Is it 'rare' that $a$ and $a+1$ are conjugate (= have the same minimal polynomial)?












2














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







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  • 1




    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.
    – Jakobian
    Jan 3 at 23:23










  • 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).
    – user237522
    Jan 3 at 23:28
















2














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







share|cite|improve this question


















  • 1




    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.
    – Jakobian
    Jan 3 at 23:23










  • 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).
    – user237522
    Jan 3 at 23:28














2












2








2







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







share|cite|improve this question













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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 3 at 23:19









user237522

2,1411617




2,1411617








  • 1




    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.
    – Jakobian
    Jan 3 at 23:23










  • 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).
    – user237522
    Jan 3 at 23:28














  • 1




    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.
    – Jakobian
    Jan 3 at 23:23










  • 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).
    – user237522
    Jan 3 at 23:28








1




1




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.
– Jakobian
Jan 3 at 23:23




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.
– Jakobian
Jan 3 at 23:23












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).
– user237522
Jan 3 at 23:28




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).
– user237522
Jan 3 at 23:28










2 Answers
2






active

oldest

votes


















6














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 ...






share|cite|improve this answer





















  • 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.
    – egreg
    Jan 3 at 23:33










  • Thank you. (I like both answers and have not decided which one to accept..)
    – user237522
    Jan 3 at 23:36












  • @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 $a-c notin k$.
    – user237522
    Jan 3 at 23:42












  • @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 $c$”.
    – egreg
    Jan 3 at 23:49












  • 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 $a^m$?
    – user237522
    Jan 3 at 23:57



















5














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?






share|cite|improve this answer





















  • Thank you. So $a$ has infinitely many conjugates, which is impossible...
    – user237522
    Jan 3 at 23:32













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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









6














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 ...






share|cite|improve this answer





















  • 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.
    – egreg
    Jan 3 at 23:33










  • Thank you. (I like both answers and have not decided which one to accept..)
    – user237522
    Jan 3 at 23:36












  • @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 $a-c notin k$.
    – user237522
    Jan 3 at 23:42












  • @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 $c$”.
    – egreg
    Jan 3 at 23:49












  • 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 $a^m$?
    – user237522
    Jan 3 at 23:57
















6














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 ...






share|cite|improve this answer





















  • 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.
    – egreg
    Jan 3 at 23:33










  • Thank you. (I like both answers and have not decided which one to accept..)
    – user237522
    Jan 3 at 23:36












  • @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 $a-c notin k$.
    – user237522
    Jan 3 at 23:42












  • @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 $c$”.
    – egreg
    Jan 3 at 23:49












  • 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 $a^m$?
    – user237522
    Jan 3 at 23:57














6












6








6






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 ...






share|cite|improve this answer












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 ...







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 3 at 23:26









Hagen von Eitzen

276k21269496




276k21269496












  • 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.
    – egreg
    Jan 3 at 23:33










  • Thank you. (I like both answers and have not decided which one to accept..)
    – user237522
    Jan 3 at 23:36












  • @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 $a-c notin k$.
    – user237522
    Jan 3 at 23:42












  • @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 $c$”.
    – egreg
    Jan 3 at 23:49












  • 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 $a^m$?
    – user237522
    Jan 3 at 23:57


















  • 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.
    – egreg
    Jan 3 at 23:33










  • Thank you. (I like both answers and have not decided which one to accept..)
    – user237522
    Jan 3 at 23:36












  • @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 $a-c notin k$.
    – user237522
    Jan 3 at 23:42












  • @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 $c$”.
    – egreg
    Jan 3 at 23:49












  • 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 $a^m$?
    – user237522
    Jan 3 at 23:57
















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.
– egreg
Jan 3 at 23:33




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.
– egreg
Jan 3 at 23:33












Thank you. (I like both answers and have not decided which one to accept..)
– user237522
Jan 3 at 23:36






Thank you. (I like both answers and have not decided which one to accept..)
– user237522
Jan 3 at 23:36














@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 $a-c notin k$.
– user237522
Jan 3 at 23:42






@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 $a-c notin k$.
– user237522
Jan 3 at 23:42














@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 $c$”.
– egreg
Jan 3 at 23:49






@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 $c$”.
– egreg
Jan 3 at 23:49














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 $a^m$?
– user237522
Jan 3 at 23:57




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 $a^m$?
– user237522
Jan 3 at 23:57











5














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?






share|cite|improve this answer





















  • Thank you. So $a$ has infinitely many conjugates, which is impossible...
    – user237522
    Jan 3 at 23:32


















5














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?






share|cite|improve this answer





















  • Thank you. So $a$ has infinitely many conjugates, which is impossible...
    – user237522
    Jan 3 at 23:32
















5












5








5






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?






share|cite|improve this answer












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?







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 3 at 23:22









Wojowu

17.2k22565




17.2k22565












  • Thank you. So $a$ has infinitely many conjugates, which is impossible...
    – user237522
    Jan 3 at 23:32




















  • Thank you. So $a$ has infinitely many conjugates, which is impossible...
    – user237522
    Jan 3 at 23:32


















Thank you. So $a$ has infinitely many conjugates, which is impossible...
– user237522
Jan 3 at 23:32






Thank you. So $a$ has infinitely many conjugates, which is impossible...
– user237522
Jan 3 at 23:32




















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