$nin mathbf{N}$ such that a solution of $X^4+nX^2 +1$ is a root of unit












1












$begingroup$


Consider $f_n(X)=X^4+nX^2 +1$ in $mathbf{Q}[X]$. I found that for all natural $n$ such that $nneq 2-m^2$ for a natural $m$, $f_n(X)$ is irreducible in $mathbf{Q}$.
Consider $K_n=mathbf{Q}(x)= mathbf{Q}[X]/(f_n(X))$.
Using Dirichlet Unit theorem, it is easy to see that $mathcal{O}^*=mu(K_n)times mathbf{Z} $, since all the roots of $f_n(X)$ are complex. It also easy to see that $xin mathcal{O}^*$. So my question is how to determine the natural $n$ such that $mathcal{O}^*/(x)$ is finite. Clearly since $x$ is a unit, we have that $x=(z,a)$, where $ain mathbf{Z}$. So, that quotient is finite for $aneq 0$, i.e., for $xnotin mu(K_n)$. For example, for the $12$th cyclotomic polynomial, i.e., $X^4-X^2+1$, $x$ is a solution for $n=-1$ and so $xin mu(K_{-1})$. But I do not know how to find the other $n$ such that it is not finite.






number-theory algebraic-number-theory class-field-theory integer-rings







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  • $begingroup$
    Welcome to Math.SE! Very nice first questions!
    $endgroup$
    – Jyrki Lahtonen
    Jan 16 at 22:16


















1












$begingroup$


Consider $f_n(X)=X^4+nX^2 +1$ in $mathbf{Q}[X]$. I found that for all natural $n$ such that $nneq 2-m^2$ for a natural $m$, $f_n(X)$ is irreducible in $mathbf{Q}$.
Consider $K_n=mathbf{Q}(x)= mathbf{Q}[X]/(f_n(X))$.
Using Dirichlet Unit theorem, it is easy to see that $mathcal{O}^*=mu(K_n)times mathbf{Z} $, since all the roots of $f_n(X)$ are complex. It also easy to see that $xin mathcal{O}^*$. So my question is how to determine the natural $n$ such that $mathcal{O}^*/(x)$ is finite. Clearly since $x$ is a unit, we have that $x=(z,a)$, where $ain mathbf{Z}$. So, that quotient is finite for $aneq 0$, i.e., for $xnotin mu(K_n)$. For example, for the $12$th cyclotomic polynomial, i.e., $X^4-X^2+1$, $x$ is a solution for $n=-1$ and so $xin mu(K_{-1})$. But I do not know how to find the other $n$ such that it is not finite.






number-theory algebraic-number-theory class-field-theory integer-rings







share|cite|improve this question











$endgroup$












  • $begingroup$
    Welcome to Math.SE! Very nice first questions!
    $endgroup$
    – Jyrki Lahtonen
    Jan 16 at 22:16
















1












1








1


1



$begingroup$


Consider $f_n(X)=X^4+nX^2 +1$ in $mathbf{Q}[X]$. I found that for all natural $n$ such that $nneq 2-m^2$ for a natural $m$, $f_n(X)$ is irreducible in $mathbf{Q}$.
Consider $K_n=mathbf{Q}(x)= mathbf{Q}[X]/(f_n(X))$.
Using Dirichlet Unit theorem, it is easy to see that $mathcal{O}^*=mu(K_n)times mathbf{Z} $, since all the roots of $f_n(X)$ are complex. It also easy to see that $xin mathcal{O}^*$. So my question is how to determine the natural $n$ such that $mathcal{O}^*/(x)$ is finite. Clearly since $x$ is a unit, we have that $x=(z,a)$, where $ain mathbf{Z}$. So, that quotient is finite for $aneq 0$, i.e., for $xnotin mu(K_n)$. For example, for the $12$th cyclotomic polynomial, i.e., $X^4-X^2+1$, $x$ is a solution for $n=-1$ and so $xin mu(K_{-1})$. But I do not know how to find the other $n$ such that it is not finite.






number-theory algebraic-number-theory class-field-theory integer-rings







share|cite|improve this question











$endgroup$




Consider $f_n(X)=X^4+nX^2 +1$ in $mathbf{Q}[X]$. I found that for all natural $n$ such that $nneq 2-m^2$ for a natural $m$, $f_n(X)$ is irreducible in $mathbf{Q}$.
Consider $K_n=mathbf{Q}(x)= mathbf{Q}[X]/(f_n(X))$.
Using Dirichlet Unit theorem, it is easy to see that $mathcal{O}^*=mu(K_n)times mathbf{Z} $, since all the roots of $f_n(X)$ are complex. It also easy to see that $xin mathcal{O}^*$. So my question is how to determine the natural $n$ such that $mathcal{O}^*/(x)$ is finite. Clearly since $x$ is a unit, we have that $x=(z,a)$, where $ain mathbf{Z}$. So, that quotient is finite for $aneq 0$, i.e., for $xnotin mu(K_n)$. For example, for the $12$th cyclotomic polynomial, i.e., $X^4-X^2+1$, $x$ is a solution for $n=-1$ and so $xin mu(K_{-1})$. But I do not know how to find the other $n$ such that it is not finite.






number-theory algebraic-number-theory class-field-theory integer-rings




number-theory algebraic-number-theory class-field-theory integer-rings






share|cite|improve this question















share|cite|improve this question













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share|cite|improve this question








edited Jan 16 at 22:13









Jyrki Lahtonen

110k13172390




110k13172390










asked Jan 16 at 15:52









Lei FeimaLei Feima

867




867












  • $begingroup$
    Welcome to Math.SE! Very nice first questions!
    $endgroup$
    – Jyrki Lahtonen
    Jan 16 at 22:16




















  • $begingroup$
    Welcome to Math.SE! Very nice first questions!
    $endgroup$
    – Jyrki Lahtonen
    Jan 16 at 22:16


















$begingroup$
Welcome to Math.SE! Very nice first questions!
$endgroup$
– Jyrki Lahtonen
Jan 16 at 22:16






$begingroup$
Welcome to Math.SE! Very nice first questions!
$endgroup$
– Jyrki Lahtonen
Jan 16 at 22:16












1 Answer
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$begingroup$

If some root $r$ of $X^4+nX^2+1$ is a root of unity, then $r^2$ is a root of unity and vanishes $X^2+nX+1$. Thus $-n$ is double the real part of $r^2$ (because $r^2$ and its complex conjugate are the roots of $X^2+nX+1$, since the product of roots is $1$), thus $|n| leq 2$.



If $|n|=2$, your polynomial is $(X^2 pm 1)^2$.
If $n=0$, your polynomial is $Phi_4$.
If $|n|=1$, $j$ or $ij$ is a root.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Now it is perfectly clear! Thanks a lot. The same noting that $n=-2 cos(4pi a/b)$
    $endgroup$
    – Lei Feima
    Jan 17 at 9:21












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1 Answer
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2












$begingroup$

If some root $r$ of $X^4+nX^2+1$ is a root of unity, then $r^2$ is a root of unity and vanishes $X^2+nX+1$. Thus $-n$ is double the real part of $r^2$ (because $r^2$ and its complex conjugate are the roots of $X^2+nX+1$, since the product of roots is $1$), thus $|n| leq 2$.



If $|n|=2$, your polynomial is $(X^2 pm 1)^2$.
If $n=0$, your polynomial is $Phi_4$.
If $|n|=1$, $j$ or $ij$ is a root.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Now it is perfectly clear! Thanks a lot. The same noting that $n=-2 cos(4pi a/b)$
    $endgroup$
    – Lei Feima
    Jan 17 at 9:21
















2












$begingroup$

If some root $r$ of $X^4+nX^2+1$ is a root of unity, then $r^2$ is a root of unity and vanishes $X^2+nX+1$. Thus $-n$ is double the real part of $r^2$ (because $r^2$ and its complex conjugate are the roots of $X^2+nX+1$, since the product of roots is $1$), thus $|n| leq 2$.



If $|n|=2$, your polynomial is $(X^2 pm 1)^2$.
If $n=0$, your polynomial is $Phi_4$.
If $|n|=1$, $j$ or $ij$ is a root.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Now it is perfectly clear! Thanks a lot. The same noting that $n=-2 cos(4pi a/b)$
    $endgroup$
    – Lei Feima
    Jan 17 at 9:21














2












2








2





$begingroup$

If some root $r$ of $X^4+nX^2+1$ is a root of unity, then $r^2$ is a root of unity and vanishes $X^2+nX+1$. Thus $-n$ is double the real part of $r^2$ (because $r^2$ and its complex conjugate are the roots of $X^2+nX+1$, since the product of roots is $1$), thus $|n| leq 2$.



If $|n|=2$, your polynomial is $(X^2 pm 1)^2$.
If $n=0$, your polynomial is $Phi_4$.
If $|n|=1$, $j$ or $ij$ is a root.






share|cite|improve this answer











$endgroup$



If some root $r$ of $X^4+nX^2+1$ is a root of unity, then $r^2$ is a root of unity and vanishes $X^2+nX+1$. Thus $-n$ is double the real part of $r^2$ (because $r^2$ and its complex conjugate are the roots of $X^2+nX+1$, since the product of roots is $1$), thus $|n| leq 2$.



If $|n|=2$, your polynomial is $(X^2 pm 1)^2$.
If $n=0$, your polynomial is $Phi_4$.
If $|n|=1$, $j$ or $ij$ is a root.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 16 at 21:58

























answered Jan 16 at 16:14









MindlackMindlack

4,900211




4,900211












  • $begingroup$
    Now it is perfectly clear! Thanks a lot. The same noting that $n=-2 cos(4pi a/b)$
    $endgroup$
    – Lei Feima
    Jan 17 at 9:21


















  • $begingroup$
    Now it is perfectly clear! Thanks a lot. The same noting that $n=-2 cos(4pi a/b)$
    $endgroup$
    – Lei Feima
    Jan 17 at 9:21
















$begingroup$
Now it is perfectly clear! Thanks a lot. The same noting that $n=-2 cos(4pi a/b)$
$endgroup$
– Lei Feima
Jan 17 at 9:21




$begingroup$
Now it is perfectly clear! Thanks a lot. The same noting that $n=-2 cos(4pi a/b)$
$endgroup$
– Lei Feima
Jan 17 at 9:21


















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