Timeline for $f(x)$ is irreducible but $f(x^n)$ is reducible
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 25, 2014 at 20:05 | answer | added | Vesselin Dimitrov | timeline score: 39 | |
| Aug 25, 2014 at 17:41 | comment | added | YCor | A stupid comment is that the question is equivalent to the same question in $\mathbf{Q}$ instead of $\mathbf{Z}$, so there is no reason to bother with integrality issues and the polynomials can be assumed monic in $\mathbf{Q}[x]$. | |
| Aug 25, 2014 at 16:31 | comment | added | user41593 | I'm pretty sure that this question appeared on some issue of Mathematical Reflections (proposed by G. Dospinescu), but I can't seem to find it. By the way, the answer should be "no", if I remember correctly. | |
| Aug 25, 2014 at 16:23 | comment | added | Stanley Yao Xiao | The polynomial $f(x) = x^2 + 1$ has the property that $f(x^n)$ is reducible except when $n$ is a power of 2, so it misses a very thin set. | |
| Aug 25, 2014 at 16:17 | review | First posts | |||
| Aug 25, 2014 at 16:26 | |||||
| Aug 25, 2014 at 16:16 | history | asked | Hesam | CC BY-SA 3.0 |