $Y$ and $Z$, and so we can write $Q=XT+YZ=XT-YZ$.
answered Mar 11 at 20:12
DaveDave
597
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
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%2f3143476%2ftransforming-quadrics-in-characteristic-2%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Having spent more time on this, I think I have solved the problem, and the result is in fact true.
Arf defines non-singularity for quadratic forms of characteristic $2$ here, explained in English here. From these papers, we see that if $Q$ is non-singular over a field of characteristic $2$, we can write $$Q=(aX^2+XT+bT^2)+(cY^2+YZ+dZ^2)$$ for some $a,b,c,din k$ and an appropriate choice of coordinates. Let’s consider $aX^2+XT+bT^2$.
If $a=b=0$ then we have $XT$ already, if say $b=0$ then we have $X(aX+T)$ so taking the inverse of the transformation sending $Xmapsto X$ and $Tmapsto aX+T$ we have $XT$.
Then we assume $a,bneq0$. Sending $Xmapstofrac1sqrtaX$ and $Tmapstofrac1sqrtbT$ we have $X^2+alpha XT+T^2$ for $alpha=frac1sqrtab$. All square roots exist since $k$ is algebraically closed, and for the same reason we can also find a root $beta$ of the polynomial $x^2+alpha x+1$. Then sending $Xmapsto X+frac1alpha^2T$ and $Tmapstobeta X+fracalpha+betaalpha^2T$ we have $XT$. This transformation is invertible since $$beginvmatrix1 & beta \frac1alpha^2 & fracalpha+betaalpha^2endvmatrix=frac1alphaneq0$$
We can repeat the same process for $Y$ and $Z$, and so we can write $Q=XT+YZ=XT-YZ$.
answered Mar 11 at 20:12
DaveDave
597
$endgroup$
add a comment |
$begingroup$
Having spent more time on this, I think I have solved the problem, and the result is in fact true.
Arf defines non-singularity for quadratic forms of characteristic $2$ here, explained in English here. From these papers, we see that if $Q$ is non-singular over a field of characteristic $2$, we can write $$Q=(aX^2+XT+bT^2)+(cY^2+YZ+dZ^2)$$ for some $a,b,c,din k$ and an appropriate choice of coordinates. Let’s consider $aX^2+XT+bT^2$.
If $a=b=0$ then we have $XT$ already, if say $b=0$ then we have $X(aX+T)$ so taking the inverse of the transformation sending $Xmapsto X$ and $Tmapsto aX+T$ we have $XT$.
Then we assume $a,bneq0$. Sending $Xmapstofrac1sqrtaX$ and $Tmapstofrac1sqrtbT$ we have $X^2+alpha XT+T^2$ for $alpha=frac1sqrtab$. All square roots exist since $k$ is algebraically closed, and for the same reason we can also find a root $beta$ of the polynomial $x^2+alpha x+1$. Then sending $Xmapsto X+frac1alpha^2T$ and $Tmapstobeta X+fracalpha+betaalpha^2T$ we have $XT$. This transformation is invertible since $$beginvmatrix1 & beta \frac1alpha^2 & fracalpha+betaalpha^2endvmatrix=frac1alphaneq0$$
We can repeat the same process for $Y$ and $Z$, and so we can write $Q=XT+YZ=XT-YZ$.
answered Mar 11 at 20:12
DaveDave
597
$endgroup$
add a comment |
$begingroup$
Having spent more time on this, I think I have solved the problem, and the result is in fact true.
Arf defines non-singularity for quadratic forms of characteristic $2$ here, explained in English here. From these papers, we see that if $Q$ is non-singular over a field of characteristic $2$, we can write $$Q=(aX^2+XT+bT^2)+(cY^2+YZ+dZ^2)$$ for some $a,b,c,din k$ and an appropriate choice of coordinates. Let’s consider $aX^2+XT+bT^2$.
If $a=b=0$ then we have $XT$ already, if say $b=0$ then we have $X(aX+T)$ so taking the inverse of the transformation sending $Xmapsto X$ and $Tmapsto aX+T$ we have $XT$.
Then we assume $a,bneq0$. Sending $Xmapstofrac1sqrtaX$ and $Tmapstofrac1sqrtbT$ we have $X^2+alpha XT+T^2$ for $alpha=frac1sqrtab$. All square roots exist since $k$ is algebraically closed, and for the same reason we can also find a root $beta$ of the polynomial $x^2+alpha x+1$. Then sending $Xmapsto X+frac1alpha^2T$ and $Tmapstobeta X+fracalpha+betaalpha^2T$ we have $XT$. This transformation is invertible since $$beginvmatrix1 & beta \frac1alpha^2 & fracalpha+betaalpha^2endvmatrix=frac1alphaneq0$$
We can repeat the same process for $Y$ and $Z$, and so we can write $Q=XT+YZ=XT-YZ$.
answered Mar 11 at 20:12
DaveDave
597
$endgroup$
Having spent more time on this, I think I have solved the problem, and the result is in fact true.
Arf defines non-singularity for quadratic forms of characteristic $2$ here, explained in English here. From these papers, we see that if $Q$ is non-singular over a field of characteristic $2$, we can write $$Q=(aX^2+XT+bT^2)+(cY^2+YZ+dZ^2)$$ for some $a,b,c,din k$ and an appropriate choice of coordinates. Let’s consider $aX^2+XT+bT^2$.
If $a=b=0$ then we have $XT$ already, if say $b=0$ then we have $X(aX+T)$ so taking the inverse of the transformation sending $Xmapsto X$ and $Tmapsto aX+T$ we have $XT$.
Then we assume $a,bneq0$. Sending $Xmapstofrac1sqrtaX$ and $Tmapstofrac1sqrtbT$ we have $X^2+alpha XT+T^2$ for $alpha=frac1sqrtab$. All square roots exist since $k$ is algebraically closed, and for the same reason we can also find a root $beta$ of the polynomial $x^2+alpha x+1$. Then sending $Xmapsto X+frac1alpha^2T$ and $Tmapstobeta X+fracalpha+betaalpha^2T$ we have $XT$. This transformation is invertible since $$beginvmatrix1 & beta \frac1alpha^2 & fracalpha+betaalpha^2endvmatrix=frac1alphaneq0$$
We can repeat the same process for $Y$ and $Z$, and so we can write $Q=XT+YZ=XT-YZ$.
answered Mar 11 at 20:12
DaveDave
597
edited Mar 11 at 20:28
answered Mar 11 at 20:12
DaveDave
597
answered Mar 11 at 20:12
DaveDave
597
answered Mar 11 at 20:12
DaveDave
597
597
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
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%2f3143476%2ftransforming-quadrics-in-characteristic-2%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
