$nin mathbf{N}$ such that a solution of $X^4+nX^2 +1$ is a root of unit
$begingroup$
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
edited Jan 16 at 22:13
Jyrki Lahtonen
109k13169372
asked Jan 16 at 15:52
Lei FeimaLei Feima
867
$endgroup$
add a comment |
$begingroup$
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
edited Jan 16 at 22:13
Jyrki Lahtonen
109k13169372
asked Jan 16 at 15:52
Lei FeimaLei Feima
867
$endgroup$
$begingroup$
Welcome to Math.SE! Very nice first questions!
$endgroup$
– Jyrki Lahtonen
Jan 16 at 22:16
add a comment |
$begingroup$
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
edited Jan 16 at 22:13
Jyrki Lahtonen
109k13169372
asked Jan 16 at 15:52
Lei FeimaLei Feima
867
$endgroup$
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
edited Jan 16 at 22:13
Jyrki Lahtonen
109k13169372
asked Jan 16 at 15:52
Lei FeimaLei Feima
867
edited Jan 16 at 22:13
Jyrki Lahtonen
109k13169372
asked Jan 16 at 15:52
Lei FeimaLei Feima
867
edited Jan 16 at 22:13
Jyrki Lahtonen
109k13169372
edited Jan 16 at 22:13
Jyrki Lahtonen
109k13169372
edited Jan 16 at 22:13
Jyrki Lahtonen
109k13169372
109k13169372
asked Jan 16 at 15:52
Lei FeimaLei Feima
867
asked Jan 16 at 15:52
Lei FeimaLei Feima
867
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
1
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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.
answered Jan 16 at 16:14
MindlackMindlack
3,85518
$endgroup$
$begingroup$
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
add a comment |
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1 Answer
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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.
answered Jan 16 at 16:14
MindlackMindlack
3,85518
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered Jan 16 at 16:14
MindlackMindlack
3,85518
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered Jan 16 at 16:14
MindlackMindlack
3,85518
$endgroup$
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.
answered Jan 16 at 16:14
MindlackMindlack
3,85518
edited Jan 16 at 21:58
answered Jan 16 at 16:14
MindlackMindlack
3,85518
answered Jan 16 at 16:14
MindlackMindlack
3,85518
answered Jan 16 at 16:14
MindlackMindlack
3,85518
3,85518
$begingroup$
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
add a comment |
$begingroup$
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
add a comment |
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$begingroup$
Welcome to Math.SE! Very nice first questions!
$endgroup$
– Jyrki Lahtonen
Jan 16 at 22:16