Mixing
Symmetric function $p_k({x_1},{x_2},…,{x_n})={x_1}^k+{x_2}^k+…+{x_n}^k$ on...
$begingroup$
This is the first time for me on symmetric functions. Let us consider the symmetric function $p_k({x_1},{x_2},...,{x_n})={x_1}^k+{x_2}^k+...+{x_n}^k$ on $GF(q)={0,a_1,a_2,...,a_{q-1}}$.
I want to show that:
a) $p_{q-1}(a_1,...,a_{q-1})=-1$
b) $p_{k}(a_1,...,a_{q-1})= 0$ for all $0 < k leq q-2$.
Some hints or procedure? $GF(q)$ is cyclic so for the first point there is some recursivity, but how can I handle this problem?
abstract-algebra finite-fields symmetric-functions
asked Jan 11 at 7:48
AlessarAlessar
308115
$endgroup$
add a comment |
$begingroup$
This is the first time for me on symmetric functions. Let us consider the symmetric function $p_k({x_1},{x_2},...,{x_n})={x_1}^k+{x_2}^k+...+{x_n}^k$ on $GF(q)={0,a_1,a_2,...,a_{q-1}}$.
I want to show that:
a) $p_{q-1}(a_1,...,a_{q-1})=-1$
b) $p_{k}(a_1,...,a_{q-1})= 0$ for all $0 < k leq q-2$.
Some hints or procedure? $GF(q)$ is cyclic so for the first point there is some recursivity, but how can I handle this problem?
abstract-algebra finite-fields symmetric-functions
asked Jan 11 at 7:48
AlessarAlessar
308115
$endgroup$
2
$begingroup$
Don't forget that the nonzero elements of $GF(q)$ form a cyclic group.
$endgroup$
– Lord Shark the Unknown
Jan 11 at 7:56
$begingroup$
This has been done on the site already. Did you search?
$endgroup$
– Jyrki Lahtonen
Jan 11 at 19:42
$begingroup$
I seem to have answered it myself here as well as here. Not voting to close as a dupe though. No reason to think that would have been the first occurrence. Also, it is surely bad form to use my dupehammer to direct attention to my own answer.
$endgroup$
– Jyrki Lahtonen
Jan 11 at 19:46
$begingroup$
Hmm. I may be feeding the search engine with wrong buzzwords...
$endgroup$
– Jyrki Lahtonen
Jan 11 at 20:10
$begingroup$
Matt Samuel came up with the same solution proving that there is nothing original with mine.
$endgroup$
– Jyrki Lahtonen
Jan 11 at 20:13
add a comment |
$begingroup$
This is the first time for me on symmetric functions. Let us consider the symmetric function $p_k({x_1},{x_2},...,{x_n})={x_1}^k+{x_2}^k+...+{x_n}^k$ on $GF(q)={0,a_1,a_2,...,a_{q-1}}$.
I want to show that:
a) $p_{q-1}(a_1,...,a_{q-1})=-1$
b) $p_{k}(a_1,...,a_{q-1})= 0$ for all $0 < k leq q-2$.
Some hints or procedure? $GF(q)$ is cyclic so for the first point there is some recursivity, but how can I handle this problem?
abstract-algebra finite-fields symmetric-functions
asked Jan 11 at 7:48
AlessarAlessar
308115
$endgroup$
This is the first time for me on symmetric functions. Let us consider the symmetric function $p_k({x_1},{x_2},...,{x_n})={x_1}^k+{x_2}^k+...+{x_n}^k$ on $GF(q)={0,a_1,a_2,...,a_{q-1}}$.
I want to show that:
a) $p_{q-1}(a_1,...,a_{q-1})=-1$
b) $p_{k}(a_1,...,a_{q-1})= 0$ for all $0 < k leq q-2$.
Some hints or procedure? $GF(q)$ is cyclic so for the first point there is some recursivity, but how can I handle this problem?
abstract-algebra finite-fields symmetric-functions
abstract-algebra finite-fields symmetric-functions
asked Jan 11 at 7:48
AlessarAlessar
308115
asked Jan 11 at 7:48
AlessarAlessar
308115
asked Jan 11 at 7:48
AlessarAlessar
308115
asked Jan 11 at 7:48
AlessarAlessar
308115
asked Jan 11 at 7:48
AlessarAlessar
308115
308115
2
$begingroup$
Don't forget that the nonzero elements of $GF(q)$ form a cyclic group.
$endgroup$
– Lord Shark the Unknown
Jan 11 at 7:56
$begingroup$
This has been done on the site already. Did you search?
$endgroup$
– Jyrki Lahtonen
Jan 11 at 19:42
$begingroup$
I seem to have answered it myself here as well as here. Not voting to close as a dupe though. No reason to think that would have been the first occurrence. Also, it is surely bad form to use my dupehammer to direct attention to my own answer.
$endgroup$
– Jyrki Lahtonen
Jan 11 at 19:46
$begingroup$
Hmm. I may be feeding the search engine with wrong buzzwords...
$endgroup$
– Jyrki Lahtonen
Jan 11 at 20:10
$begingroup$
Matt Samuel came up with the same solution proving that there is nothing original with mine.
$endgroup$
– Jyrki Lahtonen
Jan 11 at 20:13
add a comment |
2
$begingroup$
Don't forget that the nonzero elements of $GF(q)$ form a cyclic group.
$endgroup$
– Lord Shark the Unknown
Jan 11 at 7:56
$begingroup$
This has been done on the site already. Did you search?
$endgroup$
– Jyrki Lahtonen
Jan 11 at 19:42
$begingroup$
I seem to have answered it myself here as well as here. Not voting to close as a dupe though. No reason to think that would have been the first occurrence. Also, it is surely bad form to use my dupehammer to direct attention to my own answer.
$endgroup$
– Jyrki Lahtonen
Jan 11 at 19:46
$begingroup$
Hmm. I may be feeding the search engine with wrong buzzwords...
$endgroup$
– Jyrki Lahtonen
Jan 11 at 20:10
$begingroup$
Matt Samuel came up with the same solution proving that there is nothing original with mine.
$endgroup$
– Jyrki Lahtonen
Jan 11 at 20:13
2
2
$begingroup$
Don't forget that the nonzero elements of $GF(q)$ form a cyclic group.
$endgroup$
– Lord Shark the Unknown
Jan 11 at 7:56
$begingroup$
Don't forget that the nonzero elements of $GF(q)$ form a cyclic group.
$endgroup$
– Lord Shark the Unknown
Jan 11 at 7:56
$begingroup$
This has been done on the site already. Did you search?
$endgroup$
– Jyrki Lahtonen
Jan 11 at 19:42
$begingroup$
This has been done on the site already. Did you search?
$endgroup$
– Jyrki Lahtonen
Jan 11 at 19:42
$begingroup$
I seem to have answered it myself here as well as here. Not voting to close as a dupe though. No reason to think that would have been the first occurrence. Also, it is surely bad form to use my dupehammer to direct attention to my own answer.
$endgroup$
– Jyrki Lahtonen
Jan 11 at 19:46
$begingroup$
I seem to have answered it myself here as well as here. Not voting to close as a dupe though. No reason to think that would have been the first occurrence. Also, it is surely bad form to use my dupehammer to direct attention to my own answer.
$endgroup$
– Jyrki Lahtonen
Jan 11 at 19:46
$begingroup$
Hmm. I may be feeding the search engine with wrong buzzwords...
$endgroup$
– Jyrki Lahtonen
Jan 11 at 20:10
$begingroup$
Hmm. I may be feeding the search engine with wrong buzzwords...
$endgroup$
– Jyrki Lahtonen
Jan 11 at 20:10
$begingroup$
Matt Samuel came up with the same solution proving that there is nothing original with mine.
$endgroup$
– Jyrki Lahtonen
Jan 11 at 20:13
$begingroup$
Matt Samuel came up with the same solution proving that there is nothing original with mine.
$endgroup$
– Jyrki Lahtonen
Jan 11 at 20:13
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%2f3069584%2fsymmetric-function-p-kx-1-x-2-x-n-x-1kx-2k-x-nk-on%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%2f3069584%2fsymmetric-function-p-kx-1-x-2-x-n-x-1kx-2k-x-nk-on%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
2
$begingroup$
Don't forget that the nonzero elements of $GF(q)$ form a cyclic group.
$endgroup$
– Lord Shark the Unknown
Jan 11 at 7:56
$begingroup$
This has been done on the site already. Did you search?
$endgroup$
– Jyrki Lahtonen
Jan 11 at 19:42
$begingroup$
I seem to have answered it myself here as well as here. Not voting to close as a dupe though. No reason to think that would have been the first occurrence. Also, it is surely bad form to use my dupehammer to direct attention to my own answer.
$endgroup$
– Jyrki Lahtonen
Jan 11 at 19:46
$begingroup$
Hmm. I may be feeding the search engine with wrong buzzwords...
$endgroup$
– Jyrki Lahtonen
Jan 11 at 20:10
$begingroup$
Matt Samuel came up with the same solution proving that there is nothing original with mine.
$endgroup$
– Jyrki Lahtonen
Jan 11 at 20:13