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Inserted missing $\pm$ sign before $11 \bmod 30$
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Noam D. Elkies
Noam D. Elkies
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It's been noted already that in fact $h \bmod p$ has four linear factors iff $p \equiv \pm 1 \bmod 30$, and is a product of two quadratics iff $p \equiv 11 \bmod 30$$p \equiv \pm 11 \bmod 30$. This can be checked by identifying the splitting field of $h$ with the real subfield of the $15$-th cyclotomic field, generated by $c := e^{2\pi i/15} + e^{-2\pi i/15} = 2 \cos (2\pi/15)$ which is a root of $c^4 - c^3 - 4c^2 + 4c + 1 = 0$; indeed $1 + 2(c-c^2)$ is a root of $h$. The desired result soon follows from the fact that Frobenius takes $e^{2\pi i/15}$ to $e^{2p\pi i/15}$. This is consistent with a cyclic Galois group (not the Klein 4-group as some have claimed), since ${\rm Gal}({\bf Q}(c)/{\bf Q})$ is the cyclic group $({\bf Z} / 15 {\bf Z})^* / \lbrace \pm1 \rbrace$.

If for some reason you do need a quartic of this form $x^4 + ax^3 + bx^2 - ax + 1$ (i.e. with a symmetry $x \leftrightarrow -1/x$) that splits completely mod $p$ iff $p \equiv \pm 1 \bmod 20$, the first few possibilities are $(a,b) = (\pm 2,-6)$, $\pm(22, -6)$, and $(\pm 18,74)$ if I computed correctly in gp.

It's been noted already that in fact $h \bmod p$ has four linear factors iff $p \equiv \pm 1 \bmod 30$, and is a product of two quadratics iff $p \equiv 11 \bmod 30$. This can be checked by identifying the splitting field of $h$ with the real subfield of the $15$-th cyclotomic field, generated by $c := e^{2\pi i/15} + e^{-2\pi i/15} = 2 \cos (2\pi/15)$ which is a root of $c^4 - c^3 - 4c^2 + 4c + 1 = 0$; indeed $1 + 2(c-c^2)$ is a root of $h$. The desired result soon follows from the fact that Frobenius takes $e^{2\pi i/15}$ to $e^{2p\pi i/15}$. This is consistent with a cyclic Galois group (not the Klein 4-group as some have claimed), since ${\rm Gal}({\bf Q}(c)/{\bf Q})$ is the cyclic group $({\bf Z} / 15 {\bf Z})^* / \lbrace \pm1 \rbrace$.

If for some reason you do need a quartic of this form $x^4 + ax^3 + bx^2 - ax + 1$ (i.e. with a symmetry $x \leftrightarrow -1/x$) that splits completely mod $p$ iff $p \equiv \pm 1 \bmod 20$, the first few possibilities are $(a,b) = (\pm 2,-6)$, $\pm(22, -6)$, and $(\pm 18,74)$ if I computed correctly in gp.

It's been noted already that in fact $h \bmod p$ has four linear factors iff $p \equiv \pm 1 \bmod 30$, and is a product of two quadratics iff $p \equiv \pm 11 \bmod 30$. This can be checked by identifying the splitting field of $h$ with the real subfield of the $15$-th cyclotomic field, generated by $c := e^{2\pi i/15} + e^{-2\pi i/15} = 2 \cos (2\pi/15)$ which is a root of $c^4 - c^3 - 4c^2 + 4c + 1 = 0$; indeed $1 + 2(c-c^2)$ is a root of $h$. The desired result soon follows from the fact that Frobenius takes $e^{2\pi i/15}$ to $e^{2p\pi i/15}$. This is consistent with a cyclic Galois group (not the Klein 4-group as some have claimed), since ${\rm Gal}({\bf Q}(c)/{\bf Q})$ is the cyclic group $({\bf Z} / 15 {\bf Z})^* / \lbrace \pm1 \rbrace$.

If for some reason you do need a quartic of this form $x^4 + ax^3 + bx^2 - ax + 1$ (i.e. with a symmetry $x \leftrightarrow -1/x$) that splits completely mod $p$ iff $p \equiv \pm 1 \bmod 20$, the first few possibilities are $(a,b) = (\pm 2,-6)$, $\pm(22, -6)$, and $(\pm 18,74)$ if I computed correctly in gp.

Source Link
Noam D. Elkies
Noam D. Elkies
  • 79.9k
  • 15
  • 281
  • 376

It's been noted already that in fact $h \bmod p$ has four linear factors iff $p \equiv \pm 1 \bmod 30$, and is a product of two quadratics iff $p \equiv 11 \bmod 30$. This can be checked by identifying the splitting field of $h$ with the real subfield of the $15$-th cyclotomic field, generated by $c := e^{2\pi i/15} + e^{-2\pi i/15} = 2 \cos (2\pi/15)$ which is a root of $c^4 - c^3 - 4c^2 + 4c + 1 = 0$; indeed $1 + 2(c-c^2)$ is a root of $h$. The desired result soon follows from the fact that Frobenius takes $e^{2\pi i/15}$ to $e^{2p\pi i/15}$. This is consistent with a cyclic Galois group (not the Klein 4-group as some have claimed), since ${\rm Gal}({\bf Q}(c)/{\bf Q})$ is the cyclic group $({\bf Z} / 15 {\bf Z})^* / \lbrace \pm1 \rbrace$.

If for some reason you do need a quartic of this form $x^4 + ax^3 + bx^2 - ax + 1$ (i.e. with a symmetry $x \leftrightarrow -1/x$) that splits completely mod $p$ iff $p \equiv \pm 1 \bmod 20$, the first few possibilities are $(a,b) = (\pm 2,-6)$, $\pm(22, -6)$, and $(\pm 18,74)$ if I computed correctly in gp.