Timeline for Polynomial reducible modulo every integer
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 3, 2011 at 14:43 | comment | added | user18237 | KConrad, the questioner asked for $f$ reducible mod all integers, not just all primes. If the biquadratic extension had a prime $p$ with say "$e=f=2$" then $f$ would be reducible mod $p$, but irreducible modulo a sufficiently large power of $p$. | |
| Oct 3, 2011 at 13:12 | comment | added | KConrad | Concerning the last sentence in gb's answer, every biquadratic extension of ${\mathbf Q}$ is suitable since I described in my second comment how to produce $f(x)$ from any non-cyclic Galois extension of ${\mathbf Q}$. | |
| Oct 3, 2011 at 13:08 | comment | added | KConrad | In Ottem's answer, for example, $x^4 - 72x^2 + 4$ has root $\sqrt{17}+\sqrt{19}$ and ${\mathbf Q}(\sqrt{17}+\sqrt{19}) = {\mathbf Q}(\sqrt{17},\sqrt{19})$ is biquadratic over $\mathbf Q$. | |
| Oct 3, 2011 at 13:07 | comment | added | KConrad | The upshot of the previous comment is that if $K/{\mathbf Q}$ is any non-cyclic Galois extension and we take for $f(x)$ the minimal polynomial in ${\mathbf Z}[x]$ of any algebraic integer generating $K$ over ${\mathbf Q}$ then $f(x)$ is irreducible over ${\mathbf Q}$ but it's guaranteed that $f(x) \bmod p$ is reducible for every prime $p$ (if it were reducible for even one $p$ then ${\rm Gal}(K/{\mathbf Q})$ would be cyclic). The simplest such $K$ are biquadratic fields since their Galois groups over ${\mathbf Q}$ are products of two groups of order 2. | |
| Oct 3, 2011 at 13:02 | comment | added | KConrad | Expanding on Kevin's answer, if $f(x)$ is monic irreducible in ${\mathbf Z}[x]$ then there's a nonobvious consequence of $f(x) \bmod p$ being irreducible for even one prime $p$. Write $G$ for the Galois group of $f(x)$ over ${\mathbf Q}$ and $H$ for the subgroup with fixed field ${\mathbf Q}(\alpha)$ for a root $\alpha$ of $f(x)$, so the coset space $G/H$ has size $\deg f$. If $f(x) \bmod p$ is irreducible for some prime $p$ then there's an element of $G$ with order $\deg f$ whose powers represent $G/H$. In particular, if ${\mathbf Q}(\alpha)$ is Galois over ${\mathbf Q}$ then $G$ is cyclic. | |
| Oct 2, 2011 at 20:35 | comment | added | Kevin Buzzard | The field $K=\mathbf{Q}(\sqrt{13},\sqrt{17})$ would do, for example. I learnt this example in Cassels-Froehlich. The Galois group is $(\mathbf{Z}/2)^2$ of course, but the point is that only $13$ and $17$ ramify, and $13$ splits in $\mathbf{Q}(\sqrt{17})$ and $17$ splits in $\mathbf{Q}(\sqrt{13})$. | |
| Oct 2, 2011 at 19:20 | history | answered | user18237 | CC BY-SA 3.0 |