Mixing
Proof of the decomposition of an isogeny into a separable part and a frobenius.
$begingroup$
Lemma:
Let $alpha:E_1longrightarrow E_2 $ be an inseparable isogeny of elliptic
curves defined over a field $k$ of characteristic $p>0$. Then $alpha$ can be
written in the form
$$alpha(x,y)=(r_1(x^p),r_2(x^p)y^p)$$
For $r_1,r_2in K[x]$.
Now here comes the proposition that I'd like to prove.
Let $alpha$ be an isogeny of elliptic curves over a field $k$ of characteristic $p > 0$. Then
$$alpha=alpha_{sep}circpi^n$$
For some $n>0$ and some separable isogeny $alpha_n$ where $pi$ is the p-power Frobenius morphism. We have then $deg(phi)=phi^n deg(alpha_{sep})$.
Proof:
If $alpha$ is separable, then take $n=0$ and $alpha_{sep}=alpha$. Otherwise, $alpha$ is inseparable, and by the lemma, $alpha=(r_1(x^p),r_2(x^p)y^p)=(r_1(x),r_2(x)y)circpi$. We take $alpha_1=(r_1(x),r_2(x)y)$. If $alpha_1$ is separable, we are done. If not, we can repeat the procedure, eventually at $n$ steps we will have
$$alpha=(r'_1(x),r'_2(x)y)circ pi^n$$
for some $r'_1$ and $r'_2$ in $K[x]$. So far, so good. They key of this proof is that this process ends with some $alpha''=(r''_1(x),r''_2(x))$ which will necessarily (why?) be a separable. It ends because each step reduces the degree of $alpha$ in a factor of $p$, and degree of $alpha$ is finite, but I can't see how this process is decreasing the degree of $alpha$ at each step by a factor of $p$.
number-theory elliptic-curves cryptography
asked Jan 24 at 0:47
AiYagamiAiYagami
635
$endgroup$
add a comment |
$begingroup$
Lemma:
Let $alpha:E_1longrightarrow E_2 $ be an inseparable isogeny of elliptic
curves defined over a field $k$ of characteristic $p>0$. Then $alpha$ can be
written in the form
$$alpha(x,y)=(r_1(x^p),r_2(x^p)y^p)$$
For $r_1,r_2in K[x]$.
Now here comes the proposition that I'd like to prove.
Let $alpha$ be an isogeny of elliptic curves over a field $k$ of characteristic $p > 0$. Then
$$alpha=alpha_{sep}circpi^n$$
For some $n>0$ and some separable isogeny $alpha_n$ where $pi$ is the p-power Frobenius morphism. We have then $deg(phi)=phi^n deg(alpha_{sep})$.
Proof:
If $alpha$ is separable, then take $n=0$ and $alpha_{sep}=alpha$. Otherwise, $alpha$ is inseparable, and by the lemma, $alpha=(r_1(x^p),r_2(x^p)y^p)=(r_1(x),r_2(x)y)circpi$. We take $alpha_1=(r_1(x),r_2(x)y)$. If $alpha_1$ is separable, we are done. If not, we can repeat the procedure, eventually at $n$ steps we will have
$$alpha=(r'_1(x),r'_2(x)y)circ pi^n$$
for some $r'_1$ and $r'_2$ in $K[x]$. So far, so good. They key of this proof is that this process ends with some $alpha''=(r''_1(x),r''_2(x))$ which will necessarily (why?) be a separable. It ends because each step reduces the degree of $alpha$ in a factor of $p$, and degree of $alpha$ is finite, but I can't see how this process is decreasing the degree of $alpha$ at each step by a factor of $p$.
number-theory elliptic-curves cryptography
asked Jan 24 at 0:47
AiYagamiAiYagami
635
$endgroup$
add a comment |
$begingroup$
Lemma:
Let $alpha:E_1longrightarrow E_2 $ be an inseparable isogeny of elliptic
curves defined over a field $k$ of characteristic $p>0$. Then $alpha$ can be
written in the form
$$alpha(x,y)=(r_1(x^p),r_2(x^p)y^p)$$
For $r_1,r_2in K[x]$.
Now here comes the proposition that I'd like to prove.
Let $alpha$ be an isogeny of elliptic curves over a field $k$ of characteristic $p > 0$. Then
$$alpha=alpha_{sep}circpi^n$$
For some $n>0$ and some separable isogeny $alpha_n$ where $pi$ is the p-power Frobenius morphism. We have then $deg(phi)=phi^n deg(alpha_{sep})$.
Proof:
If $alpha$ is separable, then take $n=0$ and $alpha_{sep}=alpha$. Otherwise, $alpha$ is inseparable, and by the lemma, $alpha=(r_1(x^p),r_2(x^p)y^p)=(r_1(x),r_2(x)y)circpi$. We take $alpha_1=(r_1(x),r_2(x)y)$. If $alpha_1$ is separable, we are done. If not, we can repeat the procedure, eventually at $n$ steps we will have
$$alpha=(r'_1(x),r'_2(x)y)circ pi^n$$
for some $r'_1$ and $r'_2$ in $K[x]$. So far, so good. They key of this proof is that this process ends with some $alpha''=(r''_1(x),r''_2(x))$ which will necessarily (why?) be a separable. It ends because each step reduces the degree of $alpha$ in a factor of $p$, and degree of $alpha$ is finite, but I can't see how this process is decreasing the degree of $alpha$ at each step by a factor of $p$.
number-theory elliptic-curves cryptography
asked Jan 24 at 0:47
AiYagamiAiYagami
635
$endgroup$
Lemma:
Let $alpha:E_1longrightarrow E_2 $ be an inseparable isogeny of elliptic
curves defined over a field $k$ of characteristic $p>0$. Then $alpha$ can be
written in the form
$$alpha(x,y)=(r_1(x^p),r_2(x^p)y^p)$$
For $r_1,r_2in K[x]$.
Now here comes the proposition that I'd like to prove.
Let $alpha$ be an isogeny of elliptic curves over a field $k$ of characteristic $p > 0$. Then
$$alpha=alpha_{sep}circpi^n$$
For some $n>0$ and some separable isogeny $alpha_n$ where $pi$ is the p-power Frobenius morphism. We have then $deg(phi)=phi^n deg(alpha_{sep})$.
Proof:
If $alpha$ is separable, then take $n=0$ and $alpha_{sep}=alpha$. Otherwise, $alpha$ is inseparable, and by the lemma, $alpha=(r_1(x^p),r_2(x^p)y^p)=(r_1(x),r_2(x)y)circpi$. We take $alpha_1=(r_1(x),r_2(x)y)$. If $alpha_1$ is separable, we are done. If not, we can repeat the procedure, eventually at $n$ steps we will have
$$alpha=(r'_1(x),r'_2(x)y)circ pi^n$$
for some $r'_1$ and $r'_2$ in $K[x]$. So far, so good. They key of this proof is that this process ends with some $alpha''=(r''_1(x),r''_2(x))$ which will necessarily (why?) be a separable. It ends because each step reduces the degree of $alpha$ in a factor of $p$, and degree of $alpha$ is finite, but I can't see how this process is decreasing the degree of $alpha$ at each step by a factor of $p$.
number-theory elliptic-curves cryptography
number-theory elliptic-curves cryptography
asked Jan 24 at 0:47
AiYagamiAiYagami
635
asked Jan 24 at 0:47
AiYagamiAiYagami
635
asked Jan 24 at 0:47
AiYagamiAiYagami
635
asked Jan 24 at 0:47
AiYagamiAiYagami
635
asked Jan 24 at 0:47
AiYagamiAiYagami
635
635
add a comment |
add a comment |
0
active
oldest
votes
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%2f3085284%2fproof-of-the-decomposition-of-an-isogeny-into-a-separable-part-and-a-frobenius%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
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%2f3085284%2fproof-of-the-decomposition-of-an-isogeny-into-a-separable-part-and-a-frobenius%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
