Integral polynomial irreducible over $\mathbb Q$ but reducible mod $2$, $3$, and $5$ 
Possible Duplicate:
Polynomials irreducible over $\mathbb{Q}$ but reducible over $\mathbb{F}_p$ for every prime $p$ 

Anyone know an $f\in \mathbb{Z}[x]$ that is irreducible over $\mathbb{Q}$ but whose reduction mod $p$ is reducible over the first three positive primes?
 A: Hint $\rm\ \ x^2  +\, 2\cdot3\cdot 5  $
A: "Most" polynomials are irreducible. So if we can find any polynomial reducible modulo 2, 3, and 5, we should have no trouble finding one that is also irreducible over the integers.
And we can even consider the three primes independently, and use the Chinese Remainder Theorem  (CRT) to combine our results.
The simplest reducible polynomial modulo 2 is $x^2$. The same modulo 3 and 5. So applying the CRT is trivial in this case, since $x^2$ reduces to $x^2$ modulo 2 3 and 5.
Alas, $x^2$ is reducible over the integers. But "most" polynomials are irreducible, so just pick another polynomial that reduces to $x^2$ modulo 2, 3, and 5. As $30$ is equivalent to $0$ modulo 2, 3, and 5, we can add it to any coefficient without changing its residues modulo 2, 3, and 5.
We were, in some sense, incredibly unlucky with our first guess of $x^2$; any other guess is extremely likely to find a good example by choosing any other one. $x^2 + 30$ turns out to be irreducible, so there you go!
Let's do a more involved example. Suppose we choose


*

*$f(x) \equiv x(x-1) = x^2 + x\pmod 2 $

*$f(x) \equiv (x-1)(x-2) = x^2 + x + 2 \pmod 3 $

*$f(x) \equiv x^3 \pmod 5 $


Applying the CRT to each coefficient gives


*

*$f(x) \equiv 6 x^3 + 25 x^2 + 25x + 20 \pmod {30}$


Happily, $6 x^3 + 25 x^2 + 25x + 20$ is irreducible over the integers.
A: Yes. The polynomial $x^4+1$ does the trick. 
