The problem is finding an irreducible polynomial in $\mathbb{Z}[x]$ such that it is reducible modulo 2,3 and 5. I can't find anything, any help is appreciated. (Is there some general strategy for doing this?)
Asked
Active
Viewed 39 times
0
-
1Chinese remainder theorem? – Angina Seng Apr 20 '19 at 11:56
-
1Just look for a simple quadratic. Try the form $x^2+a$ . – lulu Apr 20 '19 at 11:59
-
1Have you tried some simple quadratics? – rogerl Apr 20 '19 at 11:59
1 Answers
1
Hint $:$ Take $f(x)=x^4+1.$ Then it is reducible modulo every prime $p.$ But it is irreducible in $\Bbb Z[x].$

little o
- 4,853