Mixing
Is it 'rare' that $a$ and $a+1$ are conjugate (= have the same minimal polynomial)?
$begingroup$
Let $a in bar{k}-k$, $k$ is a field of characteristic zero and $bar{k}$ is an algebraic closure of $k$.
Denote the minimal polynomial of $a$ by $m_a=m_a(t) in k[t]$.
Is it 'rare' that $m_a=m_{a+1}$? In other words, is it rare that $a$ and $a+1$ are conjugate?
For example, $k=mathbb{Q}$ and $a=sqrt{2}$. Then $m_a=t^2-2 neq m_{a+1}$.
Any hints are welcome!
field-theory extension-field minimal-polynomials
asked Jan 3 at 23:19
user237522user237522
2,1451617
$endgroup$
add a comment |
$begingroup$
Let $a in bar{k}-k$, $k$ is a field of characteristic zero and $bar{k}$ is an algebraic closure of $k$.
Denote the minimal polynomial of $a$ by $m_a=m_a(t) in k[t]$.
Is it 'rare' that $m_a=m_{a+1}$? In other words, is it rare that $a$ and $a+1$ are conjugate?
For example, $k=mathbb{Q}$ and $a=sqrt{2}$. Then $m_a=t^2-2 neq m_{a+1}$.
Any hints are welcome!
field-theory extension-field minimal-polynomials
asked Jan 3 at 23:19
user237522user237522
2,1451617
$endgroup$
1
$begingroup$
It would be better rephrase your question, to be more specific. For example, what are the conditions under what $a$ and $a+1$ are conjugate.
$endgroup$
– Jakobian
Jan 3 at 23:23
$begingroup$
Thanks for the comment. (By math.stackexchange.com/questions/118106/…, if $k subsetneq k(a)=L$ is of prime degree and a conjugate of $a$ is in $L$, then $L$ is Galois-- this result inspired my current question).
$endgroup$
– user237522
Jan 3 at 23:28
add a comment |
$begingroup$
Let $a in bar{k}-k$, $k$ is a field of characteristic zero and $bar{k}$ is an algebraic closure of $k$.
Denote the minimal polynomial of $a$ by $m_a=m_a(t) in k[t]$.
Is it 'rare' that $m_a=m_{a+1}$? In other words, is it rare that $a$ and $a+1$ are conjugate?
For example, $k=mathbb{Q}$ and $a=sqrt{2}$. Then $m_a=t^2-2 neq m_{a+1}$.
Any hints are welcome!
field-theory extension-field minimal-polynomials
asked Jan 3 at 23:19
user237522user237522
2,1451617
$endgroup$
Let $a in bar{k}-k$, $k$ is a field of characteristic zero and $bar{k}$ is an algebraic closure of $k$.
Denote the minimal polynomial of $a$ by $m_a=m_a(t) in k[t]$.
Is it 'rare' that $m_a=m_{a+1}$? In other words, is it rare that $a$ and $a+1$ are conjugate?
For example, $k=mathbb{Q}$ and $a=sqrt{2}$. Then $m_a=t^2-2 neq m_{a+1}$.
Any hints are welcome!
field-theory extension-field minimal-polynomials
field-theory extension-field minimal-polynomials
asked Jan 3 at 23:19
user237522user237522
2,1451617
asked Jan 3 at 23:19
user237522user237522
2,1451617
asked Jan 3 at 23:19
user237522user237522
2,1451617
asked Jan 3 at 23:19
user237522user237522
2,1451617
asked Jan 3 at 23:19
user237522user237522
2,1451617
2,1451617
1
$begingroup$
It would be better rephrase your question, to be more specific. For example, what are the conditions under what $a$ and $a+1$ are conjugate.
$endgroup$
– Jakobian
Jan 3 at 23:23
$begingroup$
Thanks for the comment. (By math.stackexchange.com/questions/118106/…, if $k subsetneq k(a)=L$ is of prime degree and a conjugate of $a$ is in $L$, then $L$ is Galois-- this result inspired my current question).
$endgroup$
– user237522
Jan 3 at 23:28
add a comment |
1
$begingroup$
It would be better rephrase your question, to be more specific. For example, what are the conditions under what $a$ and $a+1$ are conjugate.
$endgroup$
– Jakobian
Jan 3 at 23:23
$begingroup$
Thanks for the comment. (By math.stackexchange.com/questions/118106/…, if $k subsetneq k(a)=L$ is of prime degree and a conjugate of $a$ is in $L$, then $L$ is Galois-- this result inspired my current question).
$endgroup$
– user237522
Jan 3 at 23:28
1
1
$begingroup$
It would be better rephrase your question, to be more specific. For example, what are the conditions under what $a$ and $a+1$ are conjugate.
$endgroup$
– Jakobian
Jan 3 at 23:23
$begingroup$
It would be better rephrase your question, to be more specific. For example, what are the conditions under what $a$ and $a+1$ are conjugate.
$endgroup$
– Jakobian
Jan 3 at 23:23
$begingroup$
Thanks for the comment. (By math.stackexchange.com/questions/118106/…, if $k subsetneq k(a)=L$ is of prime degree and a conjugate of $a$ is in $L$, then $L$ is Galois-- this result inspired my current question).
$endgroup$
– user237522
Jan 3 at 23:28
$begingroup$
Thanks for the comment. (By math.stackexchange.com/questions/118106/…, if $k subsetneq k(a)=L$ is of prime degree and a conjugate of $a$ is in $L$, then $L$ is Galois-- this result inspired my current question).
$endgroup$
– user237522
Jan 3 at 23:28
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
If $a$ and $a+1$ are roots of $f(X)$, then $a$ is also a root of $g(X):=f(X+1)$, hence $g$ is a multiple of $f$. By comparing the leading coefficients (if $deg f>0$), it follows tat $g=f$. So as a function on $Bbb Z$, $f$ is periodic, hence bounded, hence constant ...
answered Jan 3 at 23:26
Hagen von EitzenHagen von Eitzen
277k22269496
$endgroup$
$begingroup$
More generally, if $a$ and $a+b$ have the same minimal polynomial, with $bin k$, $bne0$, then $f(x+b)=f(x)$, so $f$ has infinitely many roots in the algebraic closure.
$endgroup$
– egreg
Jan 3 at 23:33
$begingroup$
Thank you. (I like both answers and have not decided which one to accept..)
$endgroup$
– user237522
Jan 3 at 23:36
$begingroup$
@egreg, thank you. Please, how to guarantee that $a,c in L-k$, have the same minimal polynomial? By your observation, it is necessary that $a-c notin k$.
$endgroup$
– user237522
Jan 3 at 23:42
$begingroup$
@user237522 I don't think that there is a different answer than “$a$ and $c$ have the same minimal polynomial” or “there is an automorphism of $L$ over $k$ sending $a$ to $c$”.
$endgroup$
– egreg
Jan 3 at 23:49
$begingroup$
ok, thank you. Please, what if $c=a^m$ for some $m geq 2$? Is it 'less rare' to have an automorphism of $L$ over $k$ sending $a$ to $a^m$?
$endgroup$
– user237522
Jan 3 at 23:57
|
show 2 more comments
$begingroup$
Take an automorphism of the splitting field taking $a$ to $a+1$. Then $a+1$ goes to $a+2$, so $a+2$ is a conjugate of $a+1$, and hence of $a$. Do you see how you can reach a contradiction this way?
answered Jan 3 at 23:22
WojowuWojowu
17.3k22665
$endgroup$
$begingroup$
Thank you. So $a$ has infinitely many conjugates, which is impossible...
$endgroup$
– user237522
Jan 3 at 23:32
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%2f3061138%2fis-it-rare-that-a-and-a1-are-conjugate-have-the-same-minimal-polynomia%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If $a$ and $a+1$ are roots of $f(X)$, then $a$ is also a root of $g(X):=f(X+1)$, hence $g$ is a multiple of $f$. By comparing the leading coefficients (if $deg f>0$), it follows tat $g=f$. So as a function on $Bbb Z$, $f$ is periodic, hence bounded, hence constant ...
answered Jan 3 at 23:26
Hagen von EitzenHagen von Eitzen
277k22269496
$endgroup$
$begingroup$
More generally, if $a$ and $a+b$ have the same minimal polynomial, with $bin k$, $bne0$, then $f(x+b)=f(x)$, so $f$ has infinitely many roots in the algebraic closure.
$endgroup$
– egreg
Jan 3 at 23:33
$begingroup$
Thank you. (I like both answers and have not decided which one to accept..)
$endgroup$
– user237522
Jan 3 at 23:36
$begingroup$
@egreg, thank you. Please, how to guarantee that $a,c in L-k$, have the same minimal polynomial? By your observation, it is necessary that $a-c notin k$.
$endgroup$
– user237522
Jan 3 at 23:42
$begingroup$
@user237522 I don't think that there is a different answer than “$a$ and $c$ have the same minimal polynomial” or “there is an automorphism of $L$ over $k$ sending $a$ to $c$”.
$endgroup$
– egreg
Jan 3 at 23:49
$begingroup$
ok, thank you. Please, what if $c=a^m$ for some $m geq 2$? Is it 'less rare' to have an automorphism of $L$ over $k$ sending $a$ to $a^m$?
$endgroup$
– user237522
Jan 3 at 23:57
|
show 2 more comments
$begingroup$
If $a$ and $a+1$ are roots of $f(X)$, then $a$ is also a root of $g(X):=f(X+1)$, hence $g$ is a multiple of $f$. By comparing the leading coefficients (if $deg f>0$), it follows tat $g=f$. So as a function on $Bbb Z$, $f$ is periodic, hence bounded, hence constant ...
answered Jan 3 at 23:26
Hagen von EitzenHagen von Eitzen
277k22269496
$endgroup$
$begingroup$
More generally, if $a$ and $a+b$ have the same minimal polynomial, with $bin k$, $bne0$, then $f(x+b)=f(x)$, so $f$ has infinitely many roots in the algebraic closure.
$endgroup$
– egreg
Jan 3 at 23:33
$begingroup$
Thank you. (I like both answers and have not decided which one to accept..)
$endgroup$
– user237522
Jan 3 at 23:36
$begingroup$
@egreg, thank you. Please, how to guarantee that $a,c in L-k$, have the same minimal polynomial? By your observation, it is necessary that $a-c notin k$.
$endgroup$
– user237522
Jan 3 at 23:42
$begingroup$
@user237522 I don't think that there is a different answer than “$a$ and $c$ have the same minimal polynomial” or “there is an automorphism of $L$ over $k$ sending $a$ to $c$”.
$endgroup$
– egreg
Jan 3 at 23:49
$begingroup$
ok, thank you. Please, what if $c=a^m$ for some $m geq 2$? Is it 'less rare' to have an automorphism of $L$ over $k$ sending $a$ to $a^m$?
$endgroup$
– user237522
Jan 3 at 23:57
|
show 2 more comments
$begingroup$
If $a$ and $a+1$ are roots of $f(X)$, then $a$ is also a root of $g(X):=f(X+1)$, hence $g$ is a multiple of $f$. By comparing the leading coefficients (if $deg f>0$), it follows tat $g=f$. So as a function on $Bbb Z$, $f$ is periodic, hence bounded, hence constant ...
answered Jan 3 at 23:26
Hagen von EitzenHagen von Eitzen
277k22269496
$endgroup$
If $a$ and $a+1$ are roots of $f(X)$, then $a$ is also a root of $g(X):=f(X+1)$, hence $g$ is a multiple of $f$. By comparing the leading coefficients (if $deg f>0$), it follows tat $g=f$. So as a function on $Bbb Z$, $f$ is periodic, hence bounded, hence constant ...
answered Jan 3 at 23:26
Hagen von EitzenHagen von Eitzen
277k22269496
answered Jan 3 at 23:26
Hagen von EitzenHagen von Eitzen
277k22269496
answered Jan 3 at 23:26
Hagen von EitzenHagen von Eitzen
277k22269496
answered Jan 3 at 23:26
Hagen von EitzenHagen von Eitzen
277k22269496
277k22269496
$begingroup$
More generally, if $a$ and $a+b$ have the same minimal polynomial, with $bin k$, $bne0$, then $f(x+b)=f(x)$, so $f$ has infinitely many roots in the algebraic closure.
$endgroup$
– egreg
Jan 3 at 23:33
$begingroup$
Thank you. (I like both answers and have not decided which one to accept..)
$endgroup$
– user237522
Jan 3 at 23:36
$begingroup$
@egreg, thank you. Please, how to guarantee that $a,c in L-k$, have the same minimal polynomial? By your observation, it is necessary that $a-c notin k$.
$endgroup$
– user237522
Jan 3 at 23:42
$begingroup$
@user237522 I don't think that there is a different answer than “$a$ and $c$ have the same minimal polynomial” or “there is an automorphism of $L$ over $k$ sending $a$ to $c$”.
$endgroup$
– egreg
Jan 3 at 23:49
$begingroup$
ok, thank you. Please, what if $c=a^m$ for some $m geq 2$? Is it 'less rare' to have an automorphism of $L$ over $k$ sending $a$ to $a^m$?
$endgroup$
– user237522
Jan 3 at 23:57
|
show 2 more comments
$begingroup$
More generally, if $a$ and $a+b$ have the same minimal polynomial, with $bin k$, $bne0$, then $f(x+b)=f(x)$, so $f$ has infinitely many roots in the algebraic closure.
$endgroup$
– egreg
Jan 3 at 23:33
$begingroup$
Thank you. (I like both answers and have not decided which one to accept..)
$endgroup$
– user237522
Jan 3 at 23:36
$begingroup$
@egreg, thank you. Please, how to guarantee that $a,c in L-k$, have the same minimal polynomial? By your observation, it is necessary that $a-c notin k$.
$endgroup$
– user237522
Jan 3 at 23:42
$begingroup$
@user237522 I don't think that there is a different answer than “$a$ and $c$ have the same minimal polynomial” or “there is an automorphism of $L$ over $k$ sending $a$ to $c$”.
$endgroup$
– egreg
Jan 3 at 23:49
$begingroup$
ok, thank you. Please, what if $c=a^m$ for some $m geq 2$? Is it 'less rare' to have an automorphism of $L$ over $k$ sending $a$ to $a^m$?
$endgroup$
– user237522
Jan 3 at 23:57
$begingroup$
More generally, if $a$ and $a+b$ have the same minimal polynomial, with $bin k$, $bne0$, then $f(x+b)=f(x)$, so $f$ has infinitely many roots in the algebraic closure.
$endgroup$
– egreg
Jan 3 at 23:33
$begingroup$
More generally, if $a$ and $a+b$ have the same minimal polynomial, with $bin k$, $bne0$, then $f(x+b)=f(x)$, so $f$ has infinitely many roots in the algebraic closure.
$endgroup$
– egreg
Jan 3 at 23:33
$begingroup$
Thank you. (I like both answers and have not decided which one to accept..)
$endgroup$
– user237522
Jan 3 at 23:36
$begingroup$
Thank you. (I like both answers and have not decided which one to accept..)
$endgroup$
– user237522
Jan 3 at 23:36
$begingroup$
@egreg, thank you. Please, how to guarantee that $a,c in L-k$, have the same minimal polynomial? By your observation, it is necessary that $a-c notin k$.
$endgroup$
– user237522
Jan 3 at 23:42
$begingroup$
@egreg, thank you. Please, how to guarantee that $a,c in L-k$, have the same minimal polynomial? By your observation, it is necessary that $a-c notin k$.
$endgroup$
– user237522
Jan 3 at 23:42
$begingroup$
@user237522 I don't think that there is a different answer than “$a$ and $c$ have the same minimal polynomial” or “there is an automorphism of $L$ over $k$ sending $a$ to $c$”.
$endgroup$
– egreg
Jan 3 at 23:49
$begingroup$
@user237522 I don't think that there is a different answer than “$a$ and $c$ have the same minimal polynomial” or “there is an automorphism of $L$ over $k$ sending $a$ to $c$”.
$endgroup$
– egreg
Jan 3 at 23:49
$begingroup$
ok, thank you. Please, what if $c=a^m$ for some $m geq 2$? Is it 'less rare' to have an automorphism of $L$ over $k$ sending $a$ to $a^m$?
$endgroup$
– user237522
Jan 3 at 23:57
$begingroup$
ok, thank you. Please, what if $c=a^m$ for some $m geq 2$? Is it 'less rare' to have an automorphism of $L$ over $k$ sending $a$ to $a^m$?
$endgroup$
– user237522
Jan 3 at 23:57
|
show 2 more comments
$begingroup$
Take an automorphism of the splitting field taking $a$ to $a+1$. Then $a+1$ goes to $a+2$, so $a+2$ is a conjugate of $a+1$, and hence of $a$. Do you see how you can reach a contradiction this way?
answered Jan 3 at 23:22
WojowuWojowu
17.3k22665
$endgroup$
$begingroup$
Thank you. So $a$ has infinitely many conjugates, which is impossible...
$endgroup$
– user237522
Jan 3 at 23:32
add a comment |
$begingroup$
Take an automorphism of the splitting field taking $a$ to $a+1$. Then $a+1$ goes to $a+2$, so $a+2$ is a conjugate of $a+1$, and hence of $a$. Do you see how you can reach a contradiction this way?
answered Jan 3 at 23:22
WojowuWojowu
17.3k22665
$endgroup$
$begingroup$
Thank you. So $a$ has infinitely many conjugates, which is impossible...
$endgroup$
– user237522
Jan 3 at 23:32
add a comment |
$begingroup$
Take an automorphism of the splitting field taking $a$ to $a+1$. Then $a+1$ goes to $a+2$, so $a+2$ is a conjugate of $a+1$, and hence of $a$. Do you see how you can reach a contradiction this way?
answered Jan 3 at 23:22
WojowuWojowu
17.3k22665
$endgroup$
Take an automorphism of the splitting field taking $a$ to $a+1$. Then $a+1$ goes to $a+2$, so $a+2$ is a conjugate of $a+1$, and hence of $a$. Do you see how you can reach a contradiction this way?
answered Jan 3 at 23:22
WojowuWojowu
17.3k22665
answered Jan 3 at 23:22
WojowuWojowu
17.3k22665
answered Jan 3 at 23:22
WojowuWojowu
17.3k22665
answered Jan 3 at 23:22
WojowuWojowu
17.3k22665
17.3k22665
$begingroup$
Thank you. So $a$ has infinitely many conjugates, which is impossible...
$endgroup$
– user237522
Jan 3 at 23:32
add a comment |
$begingroup$
Thank you. So $a$ has infinitely many conjugates, which is impossible...
$endgroup$
– user237522
Jan 3 at 23:32
$begingroup$
Thank you. So $a$ has infinitely many conjugates, which is impossible...
$endgroup$
– user237522
Jan 3 at 23:32
$begingroup$
Thank you. So $a$ has infinitely many conjugates, which is impossible...
$endgroup$
– user237522
Jan 3 at 23:32
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%2f3061138%2fis-it-rare-that-a-and-a1-are-conjugate-have-the-same-minimal-polynomia%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
1
$begingroup$
It would be better rephrase your question, to be more specific. For example, what are the conditions under what $a$ and $a+1$ are conjugate.
$endgroup$
– Jakobian
Jan 3 at 23:23
$begingroup$
Thanks for the comment. (By math.stackexchange.com/questions/118106/…, if $k subsetneq k(a)=L$ is of prime degree and a conjugate of $a$ is in $L$, then $L$ is Galois-- this result inspired my current question).
$endgroup$
– user237522
Jan 3 at 23:28